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Introduction to Formal Logic
D.K. Johnston
Department of Philosophy
University of Victoria
c
2005
by D.K. Johnston
Contents
1 Some Fundamental Concepts of Logic
1.1 Inferences, validity, and soundness . . . .
1.2 Truth values and negation . . . . . . . . .
1.3 Entailment and equivalence . . . . . . . .
1.4 Consistency . . . . . . . . . . . . . . . . .
1.5 Necessity, contradiction, and contingency
2 Basic Truth Functional Logic
2.1 Logical form . . . . . . . . . . . . . . .
2.2 Truth functional operators . . . . . . .
2.3 Truth tables . . . . . . . . . . . . . . .
2.4 Logical truth . . . . . . . . . . . . . .
2.5 Logical equivalence . . . . . . . . . . .
2.6 Logical implication . . . . . . . . . . .
2.7 Validity and entailment . . . . . . . .
2.8 Some traditional rules of inference and
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equivalence
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3 Introduction to Metatheory
3.1 Definability of truth functional operators .
3.2 Truth functional completeness . . . . . . .
3.3 Disjunctive and conjunctive normal forms
3.4 Some basic concepts of set theory . . . . .
3.5 Mathematical induction . . . . . . . . . .
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4 The
4.1
4.2
4.3
4.4
4.5
4.6
Propositional Calculus
Formal syntax of TFL . . . . . . .
Formal definition of the system PC
Some theorems and metatheorems
The Deduction Theorem . . . . . .
More theorems and metatheorems
Formal semantics for TFL . . . . .
5 Metatheory of PC
5.1 Soundness of PC . . . .
5.2 Maximal consistent sets
5.3 Completeness of PC . .
5.4 Decidability of PC . . .
5.5 Compactness of PC . . .
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1
1.1
Some Fundamental Concepts of Logic
Inferences, validity, and soundness
Logic is the study of inference. An inference is a collection of two or more
sentences,1 one of which is designated as the conclusion of the inference. The
other sentences in the collection are called premises. Hence the following is an
inference:
All men are mortal.
Socrates is a man.
Therefore: Socrates is mortal.
The word ‘therefore’ is commonly used to indicate the conclusion of an inference. There are a number of words that can be used in this way: for example,
‘thus’, ‘hence’, and ‘consequently’. However, these indicator words are not considered to be part of the conclusion itself. The symbol ∴ is traditionally used
to represent these indicator words.
Definition 1.1. An inference is valid iff2 it is impossible for its premises to be
true and its conclusion false.
In other words: an inference is valid if its conclusion must be true in every
conceivable case in which its premises are true. Thus, the inference given above
is valid; for in any conceivable situation where both the sentences ‘All men are
mortal’ and ‘Socrates is a man’ are true, the sentence ‘Socrates is mortal’ must
also be true.
An invalid inference is simply one that is not valid. Hence, an inference will
be invalid if there is at least one possible case in which its premises are true and
its conclusion false. Note that it takes only one such case to render an inference
invalid. For example, the following is an invalid inference:
Socrates is a man.
∴ Socrates is mortal.
This inference is invalid because we can conceive of a situation where Socrates
is a man, but where he is not mortal. Such a situation could never occur in
the real world, because in the real world all men are mortal. But this does
not matter: all that is required is that there be some conceivable case where
the premise would be true and the conclusion false. Nor does it matter how
bizarre the circumstances of such an imaginary situation might be: as long as
the premises of an inference would be true in that situation, and its conclusion
false, that is enough to prove the inference invalid. Such situations are called
counter-examples.
An inference is sound when it fulfils two conditions: (i) it is valid; and (ii)
its premises are true. Thus the following inference is sound:
1 Some philosophers prefer to talk about propositions or statements rather than sentences.
The difference is of considerable importance, but need not concern us here.
2 ‘iff’ abbreviates the phrase ‘if and only if’.
1
2
Chapter 1
All whales are mammals.
All mammals are chordates.
∴ All whales are chordates.
This inference fulfils condition (i): there is no possible case where its premises
could be true and its conclusion false. Hence the inference is valid. But the
inference also fulfils condition (ii), because its premises are true: all whales are
in fact mammals, and all mammals have spinal chords.
Note that a sound inference will always have a true conclusion. Since a
sound inference must be valid, and its premises must be true, it follows by the
definition of validity that its conclusion must be true as well.
Although by definition a sound inference must be valid, a valid inference
need not be sound. An inference may have one or more false premises, and yet
still be valid. Here is an example:
All flowers are blue.
Roses are flowers.
∴ Roses are blue.
The first premise is false, and hence by definition the inference is not sound.
However, this inference is still valid: if the premises were true, it would be
impossible for the conclusion to be false.
As the previous example shows, a valid inference may also have a false conclusion. But if this is so, the inference must also have a false premise:
Proposition 1.1. If a valid inference has a false conclusion, then it must have
at least one false premise.
Proof. Suppose that an inference is valid. Then by definition of validity, it is
impossible for the conclusion to be false when the premises are true. Suppose
further that the conclusion is false. Therefore, it is impossible for all of the
premises to be true. Therefore, at least one of the premises must be false.
a
An inference may have both true premises and a true conclusion, yet still be
unsound. Here is an example:
All cats are mammals.
Some mammals are carnivorous.
∴ All cats are carnivorous.
Every sentence in this inference is true. However, the inference is invalid: for we
can conceive of a situation in which the premises are true, but the conclusion
is false. In the real world, all cats are carnivorous. But we can easily imagine a
world in which some of them are vegetarians; and in such a world, it could still
be true that all cats are mammals, and that some mammals are carnivorous.
Hence the inference is invalid, and so by definition it cannot be sound.
Logic is concerned solely with the validity of inferences, not with their soundness. To prove that an inference is sound, one must prove that it is valid, and
this falls within the domain of logic. However, one must also prove that the
Fundamental Concepts
3
premises of the inference are true; and whether this is so depends on the subject matter of those premises, and on the nature of the world. For example, if
the subject matter of an inference happens to be theoretical physics, determining whether the premises are true requires the relevant expertise. However, a
logician could still determine whether the inference is valid. As we will see in
later chapters, the validity of an inference has nothing to do with the subject
matter of the sentences it contains.
Exercise
1. Can removing a premise from an invalid argument make it valid?
2. Can removing a premise from a valid argument make it invalid?
3. Can adding a premise to an invalid argument make it valid?
4. Can adding a premise to a valid argument make it invalid?
5. Can a valid inference have a true conclusion if it has a false premise?
6. Can removing a premise from an unsound argument make it sound?
7. Can removing a premise from a sound argument make it unsound?
8. Can adding a premise to an unsound argument make it sound?
9. Can adding a premise to a sound argument make it unsound?
1.2
Truth values and negation
The logics that will be studied in this book obey two fundamental laws:
Law of Excluded Middle: Every sentence is either true or false.
Law of Non-Contradiction: No sentence is both true and false.
It follows from these laws that every sentence will have precisely one of two truth
values: either true or false. These values will be designated by the numerals 1
and 0 respectively.
There are some logics where these laws do not hold. Multiple-valued logics
allow sentences to have three or more truth values, and thus violate the Law of
Excluded Middle. Paraconsistent logics allow some sentences to be both true
and false, and thus violate the Law of Non-Contradiction. Although these logics
are of considerable interest, they will not be studied in this book.
Some sentences may be true at one time and false at another: for example,
‘It is raining’. In such cases, we will assume that the above laws refer to a
particular time and place at which the sentence is uttered. Thus, according
to the Law of Non-Contradiction, if the sentence ‘It is raining’ is true at a
given time and place, it cannot be false at that same time and place. Likewise,
according to the Law of Excluded Middle, this sentence must be either true or
false at any particular time and place.
4
Chapter 1
Definition 1.2. The negation of a sentence is another sentence that always has
the opposite truth value.
For example, the negation of the sentence ‘It is raining’ is ‘It is not raining’: in
any situation where the first sentence is true, the second sentence will be false;
and in any situation where the first sentence is false, the second sentence will
be true.
We will use the symbol ¬ to represent negation. The negation of a sentence α
is written ¬α, and read ‘not-alpha’. By the definition of negation, if the sentence
α is assigned truth value 1, ¬α must be assigned truth value 0. Similarly, if α
is assigned 0, ¬α must be assigned 1.
The negation of an English sentence cannot always be formed simply by
adding the word ‘not’ to the verb phrase. For example, the negation of the
sentence ‘Some mammals are carnivorous’ cannot be ‘Some mammals are not
carnivorous’: both of these sentences are in fact true, and negations must always
have opposite truth values. The negation of ‘Some mammals are carnivorous’
is ‘No mammals are carnivorous’: when the one is true, the other is false. For
the same reason, the negation of ‘Some mammals are not carnivorous’ is ‘All
mammals are carnivorous’.
Exercise
1. Suppose ¬α has truth value 1. What is the truth value of α?
2. Suppose that α has truth value 1. What is the truth value of ¬¬α?
3. Suppose that ¬α has truth value 0. What is the truth value of ¬¬α?
1.3
Entailment and equivalence
Definition 1.3. A sentence α entails a sentence β iff it is impossible for β to
be false when α is true.
In other words: α entails β when β must be true in every conceivable situation
in which α is true. Thus, the sentence ‘X is a square’ entails the sentence ‘X
has four sides’; for if X is a square, it must have four sides.
Definition 1.4. A sentence α is equivalent to a sentence β iff it is impossible
for α and β to have different truth values.
For example, the sentence ‘X is larger than Y’ is equivalent to the sentence ‘Y
is smaller than X’: in any case where one of them is true, the other must be
true; and in any case where one of them is false, the other must also be false.
These definitions of entailment and equivalence have some important consequences. For example, every sentence will entail itself:
Proposition 1.2. For any sentence α, α entails α.
Fundamental Concepts
5
Proof. Assume that α is true. Then by the Law of Non-Contradiction, it is
impossible for α to be false. Therefore, it is impossible for α to be false when α
is true. Hence, by the definition of entailment, α entails α.
a
It can be also be proved that if two sentences are equivalent, then each entails
the other:
Proposition 1.3. If α is equivalent to β, then α entails β and β entails α.
Proof. Suppose that α and β are equivalent. (i) Assume that α is true. Then by
definition of equivalence, it follows that β must be true. But then it is impossible
for β to be false, by the Law of Non-Contradiction. Therefore, it is impossible
for β to be false when α is true. Hence, by definition of entailment, α entails β.
(ii) Assume that β is true. Again, by the definition of equivalence, α must be
true; and so by the Law of Non-Contradiction, it is impossible for α to be false
when β is true. Hence β entails α by definition of entailment.
a
We extend the definition of entailment to include collections of sentences:
Definition 1.5. A set of sentences Σ entails α iff it is impossible for α to be
false when every sentence in Σ is true.
We use the symbol to designate entailment. Thus Σ α means that Σ entails
α.
It is important to notice that this definition of entailment places no restriction whatever on the number of sentences that Σ may contain. A set Σ may
contain just one sentence, or infinitely many sentences, or no sentences at all. In
each case, the same definition applies. Of special interest are those cases where
Σ α even though Σ is empty. When this is so for a sentence α, we say that
α itself is valid, and represent this by writing α. Sentences of this sort have
received a great deal of attention from logicians, and will be closely studied in
the chapters that follow.
Proposition 1.4. α is equivalent to ¬¬α.
Proof. First, assume that α is true. ∴ ¬α is false, by definition of negation.
∴ ¬¬α is true, again by definition of negation. Next, assume that α is false.
∴ ¬α is true, by definition of negation. ∴ ¬¬α is false, again by definition of
negation. ∴ ¬¬α always has the same truth value as α. ∴ α and ¬¬α are
equivalent, by definition of equivalence.
a
Exercise
1. Suppose that Σ1 α and that every member of Σ1 is also a member of
Σ2 . Is it true that Σ2 α?
2. Suppose that Σ1 α and that every member of Σ2 is also a member of
Σ1 . Is it true that Σ2 α?
3. Suppose that α is a member of Σ. Prove that Σ α.
6
Chapter 1
4. Let Σ be the premises of an inference, and α its conclusion. Prove that
the inference is valid if and only if Σ α.
1.4
Consistency
Definition 1.6. Sentences α and β are consistent iff it is possible for them both
to be true.
In other words, two sentences are consistent if there is at least one conceivable
situation in which they are both true. For example, the sentences ‘All mammals
are carnivorous’ and ‘All mice are mammals’ are consistent: even though one
of these sentences is true and the other false, there are conceivable situations in
which they would both be true.
Two sentences are inconsistent iff they are not consistent: that is, if it is
impossible for them both to be true. Thus, every sentence is inconsistent with
its own negation: for if α is true, then ¬α must be false.
Proposition 1.5. If α and β are inconsistent, then α entails ¬β.
Proof. Assume that α and β are inconsistent. ∴ There is no possible situation
in which they are both true, by definition of consistency. ∴ If α is true, then β
must be false. But by definition of negation, if β is false, then ¬β is true. ∴ If
α is true, then ¬β must be true. ∴ It is impossible for α to be true and ¬β to
be false. ∴ α entails ¬β, by definition of entailment.
a
For example, the sentence ‘John is a bachelor’ entails ‘John is not married’,
since ‘John is a bachelor’ is inconsistent with ‘John is married’.
As we will see in later chapters, much of the theory of formal logic is concerned with finding methods for proving the consistency or inconsistency of
collections of sentences. The reason for this is that consistency can be used
to define other logical properties, such as validity, entailment, and equivalence.
Therefore, an effective test for consistency will provide a test for these other
properties as well. For example, consider the following proposition:
Proposition 1.6. α entails β iff α and ¬β are inconsistent.
Proof. (i) Assume that α entails β. ∴ It is impossible for β to be false when α
is true. (definition of entailment). ∴ It is impossible for ¬β to be true when
α is true (definition of negation). ∴ α and ¬β are inconsistent (definition of
consistency).
(ii) Assume that α and ¬β are inconsistent. ∴ α entails ¬¬β by Proposition 1.5.
∴ It is impossible for ¬¬β to be false when α is true (definition of entailment).
But ¬¬β is equivalent to β by Proposition 1.4. ∴ It is impossible for β to be
false when α is true. ∴ α entails β (definition of entailment).
a
Thus, if we wish, we can replace the definition of entailment given above (i.e.
‘it is impossible for β to be false when α is true’) with the definition ‘α and ¬β
are inconsistent’.
Fundamental Concepts
7
We extend the definition of consistency to collections of sentences of any
size. A set Σ is consistent iff it is possible for all of the sentences in Σ to be true
together. Thus, Σ is consistent if there is at least one conceivable situation in
which all the sentences in Σ would be true. Otherwise, Σ is inconsistent.
When Σ contains a single sentence α, we describe the sentence itself as
being consistent or inconsistent. Thus, α is consistent iff it is possible for it to
be true: that is, if there is at least one conceivable case in which α would be
true. Similarly, α is inconsistent iff it is impossible for it to be true: that is, if
there are no conceivable cases in which α would be true.
Exercise
1. Suppose that α and β are consistent. Are ¬α and ¬β consistent?
2. Suppose that α and β are consistent. Are α and ¬β inconsistent?
3. Suppose that α and β are inconsistent. Are ¬α and ¬β inconsistent?
1.5
Necessity, contradiction, and contingency
Definition 1.7. A sentence α is necessarily true iff it is impossible for α to be
false.
Definition 1.8. A sentence α is necessarily false iff it is impossible for α to be
true.
Thus, a necessarily true sentence will be true in every conceivable situation,
and a necessarily false sentence will be false in every conceivable situation.
For example, the sentence ‘Bachelors are unmarried’ is necessarily true, while
‘Triangles have four sides’ is necessarily false. Necessarily true sentences are
often said simply to be ‘necessary’. Necessarily false sentences are often said to
be ‘contradictory’.
Definition 1.9. A sentence α is contingent iff α is neither necessarily true nor
necessarily false.
Thus, a contingent sentence will be true in at least one conceivable situation, and
false in at least one conceivable situation. For example, the sentence ‘Bachelors
are unhappy’ is contingent.
Proposition 1.7. α is necessarily true iff ¬α is necessarily false.
Proof. α is necessarily true iff it is impossible for α to be false (Definition 1.7).
It is impossible for α to be false iff α is true in every conceivable situation. α is
true in every conceivable situation iff ¬α is false in every conceivable situation
(definition of negation). ¬α is false in every conceivable situation iff it is impossible for ¬α to be true. It is impossible for ¬α to be true iff ¬α is necessarily
false (Definition 1.8). ∴ α is necessarily true iff ¬α is necessarily false.
a
8
Chapter 1
Exercise
1. Prove that every contingent sentence is consistent.
2. Prove that the negation of a contradictory sentence is necessary.
3. Prove that the negation of a contingent sentence is contingent.
4. Prove that a necessary sentence cannot entail a contingent sentence.
5. Suppose that α entails β, and that α is necessary. Prove that β is necessary.
6. Suppose that α entails β, and that β is a contradiction. Prove that α is a
contradiction.
7. Suppose that every sentence in Σ is contingent. Is Σ consistent?
2
2.1
Basic Truth Functional Logic
Logical form
Consider again this inference from Chapter 1:
All whales are mammals.
All mammals are chordates.
∴ All whales are chordates.
This inference is valid, and so too is the inference:
All cats are reptiles.
All reptiles are carnivores.
∴ All cats are carnivores.
But this second inference differs from the first only in having the terms ‘cats’,
‘reptiles’, and ‘carnivores’ in place of the terms ‘whales’, ‘mammals’, and ‘chordates’. Thus, both of these individual inferences are instances of the general
inference form:
All X are Y .
All Y are Z.
∴ All X are Z
The first inference can be constructed by substituting the terms ‘whales’, ‘mammals’, and ‘chordates’ for the variables X, Y , and Z; while the second can be
constructed by substituting ‘cats’, ‘reptiles’, and ‘carnivores’ for these same variables. In both cases, a valid inference results. It can be proved that substituting
any terms whatever for the variables in this inference form will produce a valid
inference.3
Saying that an inference form is invalid is only to say that it has at least
one invalid instance. There may well be other instances which are valid: that
is, there may be some instances whose conclusions cannot be false when their
premises are true. Thus, every instance of a valid inference form must be valid;
but not every instance of an invalid inference form need be invalid.
Inference forms are an example of logical form. The concept of a logical form
is more general than that of an inference form because individual sentences also
have logical forms. For example, the sentences ‘All mammals are reptiles’ and
‘All triangles are three-sided figures’ both have the logical form ‘All X are
Y ’: the first sentence can be generated by substituting ‘mammals’ for X and
‘reptiles’ for Y ; and the second by substituting ‘triangles’ for X and ‘three-sided
figures’ for Y . Thus, an inference form is composed of the logical forms of the
premises and conclusions of its instances.
The relationship between a sentence and its logical form is analogous to that
between arithmetical and algebraic equations. For example, the following are
3 The proof of this proposition requires the use of first order logic, which will be studied
later in this book.
9
10
Chapter 2
all true equations of arithmetic:
2+3=3+2
5+7=7+5
123 + 456 = 456 + 123
Of course, this list of equations could be extended indefinitely. However, the
entirety of this infinite list can be represented by a single algebraic equation:
a+b=b+a
Any numbers whatever can be substituted for the variables a and b, and each
time the resulting arithmetical equation will be true.
As its name suggests, formal logic is the study of logical form. In the same
way that algebraic equations reveal important properties of arithmetic, logical
forms reveal important facts about language. Using specially-developed symbolic languages, logicians describe and classify the various logical forms exhibited by the sentences of natural language. They then develop what are called
semantic theories or models for these symbolic languages. This enables them to
explain why certain logical forms are valid, why others are not; and much else
besides.
This is the process we will follow in this chapter. First, we will develop a
primitive symbolic language that will enable us to represent the logical forms
of some simple types of English sentence. We will then develop an equally
primitive semantic theory for this symbolic language. Finally, we will use this
language and its semantics to explain a number of logical properties, including
validity.
2.2
Truth functional operators
Consider the following inferences:
Roses are red and violets are blue.
∴ Roses are red.
Either you will work hard or you will fail the course.
You will not fail the course.
∴ You will work hard.
If there is smoke, then there is fire.
There is smoke.
∴ There is fire.
John is taller than Mary if and only if Mary is shorter than John.
Mary is shorter than John.
∴ John is taller than Mary.
It is easily seen that each of these inferences is valid. Their logical forms are:
11
Truth Functional Logic
A and B
∴A
Either A or B
Not-B
∴A
If A then B
A
∴B
A iff B
B
∴A
In these inference forms, the variables A, B, and C replace complete sentences,
rather than individual terms. In the first form, A is ‘Roses are red’ and B is
‘Violets are blue’. In the second, A is ‘You will work hard’ and B is ‘You will
fail the course’. In the third, A is ‘There is smoke’ and B is ‘There is fire’. In
the fourth, A is ‘John is taller than Mary’ and B is ‘Mary is shorter than John’.
It is clear that a valid inference will result whatever sentences are substituted
for the variables in each of these forms. For example, if any sentence of the form
‘A and B’ is true, then both of the component sentences represented by these
variables must be true. Thus, it will be impossible for the sentence ‘A’ to be
false when the sentence ‘A and B’ is true. It follows immediately that the
inference form is valid. Similarly, if any sentence of the form ‘A or B’ is true,
then one or the other of its component sentences must be true. Therefore, if it
so happens that one of these components is false, it will be impossible for the
other component to be false. Again, it follows that the inference form must be
valid.
The third inference form has a conditional sentence as its first premise. In
sentences of this form, the component sentence ‘A’ is called the antecedent,
and ‘B’ is called the consequent. The semantic properties of such sentences
are extremely complex: indeed, conditional sentences have been the subject of
intense logical scrutiny since ancient times. However, for present purposes, we
need only note that the antecedent of a conditional states a sufficient condition
for the truth of the consequent: that is, if a conditional ‘If A then B’ is true,
then the truth of the antecedent ‘A’ is enough to guarantee the truth of the
consequent ‘B’. Hence, if we add the antecedent as a second premise, it will be
impossible for the consequent to be false when these premises are true.4
Finally, the first premise of the fourth inference form states that the two
sentences ‘A’ and ‘B’ are sufficient conditions for one another: i.e. the truth
of either one of the component sentences guarantees the truth of the other.5
Therefore, if we add either one of these sentences as a second premise, it will be
impossible for the other sentence to be false.
Thus, in each of these inference forms, whatever particular sentences the
variables ‘A’ and ‘B’ might represent, it will be impossible for the conclusion
to be false when the premises are true. This shows that the meanings of the
component sentences have nothing to do with the validity of the inferences
in which they occur; for we can change these component sentences without
affecting the validity. Rather, the validity of these inferences is due to the
meanings of the words ‘and’, ‘not’, ‘or’, ‘if-then’, and ‘if and only if’: i.e. the
words that are left behind when the variables are introduced.
4 As always, it is essential to keep in mind the distinction between valid inferences and sound
inferences. It may well be false that the presence of smoke is a guarantee of the presence of
fire. If so, then the first premise of the example inference is false, and hence the inference
itself is unsound. However, it is still valid.
5 Note that if ‘A’ and ‘B’ are equivalent sentences, ‘A iff B’ will always be true.
12
Chapter 2
Again, there is an analogy with algebra. The fact that 2 + 3 = 3 + 2 has
nothing to do with the particular numbers 2 and 3. Rather, the truth of this
equation is due entirely to the nature of the operation of addition, which of
course is represented by the symbol +. This is demonstrated by the truth of
the algebraic equation a + b = b + a, which makes no reference to individual
numbers at all.
Words like ‘not’, ‘and’, ‘or’, ‘if-then’, and ‘iff’ are called truth functional
operators. They are combined with one or two component sentences to form
new sentences. These operators are called truth functional because the truth
values of the sentences they form can be deduced solely from the truth values
of the component sentences. For example, if we know independently that the
component sentences ‘A’ and ‘B’ are true, then we require no further information
to know that the sentence ‘A and B’ is true.
Sentences that contain operators are called compound sentences. Sentences
that contain no operators are called simple sentences. Operators that combine
two component sentences are called binary operators. Those that attach themselves to only one component sentence are called unary operators. For example,
‘not’ is a unary operator, while ‘or’ and ‘if-then’ are binary operators.
We are now ready to define a basic symbolic language. In this language,
simple sentences are represented by the upper-case letters A, B, C, D, etc.
which are called sentential variables.6 The symbols that represent the truth
functional operators are as follows, along with their traditional names:
not
or
and
if-then
iff
¬
∨
∧
→
↔
negation
disjunction
conjunction
material implication
material equivalence
The sentences of this symbolic language are called formulas, to distinguish them
from the sentences of a natural language like English.
Using the formulas of this symbolic language, the logical forms of the inferences given on page 11 above can be represented as follows:
A∧B
∴A
A∨B
¬A
∴B
A→B
A
∴B
A↔B
B
∴A
Despite its simplicity, this symbolic language is capable of representing the
logical forms of some rather complicated sentences. Consider these examples:
(a) If the defendant was not convicted by a jury, and he pleaded innocent,7
then an appeal is allowed if and only if the statute does not forbid it.
6 If we should ever need more than the twenty-six letters of the alphabet, we simply go
back to the beginning and apply subscripts: e.g. A1 , B1 , etc.
7 No one ever pleads ‘innocent’ in a Canadian court; rather, they plead ‘not guilty’. The
difference is of considerable legal importance, but need not concern us here.
13
Truth Functional Logic
(b) If the defendant was convicted by a judge, then he was not convicted by a
jury; and if the defendant did not plead guilty, then he pleaded innocent.
(c) The defendant was convicted by a judge, and he did not plead guilty.
(d) If the statute does not forbid it, then an appeal is allowed.
First we isolate the simple sentences, and assign a distinct sentential variable
to each:
A = The defendant was convicted by a jury.8
B = The defendant pleaded innocent.9
C = An appeal is allowed.
D = The statute forbids an appeal.
E = The defendant was convicted by a judge.
F = The defendant pleaded guilty.
We can now use these sentential variables, along with the truth functional operators, to represent the logical forms of the four compound sentences (a) through
(d):
(a)
(b)
(c)
(d)
(¬A ∧ B) → (C ↔ ¬D)
(E → ¬A) ∧ (¬F → B)
E ∧ ¬F
¬D → C
To see how useful symbolic translations of natural language sentences can
be, consider this question: Given that the sentences (a), (b), and (c) are true,
does sentence (d) also have to be true? This question is not easily answered by
examining the English sentences themselves. But there is a simple mechanical
method that can be applied to their symbolic translations that will always provide a correct answer to such questions. However, in order to use this method,
we must first provide a semantic theory for the symbolic language. This is the
task of the next section.
Exercise
Let A = ‘The ticket holder works for the company’, B = ‘The ticket is drawn’,
C = ‘The ticket holder wins the prize’, and D = ‘The ticket has expired’. Given
these values, write out the English sentence that is represented by formula (a)
above.
8 Since a simple sentence contains no truth functional operators, ‘The defendant was not
convicted by a jury’ is a compound sentence, rather than a simple one, because it contains
the operator ‘not’.
9 Simple sentences should be able to stand on their own, and so pronominal expressions
such as ‘he’, ‘she’, and ‘it’ are replaced by the words that they refer to.
14
2.3
Chapter 2
Truth tables
Recall that an operator is said to be truth functional when the truth values of
the compound sentences it forms can be calculated solely on the basis of the
truth values of the component sentences that it joins. Thus, ∧ (conjunction)
is a truth functional operator, because to find out the truth value of a formula
of the form α ∧ β,10 one need only know the truth values of the component
formulas α and β. If both α and β are true, then the compound formula α ∧ β
must also be true. On the other hand, if one or both of the component formulas
is false, then α ∧ β itself must be false.
This information about the meaning of the operator ∧ can be summarised
in the following truth table:
∧ 0 1
0 0 0
1 0 1
The truth values of the first component formula α are listed on the left-hand
side of the table. The truth values of the second component formula β are listed
across the top of the table. The remaining rows and columns give the truth
value of the compound formula α ∧ β for each combination of truth values of its
components.
We provide a semantics for our symbolic language by assigning each truth
functional operator its own truth table. In this way, each operator is given a
definite ‘meaning’. As we will see, this is enough to give every formula of our
symbolic language a meaning.
Since negation is a unary operator, the truth table for ¬ requires only two
columns:
¬
0 1
1 0
This table shows that the truth value of the compound formula ¬α is always
the opposite of that of the component formula α.
The truth tables for the remaining operators in our symbolic language are
as follows:
∨ 0
0 0
1 1
1
1
1
→ 0
0 1
1 0
1
1
1
↔ 0
0 1
1 0
1
0
1
The reasoning behind the values assigned in these truth tables is somewhat more
complex than in the case of negation and conjunction. For example, the table for
∨ (disjunction) has the compound formula true when both of the components
10 In our symbolic language, the variables A, B, C, etc. are reserved for simple sentences.
However, the truth functional operators can be used to combine both sentential variables and
compound formulas. Hence we will use lower-case Greek letters α, β, γ, etc. to represent
formulas of either type. The distinction between these two kinds of variable will be made
more explicit in Chapter 4, where a rigorous syntactic definition of our symbolic language will
be presented.
15
Truth Functional Logic
are true. Yet when we say things like ‘You can have either Jell-O or ice-cream’,
we do not usually mean that one can have both.11
Similarly, when both α and β are true, the truth table for material implication tells us that the formula α → β is also true. But this would seem to
imply that the conditional sentence ‘If snow is white then grass is green’ is true,
since both its antecedent and consequent are true. Likewise, this table indicates
that the conditional ‘If snow is blue then roses are red’ is true, since here the
antecedent is false and the consequent is true.
Many of these apparent oddities are due to the Law of Excluded Middle.
According to this law, every sentence must be either true or false. Therefore,
since we would not say that the sentence ‘If snow is white then grass is green’ is
false, it follows that it must be true. Fortunately, questions of how accurately
the truth functional operators reflect the meanings of English words like ‘and’,
‘or’, and ‘if’, belong to the philosophy of logic, rather than to formal logic itself.
For our purposes here, it is enough to note that the truth tables given above are
probably the best that can be done within the two-valued system imposed by the
Law of Excluded Middle.12 Nevertheless, despite its shortcomings, our symbolic
language gives us the means to explain a number of important concepts.
Truth tables allow us to calculate the truth value of any formula in our
symbolic language, provided only that we know the truth value of each of the
sentential variables that the formula contains. For example, consider the following formula:
(A ∨ B) ∧ (¬B ∨ A)
(2.1)
Assume that the variable A is true, and the variable B is false. This assumption
may be represented as follows:
(A ∨ B ) ∧ (¬B ∨ A)
1
0
0
1
Applying the truth table for ∨ to the component A ∨ B gives us:
(A ∨ B) ∧ (¬B ∨ A)
1
0
1
Next we apply the truth table for ¬ to the component ¬B:
(A ∨ B) ∧ (¬B ∨ A)
1
1
1
Now we are able to use the table for ∨ again to calculate a value for the component ¬B ∨ A:
(A ∨ B) ∧ (¬B ∨ A)
1
1
Finally, we apply the table for ∧ to obtain a truth value for the whole formula:
(A ∨ B) ∧ (¬B ∨ A)
1
Thus, we have shown that when A = 1 and B = 0, the formula itself has a truth
value of 1.
11 In
those cases where one cannot have both, ice-cream is recommended.
of the multiple-valued logics mentioned on page 3 were originally formulated in
order to provide a more accurate semantics for conditional sentences.
12 Many
16
Chapter 2
Exercise
Suppose that A = 1, B = 0, and C = 0. Use the truth tables to determine the
truth value of each of the following formulas:
1. (A → B) ∨ (B → C)
2. (A ∨ B) ∧ (A ∨ C)
3. (A ∧ B) ∨ (A ∧ C)
4. ¬(A ∨ ¬(B → C))
5. (A ↔ C) ∨ ((¬A → B) ∧ (A → ¬C))
2.4
Logical truth
Each distinct way of assigning truth values to the sentential variables of a formula is called a valuation. For any formula, the number of valuations is equal
to 2n , where n is the number of distinct sentential variables in the formula. For
example, the formula:
A → (B → A)
(2.2)
contains two different sentential variables, and so there are 22 = 4 valuations.
These can be listed as follows:
A B
0 0
0 1
1 0
1 1
A formula with three distinct sentential variables has 23 = 8 valuations; one
with four such variables has 24 = 16 valuations; and so on.
Definition 2.1. A formula is logically true iff it has the truth value 1 on every
valuation.
It is usually not necessary to examine all of these valuations in order to show
that a given formula is a logical truth. Suppose that formula (2.2) above is not
a logical truth. Then for some valuation, the formula must have a value of 0.
We represent this assumption as follows:
A → (B → A)
0
But from the truth table for →, we see that this formula can have the value 0
only when its antecedent is 1 and its consequent is 0:
A → (B → A)
1
0
0
17
Truth Functional Logic
Applying the → table to the consequent gives us:
A → (B → A)
1
∗
0
1
0
0
∗
But no valuation can assign the variable A both 0 and 1. Therefore, the assumption that the formula has a value of 0 on some valuation cannot be true. It
follows that formula (2.2) must have the value 1 on every valuation, and hence
that it is a logical truth.
If the assumption that a formula has a value of 0 does not lead to an absurdity, then we know that the formula is not a logical truth. In effect, we will
have discovered a valuation on which the formula is false. Consider again the
example of formula (2.1):
(A ∨ B) ∧ (¬B ∨ A)
0
The truth table for ∧ tells us that it takes a value of 0 whenever one or both of
its conjuncts has 0. Assume that the first conjunct is 0:
(A ∨ B) ∧ (¬B ∨ A)
0
0
But ∨ takes the value 0 only when both disjuncts are 0:
(A ∨ B ) ∧ (¬B ∨ A)
0 0 0
0
Now we have calculated truth values for all of the sentential variables in the
formula: i.e. we have produced a valuation. We can complete the calculation as
follows:
(A ∨ B ) ∧ (¬B ∨ A)
0 0 0
0
1 0 1 0
This distribution of truth values is consistent with the truth tables of the truth
functional operators. Thus, we have shown that there is a valuation where this
formula has a truth value of 0. It follows that formula (2.1) is not a logical
truth.
Definition 2.2. A formula is logically false iff it has the truth value 0 on every
valuation.
Definition 2.3. A formula is logically indeterminate iff it is neither logically
true nor logically false.
Hence, α is logically indeterminate when there is at least one valuation where
α = 0, and at least one valuation where α = 1.
If the assumption that a formula α has the truth value 1 leads to an absurdity, then α must be a logical falsehood; for there can be no valuation on
which the formula has the value 1. But if no such absurdity results from the
assumption, then we know that α is either logically true or logically indeterminate. And if no absurdity results from the further assumption that α = 0, then
the only possibility is that α is logically indeterminate. Thus, this method of
‘backward calculation’ allows us to determine whether any formula is logically
true, logically false, or logically indeterminate.
18
Chapter 2
Exercise
1. Determine whether each of the following formulas is logically true, logically
false, or logically indeterminate:
(a) (A ∧ (B ∨ C)) → (A ∧ C)
(b) (¬B ∨ A) → ¬B
(c) ¬(¬B → (B → A))
(d) (A ∨ (B ∧ C)) → (A ∨ C)
(e) (A ∧ B) ∨ (¬A ∧ ¬B)
2. Assume that α is logically true and that β is logically false. Determine
whether each of the following is logically true, logically false, or logically
indeterminate:
α∨β
2.5
α∧β
α→β
β→α
α↔β
Logical equivalence
Definition 2.4. Formulas α and β are logically equivalent iff the truth value of
α is identical with the truth value of β on every valuation.
Thus, if the assumption that α and β have different truth values on some valuation leads to an absurdity, we may infer that they are logically equivalent. For
example, make this assumption about the following formulas:
¬(A ∧ B)
1
(¬A ∨ ¬B)
0
The truth tables for ¬ and ∨ then imply:
¬(A ∧ B)
1
0
(¬A ∨ ¬B)
0
0 0
We can now calculate values for both A and B in the second formula, and
rewrite them in the first:
¬(A ∧ B )
1 1 0 1
∗ ∗ ∗
(¬A ∨ ¬B )
0 1 0 0 1
The values marked * are inconsistent with the truth table for ∧. Thus, it is
impossible for the first formula to be 1 and the second 0.
But we do not yet know that these formulas are logically equivalent, for we
have not yet ruled out the possibility that the first is 0 and the second is 1:
¬(A ∧ B)
0
(¬A ∨ ¬B)
1
However, once the calculation is complete, we again find a contradiction:
¬(A ∧ B )
0 1 1 1
(¬A ∨ ¬B )
0 1 1 0 1
∗
∗ ∗
19
Truth Functional Logic
In this case, the values marked * are inconsistent with the truth table for ∨.
Thus, it is impossible for either of these formulas to be 1 when the other is 0.
We may conclude that they are logically equivalent.
Proposition 2.1. If α and β are logically equivalent, then α ↔ β is logically
true.
Proof. Suppose α and β are logically equivalent. ∴ α and β must have identical
truth values on every valuation (Definition 2.4). ∴ α = 1 on every valuation
where β = 1, and α = 0 on every valuation where β = 0. But (α ↔ β) = 1
whenever α = 1 = β and whenever α = 0 = β. ∴ (α ↔ β) = 1 on every
valuation. ∴ α ↔ β is a logical truth (Definition 2.1).
a
Exercise
1. Determine which of the following pairs of formulas are logically equivalent:
(a) ¬A ∧ ¬B, ¬(A ∨ B)
(b) ¬A ∨ ¬B, ¬(A ∨ B)
(c) A → ¬B, B → ¬A
(d) A → B, ¬B → ¬A
(e) A ∨ (B ∧ C), (A ∧ B) ∨ (A ∧ B)
(f) A ∨ (B ∧ C), (A ∨ B) ∧ (A ∨ B)
2. Prove that if α and β are both logically true, then they are logically equivalent.
3. Prove that α and β are logically equivalent iff ¬α and ¬β are logically
equivalent.
2.6
Logical implication
Definition 2.5. The formula α logically implies the formula β iff there is no
valuation on which α = 1 and β = 0.
Thus, if α logically implies β, the assumption that there is some valuation where
α = 1 and β = 0 will lead to an absurdity. For example, assume the following:
¬(A ∨ B)
¬A ∨ ¬B
¬(A ∨ B)
¬A ∨ ¬B
¬(A ∨ B )
¬A ∨ ¬B
1
0
This implies:
0
0
0
which in turn implies:
0 0 0
∗
∗
1
∗
1
∗
20
Chapter 2
The values marked * are inconsistent; for A and B cannot be both true and
false simultaneously. Therefore, the original assumption must be false.
On the other hand, the following values are consistent:
¬A ∨ ¬B
0 1 1 1 0
¬(A ∨ B )
0 1 1 0
Thus, on a valuation where A = 1 and B = 0, (¬A ∨ ¬B) = 1 and ¬(A ∨ B) = 0.
Consequently, the former does not logically imply the latter.
Proposition 2.2. α logically implies β iff α → β is logically true.
Proof. (⇒) Assume that α → β is not logically true. Then by Definition 2.1,
there is a valuation where (α → β) = 0. But the truth table for → has 0 only
where α = 1 and β = 0. ∴ By Definition 2.5, α does not logically imply β.
(⇐) Assume that α does not logically imply β. Then by Definition 2.5, there
is a valuation where α = 1 and β = 0. But the truth table for → has 0 when
α = 1 and α = 0. ∴ There is a valuation where (α → β) = 0. ∴ α → β is not
logically true.
a
Exercise
1. Prove that α logically implies β iff ¬β logically implies ¬α.
2. Prove that if α is logically true, then any formula whatever logically implies
α.
3. Prove that if α is logically false, then α logically implies any formula whatever.
4. Assume that α logically implies β. Does ¬α logically imply ¬β?
2.7
Validity and entailment
Recall that an inference is valid iff there is no conceivable situation where its
premises are true and its conclusion false. Recall also that an inference is valid
iff its premises entail its conclusion.13 These concepts of validity and entailment apply equally to inferences formed in our symbolic language. But for the
sentences (i.e. formulas) of this artificial language, a ‘conceivable situation’ is
simply a valuation. Therefore, in this language an inference will be valid iff
there is no valuation where all of its premises are 1, and its conclusion is 0.
The following proposition demonstrates a close relationship between validity
and logical truth:
Proposition 2.3. An inference with premises α1 . . . αn and conclusion β is
valid iff the formula α1 ∧ . . . ∧ αn → β is logically true.
13 See
the Exercise on page 5.
21
Truth Functional Logic
Proof. (⇒) Assume that α1 ∧ . . . ∧ αn → β is not logically true. ∴ There is a
valuation where (α1 ∧ . . . ∧ αn ) = 1 and β = 0. (truth table for →) ∴ On that
valuation, each αi = 1 (truth table for ∧). ∴ There is a valuation where the
premises are 1 and the conclusion is 0. ∴ The inference is invalid.
(⇐) Assume that the inference is invalid. ∴ There is a valuation where each αi
= 1 and β = 0. ∴ (α1 ∧ . . . ∧ αn ) = 1 on that valuation (truth table for ∧). ∴
The formula (α1 ∧ . . . ∧ αn → β) = 0 on that valuation (truth table for →). ∴
That formula is not a logical truth.
a
Proposition 2.3 provides us with an effective test of the validity of an inference.
To see whether a given inference is valid, test the corresponding → formula for
logical truth. If the formula is a logical truth, the inference is valid; and if the
formula is not a logical truth, the inference is invalid.
However, this procedure does not provide an effective test for entailment.
Recall that the definition of Σ α allows Σ to contain an infinite number of
sentences. But in such cases, no formula of the form α → β can be constructed,
because the antecedent α would be an infinitely long conjunction. Entailment
requires more powerful semantic techniques, which will be developed in Chapter
5.
Exercise
Determine whether the following inferences are valid:
1. A ∧ ¬B; ∴ B → A
2. A → (B ∧ C), C ↔ B; ∴ ¬C → ¬A
3. (A → B) → (B → C), B; ∴ A → C
4. (A ∨ B) → C, ¬A; ∴ ¬B → ¬C
5. (A ∨ B) → C, ¬C; ∴ B → ¬A
2.8
Some traditional rules of inference and equivalence
In sections 2.1 and 2.2 above we saw that the validity of some inferences depends
solely on the logical forms of their premises and conclusions, and has nothing
to do with the specific content of these sentences. The same is true for our
symbolic language. Consider the following inference form:
α→β
α
∴β
It is easily demonstrated that whatever formulas are substituted for α and β,
the resulting inference will be valid: for if both α and α → β are true, the truth
table for → tells us that the consequent β must also be true. Since this is a
general fact about the logical system, it constitutes a metatheorem.
22
Chapter 2
Since an inference is valid iff the premises of that inference entail its conclusion, we can rewrite the above inference form like this:
{α → β, α} β
This says that the formulas α → β and α together entail the formula β.
A number of these valid inference forms have been singled out by logicians
as being of particular interest, and have been given individual names. Here is a
list of the ones most commonly found in introductory logic texts:
Modus Ponens
Modus Tollens
Simplification
Addition
Conjunction
Hypothetical Syllogism
Disjunctive Syllogism
Constructive Dilemma
Destructive Dilemma
{α → β, α} β
{α → β, ¬β} ¬α
{α ∧ β} α
{α} α ∨ β
{α, β} α ∧ β
{α → β, β → γ} α → γ
{α ∨ β, ¬α} β
{α → β, γ → δ, α ∨ γ} β ∨ δ
{α → β, γ → δ, ¬β ∨ ¬δ} ¬α ∨ ¬γ
These are called ‘rules of inference’. In each case, it can be shown that any
inference that whose premises and conclusion have the indicated forms will be
valid within our logical system. When this is so, we will say that the inference
rule itself is valid.
In a similar manner, logicians have also given names to certain pairs of equivalent formulas. Here is a list of some of the more common ‘rules of equivalence’:
De Morgan’s Laws
Commutation
Association
Distribution
Double Negation
Implication
Equivalence
Transposition
Exportation
Tautology
¬(α ∨ β) ≡ (¬α ∧ ¬β)
¬(α ∧ β) ≡ (¬α ∨ ¬β)
(α ∨ β) ≡ (β ∨ α)
(α ∧ β) ≡ (β ∧ α)
((α ∨ β) ∨ γ) ≡ (α ∨ (β ∨ γ))
((α ∧ β) ∧ γ) ≡ (α ∧ (β ∧ γ))
(α ∨ (β ∧ γ)) ≡ ((α ∨ β) ∧ (α ∨ γ))
(α ∧ (β ∨ γ)) ≡ ((α ∧ β) ∨ (α ∧ γ))
α ≡ ¬¬α
(α → β) ≡ (¬α ∨ β)
(α ↔ β) ≡ ((α → β) ∧ (β → α))
(α → β) ≡ (¬β → ¬α)
((α ∧ β) → γ) ≡ (α → (β → γ))
α ≡ (α ∨ α)
α ≡ (α ∧ α)
Each of these equivalence rules represents an infinite number of individual pairs
of formulas, just as each of the inference rules represents an infinite number
of individual inferences. For example, the Rule of Tautology tells us that A is
equivalent to (A∧A), that B is equivalent to (B ∧B), that (A → B) is equivalent
Truth Functional Logic
23
to ((A → B) ∧ (A → B)), and so on. This is because the metavariables α, β, γ,
etc. can represent any formula whatever.
These combinations of metavariables with operators are called formula schemas. A formula is said to be an instance of a schema if the formula can be
generated from the schema by replacing the metavariables. Thus, the formulas
(A ∧ A), (B ∧ B), and ((A → B) ∧ (A → B)) are all instances of the single
formula schema (α ∧ α): the first formula is generated by replacing α with A;
the second by replacing α with B; and the third by replacing α with (A → B).
The relation between a formula and a formula schema is precisely analogous to
the relation between a sentence of natural language and the logical form of that
sentence. It is essential to realise that, although every instance of a formula
schema is part of the symbolic language, the schema itself is not.
We will say that a rule of equivalence is valid when every one of its instances
constitutes an equivalent pair of formulas. Thus, to prove that an equivalence
rule is invalid, one need only find a single pair of formulas that are instances of
the schemas of that rule, but which are not equivalent.
These rules of inference and equivalence are used in what are called ‘natural deduction systems’. The validity of an inference is proved within a natural
deduction system by repeatedly applying the rules of inference and equivalence
to the premises, and thereby transforming these formulas until the desired conclusion is produced. For example, the following demonstrates that the premises
(A ∨ B) → C and ¬C entail the conclusion ¬A:
(1)
(2)
(3)
(4)
(5)
(A ∨ B) → C
¬C
¬(A ∨ B)
¬A ∧ ¬B
¬A
(premise)
(premise)
(from 1,2 by Modus Tollens)
(from 3 by De Morgan’s Rules)
(from 4 by Simplification)
A derivation is an ordered list of formulas that meets certain conditions. The
conditions that govern derivations in a system of natural deduction include the
requirement that every formula in the list (except those designated as premises)
must come from previous formulas in the list, through the application of a rule of
inference or a rule of equivalence. As the annotations show, the list of formulas
given above meets this requirement, and therefore qualifies as a derivation.
Natural deduction systems will not be further studied in this book. However,
the concept of formal derivability that will be examined in Chapter 4 has many
similar features.
Exercise
1. Prove the validity of each of the rules of inference listed above.
2. Prove the validity of each of the rules of equivalence listed above.
3
Introduction to Metatheory
Our symbolic language and its truth table semantics together constitute a logical
system. As we have seen, certain formulas within this system can be shown to
be logically true. Such formulas are usually called ‘theorems’ of the system.
What theorems a given system contains is a matter of great importance. But
equally important are the numbered propositions that we so laboriously proved
in the preceding chapter. These propositions state facts about the system, and
are properly known as metatheorems. Thus, the theorems of a system are part
of the system itself, while its metatheorems are not.14
In this chapter, we will prove a few simple metatheorems about the formal
system that we have devised. We will also introduce a number of basic mathematical concepts that will be needed to prove the more advanced metatheorems
of the chapters that follow.
3.1
Definability of truth functional operators
The equivalence rule:
α ∧ β ≡ ¬(¬α ∨ ¬β)
is easily proved to be valid. By definition of logical equivalence (Definition 2.4
on page 18), this means that the truth value of the first formula will be identical
to the truth value of the second formula on every valuation. But this means
that substituting the one formula for the other will never change the truth value
of any other formula in which they occur. For example, the formulas:
(A ∧ B) → C
¬(¬A ∨ ¬B) → C
will have identical truth values on every valuation. Therefore, adding the operator ∧ to a formal language that already has the operators ¬ and ∨ will add
nothing to the expressive power of that language: every formula in the expanded
language can be matched with a logically equivalent formula in the unexpanded
language. We say that the operator ∧ is definable by the operators ¬ and ∨.
The operator → is also definable by ¬ and ∨, as the following equivalence
shows:
α → β ≡ ¬α ∨ β
Thus, the expressive power of a language that already contains ¬ and ∨ will not
be increased by the addition of →. But → and ∧ are themselves sufficient to
define the operator ↔:
α ↔ β ≡ (α → β) ∧ (β → α)
Since we already know that ¬ and ∨ are sufficient to define both ∧ and →, it
follows that ¬ and ∨ are also sufficient by themselves to define ↔. Thus, a
14 Logicians often use the word ‘theorem’ to refer to metatheorems as well, and we will do
the same here. Since the theorems of a system are always written in the symbolic language
of that system, this somewhat imprecise terminology should not lead to confusion.
24
25
Metatheory
symbolic language that contains only the two operators ¬ and ∨ can express
everything that a language containing the five operators ¬, ∨, ∧, →, and ↔ can
express.
However, the pair of operators ¬, ∨ is not unique in this respect: both the
pairs ¬, ∧ and ¬, → have expressive power equal to ¬, ∨. This can be shown
simply by verifying the validity of the equivalences:
α ∨ β ≡ ¬(¬α ∧ ¬β)
α ∨ β ≡ ¬α → β
Since we already know that ¬ and ∨ are sufficient to define ∧ and →, these two
equivalences are all that is needed to show that → is definable by ¬ and ∧, and
that ∧ is definable by ¬ and →.
Exercise
1. For each of the following formulas, construct a logically equivalent formula
that contains only the operators ¬ and ∨:
A↔B
(A ∧ ¬B) ∨ C
((A → B) ∧ ¬B) → ¬A
2. For each of the following formulas, construct a logically equivalent formula
that contains only the operators ¬ and ∧:
A→B
(A ∧ ¬B) ∨ C
((A → B) ∧ ¬B) → ¬A
3. For each of the following formulas, construct a logically equivalent formula
that contains only the operators ¬ and →:
A∧B
3.2
(A ∧ ¬B) ∨ C
((A ∨ B) ∧ ¬B) → A
Truth functional completeness
Recall that the truth table for a unary truth functional operator needs to specify
values for only two cases: one value for when the component formula is 0, and
another value for when the component formula is 1. Each of these cases will have
two possible values: 0 and 1. Thus, there are precisely 2 × 2 = 4 different truth
tables for a unary operator. The truth tables for these four distinct operators
can be set out as follows:
0
1
⊗1
0
0
⊗2
0
1
⊗3
1
0
⊗4
1
1
The truth values for the component formula are listed in the left-hand column.
The truth values for the compound formula are listed under the corresponding
operator. Our negation operator ¬ is identical with ⊗3 .
26
Chapter 3
A truth table for a binary truth functional operator has four entries. Since
there are two possible values for each entry, there is a total of 24 = 16 distinct
binary operators:
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
⊗9
1
0
0
0
⊗1
0
0
0
0
⊗2
0
0
0
1
⊗10
1
0
0
1
⊗3
0
0
1
0
⊗11
1
0
1
0
⊗4
0
0
1
1
⊗5
0
1
0
0
⊗6
0
1
0
1
⊗12
1
0
1
1
⊗13
1
1
0
0
⊗14
1
1
0
1
⊗7
0
1
1
0
⊗15
1
1
1
0
⊗8
0
1
1
1
⊗16
1
1
1
1
In these tables, ∧ is ⊗2 , ∨ is ⊗8 , ↔ is ⊗10 , and → is ⊗14 .
We have already seen that the operators ¬ and ∨ are capable of defining ∧,
→, and ↔. In fact, ¬ and ∨ by themselves are capable of defining every binary
truth functional operator. For each one of the above truth tables, it is possible
to construct a formula that has those same values, but that contains only the
operators ¬ and ∨.
For example, consider ⊗7 . According to the truth table for this operator, a
formula of the form α ⊗7 β will have the truth value 1 in two cases: where α = 0
and β = 1, and also where α = 1 and β = 0. For the first case, we construct
the conjunction ¬α ∧ β. Clearly, this conjunction will have the value 1 iff α = 0
and β = 1. For the second case, we construct the conjunction α ∧ ¬β. This
conjunction will be 1 iff α = 1 and β = 0. Consequently, the disjunction:
(¬α ∧ β) ∨ (α ∧ ¬β)
(3.1)
will be 1 iff either α = 0 and β = 1 or α = 1 and β = 0, and will be 0 in all
other cases.
Thus, formula 3.1 will be logically equivalent to the formula α ⊗7 β.15 But
we already know that ∧ is definable by ¬ and ∨, and so 3.1 is logically equivalent
to a formula that contains only ¬ and ∨:
¬(α ∨ ¬β) ∨ ¬(¬α ∨ β)
(3.2)
Thus, ⊗7 is definable by ¬ and ∨ alone.
The same procedure can be applied to any binary operator. Each row in
the truth table can be represented by a conjunction of the component formulas.
The first row is represented by the conjunction ¬α ∧ ¬β, because on this row
both components are 0. The second row is represented by ¬α ∧ β, because there
α = 0 and β = 1; and so on. To define a given operator, one forms a disjunction
15 Strictly speaking, these are formula schemas rather than formulas, since they are not part
of the symbolic language.
27
Metatheory
of those conjunctions that represent the truth table rows where the operator has
1.16 Thus, each binary truth functional operator can be defined by a formula
that contains only the operators ¬, ∨, and ∧. But each of these formulas is
logically equivalent to a formula that contains only ¬ and ∨. Consequently, we
say that the pair of operators {¬, ∨} is truth functionally complete.
Exercise
1. Prove that the pair of operators {¬, ∧} is truth functionally complete.
2. Prove that the pair of operators {¬, →} is truth functionally complete.
3. Prove that the operator ⊗9 is truth functionally complete.
4. Prove that the operator ⊗15 is truth functionally complete.
3.3
Disjunctive and conjunctive normal forms
A formula is in disjunctive normal form (DNF) iff it is a disjunction, in which
each disjunct is a conjunction that contains only sentential variables or the
negations of sentential variables. For example, each of the following formulas is
in disjunctive normal form:
(A ∧ B) ∨ (¬A ∧ B)
(A ∧ B ∧ ¬C) ∨ (C ∧ D ∧ B) ∨ (¬B ∧ ¬D)
(A ∧ B) ∨ (B ∧ A)
Note that we have omitted some of the parentheses that would normally occur
in the second formula. We can do this because the Rule of Association (see page
22) shows that the placement of these parentheses makes no difference to the
truth value of disjunctions or conjunctions.
We allow for disjunctions that contain a single disjunct, and conjunctions
that contain a single conjunct. Thus, each of the following are also considered
to be in DNF:
A ∨ (¬B ∧ C)
A ∧ ¬B ∧ C
¬B
In the first case, the disjunct A is considered to be a conjunction that contains
a single conjunct. In the second case, we have a disjunction that contains a
single conjunction. In the third case, we again have a disjunction that contains
a single conjunction, which itself contains a single conjunct.
Let α be a formula that contains only the sentential variables A1 . . . An ,
arranged in some fixed (e.g. alphabetical) order. Then, as we saw in section 2.4
16 The ⊗ operator is a special case, since it has the value 0 in all cases. But here we can
1
select the contradictory formula α ∧ ¬α to represent the operator.
28
Chapter 3
above, α will have 2n valuations. For each valuation that assigns α the truth
value 1, form a conjunction of the sentential variables, using Ai if the valuation
assigns 1 to that variable, and using ¬Ai if the valuation assigns 0. Finally, form
a disjunction of all of these conjunctions. This new formula will be in DNF.
In cases where no valuation assigns 1 to α (i.e. where α is logically false),
form the conjunction:
A1 ∧ ¬A1 ∧ A2 ∧ . . . ∧ An
Again, this formula will be in DNF.
As an example, consider the formula:
(A ∨ B) → A
(3.3)
Applying the above procedure, we generate the formula:
(¬A ∧ ¬B) ∨ (A ∧ ¬B) ∨ (A ∧ B)
(3.4)
which is in DNF.
A formula α will always be logically equivalent to the DNF formula that is
generated by this procedure. When a valuation makes α true, the conjunction
corresponding to that valuation will also be true; and consequently the entire
disjunction will be true. But when a valuation makes α false, it will have no
corresponding conjunction. Furthermore, such a valuation will make all the
conjunctions in the DNF false; for it will make at least one conjunct false in
each. Consequently, the entire disjunction will be false.
As an illustration, again consider formula (3.3). Given a valuation on which
A = 1 and B = 0, this formula will be true. But this valuation also makes the
second disjunct of (3.4) true, which suffices to make the entire formula true.
Next, consider a valuation on which A = 0 and B = 1. This will make formula
(3.3) false. But it will also make each of the disjuncts in (3.4) false: since B = 1,
¬B = 0, making the first conjunction false; and since A = 0, the second and
third conjunctions are false. This suffices to make the entire disjunction false.
Thus, the DNF formula will be true on any valuation that makes α true; and
it will be false on any valuation that makes α false. It follows from Definition
2.4 that α is logically equivalent to this formula.
Formulas often have more than one equivalent DNF. Let α be a formula,
and let ‘β ⇒ γ’ mean ‘Replace any subformula of the form β that occurs in α
with a subformula of the form γ’.17 Then repeated applications of the following
syntactic transformations will produce a DNF formula that is equivalent to α:
[1] (β ↔ γ) ⇒ ((β → γ) ∧ (γ → β))
[2] (β → γ) ⇒ (¬β ∨ γ)
[3] ¬(β ∨ γ) ⇒ (¬β ∧ ¬γ)
[4] ¬(β ∧ γ) ⇒ (¬β ∨ ¬γ)
17 α is a subformula of ¬α; α and β are subformulas of α ∧ β; and so on. Every formula is
considered to be a subformula of itself.
29
Metatheory
[5] ¬¬β ⇒ β
[6] (β ∧ (γ ∨ δ)) ⇒ ((β ∧ γ) ∨ (β ∧ δ))
[7] ((β ∨ γ) ∧ δ) ⇒ ((β ∧ δ) ∨ (γ ∧ δ))
The resulting DNF will be equivalent to α because each of the above transformations corresponds to one of the rules of equivalence given in section 2.8: [1]
corresponds to Equivalence; [2] to Implication; [3] and [4] to De Morgan’s Laws;
and so on.
For example, when applied to formula (3.3), this procedure produces the
DNF formula:
(¬A ∧ ¬B) ∨ A
(3.5)
which obviously differs from the DNF (3.4). Nevertheless, this formula is equivalent to (3.3).
Although the preceding discussion does not constitute a formal proof, we
can nevertheless be confident in asserting:
Theorem 3.1. Every formula has a logically equivalent disjunctive normal
form.
A formula is in conjunctive normal form (CNF) iff it is a conjunction, in
which each conjunct is a disjunction that contains only sentential variables or
the negations of sentential variables. For example, each of the following formulas
is in conjunctive normal form:
(A ∨ B) ∧ (¬A ∨ B)
(A ∨ B ∨ ¬C) ∧ (C ∨ D ∨ B) ∧ (¬B ∨ ¬D)
(A ∨ B) ∧ (B ∨ A)
A ∧ ¬B ∧ C
¬B
Again, suppose that α contains the sentential variables A1 . . . An . To produce an equivalent CNF formula using valuations, proceed as follows: For every
valuation on which α = 0, form a disjunction of the sentential variables, using
Ai if Ai = 0 on that valuation, and using ¬Ai otherwise. Then form a conjunction of all of these disjunctions. The resulting formula will be in CNF. And in
those cases where there is no valuation on which α = 0 (i.e. where α is logically
true), form the disjunction:
A1 ∨ ¬A1 ∨ A2 ∨ . . . ∨ An
The corresponding transformation procedure is the same as the one for disjunctive normal forms, except that clauses [6] and [7] are replaced by:
[6] (β ∨ (γ ∧ δ)) ⇒ ((β ∨ γ) ∧ (β ∨ δ))
[7] ((β ∧ γ) ∨ δ) ⇒ ((β ∨ δ) ∧ (γ ∨ δ))
30
Chapter 3
When applied to formula (3.3), these two procedures produce the equivalent
CNF formulas:
A ∨ ¬B
(¬A ∨ A) ∧ (¬B ∨ A)
(3.6)
(3.7)
Finally, corresponding to Theorem 3.1, we have:
Theorem 3.2. Every formula has a logically equivalent conjunctive normal
form.
Exercise
Construct equivalent disjunctive and conjunctive normal forms for each of the
following formulas:
1. A → (B → A)
2. (A ∨ ¬B) ∧ (A → B)
3. ¬(A ∨ (B ∨ C))
4. B → (¬B → A)
5. A ∧ (¬B ∧ C)
6. ¬((A → B) ∨ (B → A))
7. (A ∧ B) ∨ (A ∧ C)
3.4
Some basic concepts of set theory
A set is any collection of things. The elements (or ‘members’) of a set are simply
those things that the set contains. We use the set braces { and } to enclose the
names of the elements of a particular set. Thus, the following are examples of
sets:
X = {1, 2, 3, 4}
Y = {4, 5, 6}
Z = {3, 4}
Note that the order of the elements is not significant: for example, {1, 2, 3},
{3, 2, 1}, {2, 1, 3} are all the same set. We use the symbol ∈ to indicate set
membership: x ∈ B means that the element x is contained in the set B. Thus,
1 ∈ X and 4 ∈ Y in the above examples.
A set A is a subset of a set B iff every element of A is also an element of B.
We write A ⊆ B when A is a subset of B. Thus, Z ⊆ X in the above examples.
It follows from the definition of ⊆ that every set is a subset of itself.
A set A is said to be a proper subset of a set B iff A is a subset of B, but B
is not a subset of A. Thus, if A is a proper subset of B, every element of A is
also an element B; but there is at least one element of B that is not an element
of A.
Metatheory
31
The union of two sets A and B is the set that contains all the elements of A
and all the elements of B. We write A ∪ B to designate the union of the sets A
and B. Thus, for any element x, x ∈ A ∪ B iff either x ∈ A or x ∈ B (or both).
For example:
X ∪ Y = {1, 2, 3, 4, 5, 6}
Every element of this set can be found in either X or in Y (or both).
We write A ∩ B to designate the intersection of the sets A and B. The
intersection of two sets contains only those elements that are in both of the
sets. Thus, X ∩ Y = {4}, since 4 is the only element that is in both X and Y .
The special symbol ∅ indicates the empty or null set. This is the unique set
that contains no elements at all. Thus, if two sets A and B have no elements
in common, A ∩ B = ∅. Note that by definition of ⊆, it follows trivially that
∅ ⊆ A for any set A.
Specifying a set simply by listing its elements is impractical when the set is
large, and impossible when the set is infinite. But sets can also be specified by
a set definition. For example, the set of all positive integers can be specified by
the definition:
{x | x > 0}
We read this as ‘the set of x such that x > 0’. Every element of the set will
have the property stated by the definition ‘x > 0’; and anything that has this
property will be an element of the set. Similarly, the set of all triangles can be
defined as:
{x | x is a plane figure with three straight sides}
A set is said to be well-ordered by a relation R iff every non-empty subset
of that set has a least element with respect to R. An element x is the least
element of a set A iff xRy for every y ∈ A. Thus, the set of positive integers is
well-ordered by the relation ≤, because any of its non-empty subsets will contain
a number that is less than or equal to every number in the subset.
A fundamental result of set theory is the Well-Ordering Theorem, which
states that every set can be well-ordered. Thus, for any set A, there is some
relation R such that every non-empty subset of A has a least element with
respect to R: that is, for any non-empty B ⊆ A, there will be an x ∈ B such
that xRy for every y ∈ B. However, there is no guarantee that a given set can
be well-ordered by any relation one chooses. For example, the set of all people
is not well-ordered by the relation ‘shorter than’: since some people are the
same height, not every non-empty subset will have a least element.18 All the
Well-Ordering Theorem guarantees is that there is some relation whereby the
set of all people can be well-ordered.
The elements of ordered sets are listed between angle brackets: for example,
h1, 2, 3i is an ordered set containing three elements. Ordered sets that contain
18 Furthermore, no person can be shorter than themselves; and the definition of a wellordering requires that the least element x of a subset be related to every element of the
subset, which includes x itself.
32
Chapter 3
the same elements need not be identical: the order of the elements must also be
the same. Thus, h1, 2, 3i, h3, 2, 1i, and h2, 1, 3i are all different sets.
Given two sets X and Y , f is a function from X to Y iff f assigns an element
of Y to every element of X. When x ∈ X, f (x) designates the element of Y
that f assigns to x. Thus, for every x ∈ X, there will be a y ∈ Y such that
f (x) = y. However, it need not be the case that every element of Y is assigned
to some element of X: there may be some of Y ‘left over’. This may be so even
when X is larger than Y , because nothing prevents the same element of Y from
being assigned to a number of different x’s. When f is a function from X to Y ,
we write f : X → Y .
Exercise
1. Suppose that A = {1, 2, 3, 4} and B = {3, 4, 5}. List the elements of A ∪ B
and A ∩ B.
2. Prove that A ∪ A = A ∩ A = A for any set A.
3. Prove that for any sets A and B, A ⊆ A ∪ B.
4. Prove that for any sets A and B, A ∩ B ⊆ A.
5. Prove that for any sets A and B, A ∩ B ⊆ A ∪ B.
6. List all subsets of the set {x, y, z}. (Hint: If a set contains n elements, it
has exactly 2n subsets.)
3.5
Mathematical induction
Suppose that A is a set, and that x0 . . . xn . . . is a well-ordering of the elements
of A. (It makes no difference whether A is finite or infinite.) Suppose that we
want to prove that every element of A has a certain property F. The Principle
of Mathematical Induction allows us to draw this conclusion, provided that we
can prove two preliminary premises:
(i) x0 is F.
(ii) If xm is F for every m < n, then xn is F.
It is easily seen why these two premises imply that every x ∈ A will be F. Since
x0 is F by premise (i), x1 must be F by premise (ii); but then both x0 and x1
are F, so premise (ii) implies that x2 is F; and so on for all the elements of A.
Premise (i) is called the base step of the proof, and premise (ii) is called the
induction step. To prove the induction step, we usually assume that xm is F for
every m < n, and then derive ‘xn is F’ from this assumption. The assumption
that xm is F for every m < n is called the induction hypothesis.
The Principle of Mathematical Induction will be used to prove several important metatheorems in the chapters to follow. For the purposes of illustration,
we will now prove a trivial result in the theory of numbers. Let A be the set
Metatheory
33
of all integers greater than 1: i.e. A = {x | x > 1}. It is obvious that A is
well-ordered by the ≤ relation, and that x0 = 2 in this ordering. We will now
use mathematical induction to prove:
Theorem 3.3. Any number greater than 1 is divisible by a prime number.19
Proof. By induction:
Base step: x0 = 2 is a prime number, and any number is divisible by itself.
Thus x0 is divisible by a prime number.
Induction step: Suppose that for every m < n, xm is divisible by a prime
number. (This is the induction hypothesis.) We need to show that there is a
prime number p that divides xn . If xn is itself prime, then p = xn , since every
number is divisible by itself. Suppose that xn is not prime. Then xn is divisible
by some xm , where 1 < xm < xn (since a prime number is divisible only by
itself and 1). Since xm < xn , m < n, and so the induction hypothesis will apply
to xm : i.e. there will be a prime number p which divides xm . But since xm
divides xn , p will also divide xn . Thus, xn is divisible by a prime number. a
Thus we have proved both the base step and the induction step. The theorem
itself follows immediately by the Principle of Mathematical Induction.
19 The
number 1 itself is not considered to be prime.
4
The Propositional Calculus
4.1
Formal syntax of TFL
In this section, we will give a precise description of the symbolic language that
was introduced informally in section 2.2 above. The language is called TFL
(‘Truth Functional Logic’), and like all formal languages, its definition consists
of two parts: a vocabulary and a syntax. The vocabulary is simply a specification of the primitive symbols of the language, while the syntax consists of
formation rules that determine how these symbols are combined into formulas
of the language.
In addition, we will introduce a number of non-primitive symbols by definition. The sole purpose of these defined symbols is to make the formulas of TFL
shorter and more readily understood. The defined symbols are to be considered
simply as abbreviations, and as such they make no difference whatever to the
logical properties of TFL.
The vocabulary of TFL consists of an infinite set of atomic sentences:
At = {pn | n ≥ 0}
along with two truth functional operators → and ⊥, and the left and right
parentheses ( and ). The connective → is already familiar. The set At takes the
place of the variables A, B, C, etc. that were used in the informal version of
our symbolic language. The symbol ⊥ is called ‘the false’, and its function will
become apparent later, when we specify the abbreviating definitions for TFL.
The formation rules (syntax) of TFL are also very simple:
1. Every member of At is a formula of TFL.
2. ⊥ is a formula of TFL.
3. If α and β are formulas of TFL, then (α → β) is a formula of TFL.
These rules determine, for any sequence of primitive symbols, whether or not
that sequence constitutes a formula of TFL. For example, these rules determine
that the sequence of symbols
(p1 → (p2 → ⊥))
is a formula. We can demonstrate this by showing how to construct the formula
from the primitive symbols, by repeated applications of the formation rules:
p2 is a formula by (1).
⊥ is a formula by (2).
Hence (p2 → ⊥) is a formula by (3).
p1 is a formula by (1).
Hence (p1 → (p2 → ⊥)) is a formula by (3).
34
35
The Propositional Calculus
The symbolic language TFL can now be defined as the set consisting of all and
only those sequences of symbols that can be generated by a finite number of
applications of the rules (1), (2), and (3). Since TFL is defined as a set, ‘α is a
formula of TFL’ means simply that α ∈ TFL.
We adopt the informal convention that outermost pairs of parentheses can
be omitted. Thus:
(p1 → (p2 → ⊥))
p1 → (p2 → ⊥)
will count as the same formula.
By definition, every symbol of every formula of TFL will be a member of
the set:
{pn | n ≥ 0} ∪ {⊥, →, (, )}
This set represents the complete vocabulary of the language TFL. However, we
will now introduce a number of abbreviating conventions, for the sole purpose
of allowing the formulas of TFL to be presented in shorter and more readable
forms:
TFL formula
(α → ⊥)
(α → ⊥) → β
(α → (β → ⊥)) → ⊥
((α → β) → ((β → α) → ⊥)) → ⊥
Abbreviation
¬α
α∨β
α∧β
α↔β
The abbreviating symbols ¬, ∨, ∧, and ↔ are intended to represent the same
concepts as when they were introduced informally in Chapter 2. However, it is
essential to remember that truth tables play no part in the formal definition of
the symbolic language TFL. The formal definition of a symbolic language does
not involve any semantic information about that language. All that is required
for a formal definition is a set of primitive symbols, along with a collection of
formation rules. It does not matter whether the resulting symbolic language
can subsequently be given any interesting or useful semantic interpretation.
Exercise
Use the above list of abbreviations to verify the following identities:
(α ∨ β) = (¬α → β)
(α ∧ β) = ¬(α → ¬β)
(α ↔ β) = ((α → β) ∧ (β → α))
4.2
Formal definition of the system PC
The system PC (Propositional Calculus) consists of the formal language TFL,
together with a set of axioms, and a single inference rule. The axiom set contains
every instance of the formula schemas:
36
Chapter 4
[A1] α → (β → α)
[A2] (α → (β → γ)) → ((α → β) → (α → γ))
[A3] (¬α → ¬β) → (β → α)
Thus, PC has infinitely many axioms, each one of which is a member of TFL.
For example, each of the formulas:
p1 → (p2 → p1 )
p10 → (p22 → p10 )
(p3 → p1 ) → (p2 → (p3 → p1 ))
will be an axiom, because each one is an instance of the schema [A1]. Similarly,
the formula:
(¬(p1 ∧ p2 ) → ¬(p3 ∨ p4 )) → ((p3 ∨ p4 ) → (p1 ∧ p2 ))
will be an axiom, because it is an instance of [A3] in which α = (p1 ∧ p2 ) and
β = (p3 ∨ p4 ).
The single inference rule of PC is Modus Ponens:
α → β, α β
We say that the formula β follows from the formulas α → β and α by Modus
Ponens. We say that a set of formulas Σ is closed under Modus Ponens iff β ∈ Σ
whenever both α → β ∈ Σ and α ∈ Σ.
Where α ∈ TFL and Σ ⊆ TFL, we define the metatheoretical symbol ` as
follows:
Definition 4.1. Σ ` α iff there is a finite series of TFL formulas σ1 . . . σn such
that σn = α, and for each σi , either:
(i) σi ∈ Σ, or
(ii) σi is an axiom, or
(iii) σi follows from two previous members of the series by Modus Ponens.
When Σ ` α, we say that α is derivable from Σ, or that Σ proves α. When
∅ ` α, we write ` α, and say that α is a theorem of PC. Thus, it follows trivially
from Definition 4.1 that every axiom of PC is a theorem of PC: the required
series of formulas will contain just the axiom itself.
Lemma 4.1 (MP). If Σ ` α and Σ ` α → β then Σ ` β.
This metatheorem is a straightforward consequence of Definition 4.1, and will
be referred to hereafter as [MP].
Definition 4.2. A set Σ of TFL formulas is consistent iff Σ 0 ⊥. Otherwise, Σ
is inconsistent.
The Propositional Calculus
37
Exercise
Determine which of the following formulas are axioms of PC. If a formula is an
axiom, identify the axiom schema associated with it.
1. p2 → (p2 → p2 )
2. (p3 → (p2 → p3 )) → ((p3 → p2 ) → (p3 → p3 ))
3. (¬¬p1 → ¬¬p2 ) → (p2 → p1 )
4. (p1 → (p2 → (p1 → p2 ))) → ((p1 → p2 ) → (p1 → p2 ))
5. (¬¬p1 → ¬p2 ) → (p2 → ¬p1 )
4.3
Some theorems and metatheorems
The following metatheorem proves that every instance of the formula schema
α → α is a theorem of PC:
[T1] ` α → α
1.
2.
3.
4.
5.
` (α → ((β → α) → α)) → ((α → (β → α)) → (α → α))
` α → ((β → α) → α)
` (α → (β → α)) → (α → α)
` α → (β → α)
`α→α
[A2]
[A1]
1,2 [MP]
[A1]
3,4 [MP]
As we noted above, the definition of ` implies that every instance of a PC
axiom is a theorem of PC. Thus, every instance of the schema in line 1 will be
a theorem, because every such instance will also be an instance of [A2]. This is
indicated in the annotation to the line. Similarly, every instance of the schema
in line 2 will be an instance of [A1], as indicated in the annotation. Given lines
1 and 2, line 3 follows by [MP] (i.e. Lemma 4.1). Line 4 follows simply because
every instance of [A1] will be a theorem. Given lines 3 and 4, [MP] is again
invoked to establish line 5.
Thus we have shown that every instance of the schema α → α is a theorem of
PC. For future reference, we assign the name [T1] to this theorem schema. This
type of metatheorem, in which a formula or schema is proved to be a theorem
of the logical system, is called a deduction.
It is common practice amongst logicians to extend the term ‘theorem’ to
formula schemas. A formula schema is called a theorem when every one of its
instances is a theorem. Thus, the schema [T1] will often be referred to simply
as a theorem, even though the schema itself is not a member of TFL, and hence
cannot be a member of the set of theorems of PC. As long as the distinction
between a formula and a formula schema is kept in mind, this somewhat lax use
of the word ‘theorem’ should cause no difficulty.
Lemma 4.2. If Σ1 ` α and Σ1 ⊆ Σ2 , then Σ2 ` α.
38
Chapter 4
Proof. Suppose that Σ1 ` α. Let σ1 . . . σn be the series required by Definition
4.1. But if Σ1 ⊆ Σ2 , then every σi ∈ Σ1 will also be in Σ2 . Thus, the existence
of the series σ1 . . . σn implies that Σ2 ` α.
a
As a corollary to this result, note that if ` α, then Σ ` α for any set Σ ⊆ TFL,
since ∅ ⊆ Σ for any set Σ.
Lemma 4.3. If Σ ` α, then there is a finite ∆ ⊆ Σ such that ∆ ` α.
Proof. Suppose that Σ ` α, and let σ1 . . . σn be the required series. Let ∆ =
Σ ∩ {σ1 . . . σn }. Thus ∆ is finite. But the existence of the series σ1 . . . σn also
guarantees that ∆ ` α.
a
Exercise
1. Prove that {α, ¬β} is inconsistent if ` α → β.
2. Prove that {α} ` β → α.
3. Prove that {¬α} ` α → β.
4.4
The Deduction Theorem
The following metatheorem is traditionally known as the ‘Deduction Theorem’,
and will be referred to hereafter as [DT]. It plays an essential part in the proof
of many of the theorems and metatheorems that follow.
Theorem 4.4 (DT). Σ ∪ {α} ` β iff Σ ` α → β
Proof. (⇒) Assume that Σ ∪ {α} ` β, and let σ1 . . . σn be the required series.
The proof is by mathematical induction on the size of n.
Base step: If n = 1, then either β is an axiom, or β ∈ Σ, or β = α.
If β is an axiom or β ∈ Σ, then Σ ` β by definition of `.
But Σ ` β → (α → β) by [A1]. ∴ Σ ` α → β by [MP].
And if α = β, then Σ ` α → β because Σ ` α → α by [T1].
Induction step: We need only consider the case where β follows by [MP].
But then some σi and σj = σi → β must occur earlier in the series.
The series σ1 . . . σi guarantees that Σ ∪ {α} ` σi by Definition 4.1.
Likewise, the series σ1 . . . σj gives us Σ ∪ {α} ` σi → β.
∴ Σ ` α → σi and Σ ` α → (σi → β) by the induction hypothesis.
But Σ ` (α → (σi → β)) → ((α → σi ) → (α → β)) by [A2].
∴ Σ ` (α → σi ) → (α → β) by [MP].
∴ Σ ` α → β by [MP].
(⇐) Assume that Σ ` α → β. But Σ ⊆ Σ ∪ {α}.
∴ Σ ∪ {α} ` α → β by Lemma 4.2. But Σ ∪ {α} ` α.
∴ Σ ∪ {α} ` β by [MP].
a
39
The Propositional Calculus
Exercise
1. Prove that if {¬β} ` ¬α, then {α} ` β.
2. Prove that {α} ` β → γ iff {β} ` α → γ.
4.5
More theorems and metatheorems
[T2] ` ¬α → (α → β)
1.
2.
3.
4.
5.
` ¬α → (¬β → ¬α)
{¬α} ` ¬β → ¬α
{¬α} ` (¬β → ¬α) → (α → β)
{¬α} ` α → β
` ¬α → (α → β)
[A1]
1 [DT]
[A3]
2,3 [MP]
4 [DT]
[T3] ` α → ¬¬α
1.
2.
3.
4.
5.
6.
{α, α → ⊥} ` α
{α, α → ⊥} ` α → ⊥
{α, α → ⊥} ` ⊥
{α} ` (α → ⊥) → ⊥
` α → ((α → ⊥) → ⊥)
` α → ¬¬α
[Df.`]
[Df.`]
1,2 [MP]
3 [DT]
4 [DT]
5 [Df.¬]
[T4] ` ¬¬α → α
1. ` ¬α → ¬¬¬α
2. ` (¬α → ¬¬¬α) → (¬¬α → α)
3. ` ¬¬α → α
[T3]
[A3]
2,3 [MP]
Lemma 4.5. Σ ` α iff Σ ∪ {¬α} ` ⊥
Proof. (⇒) Assume that Σ ` α. ∴ Σ ∪ {¬α} ` α by Lemma 4.2.
But Σ ∪ {¬α} ` ¬α by [Df.`]. ∴ Σ ∪ {¬α} ` α → ⊥ by [Df.¬].
∴ Σ ∪ {¬α} ` ⊥ by [MP].
(⇐) Assume that Σ ∪ {¬α} ` ⊥. ∴ Σ ` ¬α → ⊥ by [DT].
∴ Σ ` ¬¬α by [Df.¬]. But Σ ` ¬¬α → α by [T4].
∴ Σ ` α by [MP].
Exercise
Prove that the following formula schemas are theorems of PC.
1. (α → β) → ((γ → α) → (γ → β))
2. (α → β) → ((β → γ) → (α → γ))
a
40
Chapter 4
3. ((α → β) → γ) → (β → γ)
4. ((α → β) → γ) → (β → (α → γ))
5. (α → β) → (¬β → ¬α)
6. (¬α → α) → α
4.6
Formal semantics for TFL
A function V : TFL → {0, 1} is a valuation iff it fulfils the following conditions:
1. V (⊥) = 0
2. V (α → β) = 1 iff V (α) ≤ V (β)
If Σ ⊆ TFL and V is a valuation, we write V (Σ) = 1 when V (α) = 1 for every
α ∈ Σ. Note that the valuations introduced in section 2.4 above assigned truth
values only to atomic sentences. The valuations defined here assign a truth
value to every formula of the language.
We can now provide more rigorous definitions of some of the semantic concepts introduced in Chapters 1 and 2:
Definition 4.3. α ∈ TFL is valid iff V (α) = 1 for every valuation.
Definition 4.4. Σ ⊆ TFL entails α iff there is no valuation V such that V (Σ) =
1 and V (α) = 0.
When Σ entails α, we write Σ α. When ∅ entails α, we write α.
Lemma 4.6. If Σ1 α and Σ1 ⊆ Σ2 , then Σ2 α.
Proof. Suppose that Σ1 α. Then for all valuations V , V (Σ1 ) = 1 implies
V (α) = 1. But if Σ1 ⊆ Σ2 , then V (Σ2 ) = 1 implies V (Σ1 ) = 1. ∴ If Σ1 ⊆ Σ2 ,
V (Σ2 ) = 1 implies V (α) = 1, and so Σ2 α.
a
Lemma 4.7. α is valid iff Σ α for all Σ ⊆ TFL.
Proof. (⇒) If V (α) = 1 for all valuations, then trivially there is no valuation
where V (α) = 0. Thus by Definition 4.4, Σ α for any Σ ⊆ TFL.
(⇐) Suppose that α is not valid. ∴ V (α) = 0 for some valuation V . But
V (α → α) = 1. Thus {α → α} 2 α by Definition 4.4.
a
As a corollary to these lemmas, note that α is valid iff α.
Definition 4.5. Σ is satisfiable iff there is at least one valuation V such that
V (Σ) = 1.
Lemma 4.8. Σ is satisfiable iff Σ 2 ⊥.
The Propositional Calculus
41
Proof. (⇒) Suppose that Σ ⊥. Then for any V , V (Σ) = 1 ⇒ V (⊥) = 1. But
V (⊥) = 0 for every V . ∴ V (Σ) 6= 1 for every V . ∴ Σ is not satisfiable.
(⇐) Suppose that Σ 2 ⊥. ∴ There is a V where V (Σ) = 1 and V (⊥) = 0. ∴ Σ
is satisfiable.
a
The semantic concept of entailment corresponds to the syntactic concept of
derivability defined in the last chapter (Definition 4.1). Similarly, the semantic
notion of satisfiability corresponds to the syntactic notion of consistency (Definition 4.2). Just how close these correspondences are will be seen in the next
chapter.
Exercise
1. Prove that V (¬α) = 1 iff V (α) = 0.
2. Prove that V (α ∧ β) = 1 iff V (α) = 1 and V (β) = 1.
3. Prove that V (α ∨ β) = 1 iff V (α) = 1 or V (β) = 1.
4. Prove that V (α ↔ β) = 1 iff V (α) = V (β).
5. Suppose that Σ ⊥. Prove that Σ entails every α ∈ TFL.
6. Prove that Σ α iff Σ ∪ {¬α} is not satisfiable.
5
Metatheory of PC
5.1
Soundness of PC
A logical system is said to be semantically sound iff every derivation in that
system is represented by an entailment in its semantics. Thus, we can show
that the system PC is sound by proving that, for every α ∈ TFL and every
Σ ⊆ TFL, Σ α whenever Σ ` α.
Lemma 5.1. Every PC axiom is valid.
Proof. Assume that [A1] is not valid.
∴ V (α → (β → α)) = 0 for some valuation V .
∴ V (α) = 1 and V (β → α) = 0.
∴ V (β) > V (α). ∴ V (β) > 1, which is absurd.
∴ V (α → (β → α)) = 1 for every valuation.
a
The proofs for [A2] and [A3] are left as an exercise.
Lemma 5.2. For any valuation V , if V (α) = 1 and V (α → β) = 1, then
V (β) = 1
The proof of this Lemma is also left as an exercise.
Theorem 5.3 (Soundness). For all Σ ⊆ TFL and α ∈ TFL,
Σ`α⇒Σα
Proof. Assume that Σ ` α. Let σ1 . . . σn be the series required by Definition 4.1.
Assume that V (Σ) = 1 for an arbitrary valuation V . We show that V (α) = 1
by mathematical induction on the size of n.
Base step:
Since n = 1, either α ∈ Σ, or α is an axiom.
If α ∈ Σ, then V (α) = 1 because V (Σ) = 1.
If α is an axiom, then V (α) = 1 by Lemma 5.1
Induction step:
Assume that, for some i, σi and σi → α occur earlier in the series.
Then Σ ` σi and Σ ` σi → α.
∴ Σ σi and Σ σi → α by the induction hypothesis.
But V (Σ) = 1. ∴ V (σi ) = 1 and V (σi → α) = 1 by Definition 4.4.
∴ V (α) = 1 by Lemma 5.2.
a
Exercise
1. Prove that the theorems [T1] through [T4] are valid.
2. Prove that all of the formulas listed in the Exercise on page 39 are valid.
3. Prove that Σ ∪ {α} β iff Σ α → β.
42
Metatheory of PC
5.2
43
Maximal consistent sets
Definition 5.1. Σ ⊆ TFL is maximal iff, for all α ∈ TFL, either α ∈ Σ or
¬α ∈ Σ.
Definition 5.2. Σ ⊆ TFL is maximal consistent iff Σ is both maximal and
consistent.
Intuitively, a maximal consistent set can be considered to be a set that is a large
as it can be without becoming inconsistent.
Lindenbaum’s Lemma. Every consistent set is included in a maximal consistent set.
Assume that Σ is a consistent set of TFL formulas. Let α0 . . . αn . . . be a wellordering of TFL. Let Σ0 = Σ, and for every n ≥ 0, define:
(
Σn ∪ {αn } if Σn 0 ¬αn
Σn+1 =
Σn otherwise.
Finally, let Σ+ = Σ0 ∪ Σ1 ∪ Σ2 . . .
Lemma 5.4. For every i ≥ 0, Σi is consistent.
Proof. The proof is by mathematical induction.
Base step: Σ0 is consistent because Σ0 = Σ.
Induction step: Assume that Σn is consistent.
∴ If Σn+1 = Σn , then Σn+1 is consistent.
Suppose Σn+1 is inconsistent.
∴ Σn ∪ {αn } ` ⊥ by Definition 4.2.
∴ Σn ` αn → ⊥ by [DT].
∴ Σn ` ¬αn , contrary to the definition of Σn+1 .
∴ Σn+1 must be consistent.
a
Lemma 5.5. Σ+ is consistent.
Proof. Suppose that Σ+ is inconsistent. ∴ Σ+ ` ⊥ by Definition 4.2.
Then by Lemma 4.3, there is a finite ∆ ⊆ Σ+ such that ∆ ` ⊥.
But ∆ ⊆ Σi for some i. ∴ Σi ` ⊥, contrary to Lemma 5.4.
∴ Σ+ is consistent.
a
Lemma 5.6. Σ+ is maximal.
Proof. Assume that α 6∈ Σ+ and ¬α 6∈ Σ+ for some α ∈ TFL.
Let i and j be the indices of α and ¬α in the ordering of TFL.
∴ Σi ` ¬α and Σj ` ¬¬α by definition of Σ+ . ∴ Σj ` ¬α → ⊥.
But both Σi and Σj are subsets of Σ+ .
∴ Σ+ ` ¬α and Σ+ ` ¬α → ⊥ by Lemma 4.2.
∴ Σ+ ` ⊥ by [MP], contrary to Lemma 5.5.
∴ Either α ∈ Σ+ or ¬α ∈ Σ+ .
a
44
Chapter 5
Thus, Σ+ is both maximal and consistent, and includes Σ. This completes the
proof of Lindenbaum’s Lemma.
To conclude this section, we establish an important fact about maximal
consistent sets:
Lemma 5.7. If Σ is maximal consistent, then Σ ` α ⇒ α ∈ Σ
Proof. Assume that Σ is maximal consistent, and that Σ ` α.
Suppose α 6∈ Σ. ∴ ¬α ∈ Σ. ∴ α → ⊥ ∈ Σ.
∴ Σ ` α → ⊥. ∴ Σ ` ⊥ by [MP].
∴ Σ is inconsistent, contrary to assumption.
∴ α ∈ Σ.
a
Thus, maximal consistent sets are said to be deductively closed.
5.3
Completeness of PC
A logical system is said to be semantically complete iff every entailment of the
system is represented by a derivation in the system itself.
Theorem 5.8 (Completeness). For all Σ ⊆ TFL and α ∈ TFL,
Σα⇒Σ`α
Proof. Assume that Σ 0 α. ∴ Σ ∪ {¬α} is consistent by Lemma 4.5. ∴ Σ ∪ {¬α}
is a subset of some maximal consistent Σ+ by Lindenbaum’s Lemma. For all
β ∈ TFL, define V (β) = 1 if β ∈ Σ+ , V (β) = 0 otherwise. We show that V is a
valuation.
(i) Since ⊥ 6∈ Σ+ , V (⊥) = 0.
(ii) Assume that V (β → γ) = 1 and that V (β) > V (γ).
∴ V (β) = 1 and V (γ) = 0. ∴ β ∈ Σ+ and γ 6∈ Σ+ .
But (β → γ) ∈ Σ+ . ∴ γ ∈ Σ+ by [MP].
∴ γ ∈ Σ+ and γ 6∈ Σ+ , which is absurd.
∴ V (β → γ) = 1 ⇒ V (β) ≤ V (γ).
(iii) Assume that V (β) ≤ V (γ). ∴ Either V (β) = 0 or V (γ) = 1.
∴ Either ¬β ∈ Σ+ or γ ∈ Σ+ . ∴ Either Σ+ ` ¬β or Σ+ ` γ.
But Σ+ ` ¬β → (β → γ) by [T2] and Σ+ ` γ → (β → γ) by [A1].
∴ Σ+ ` β → γ by [MP]. ∴ (β → γ) ∈ Σ+ by Lemma 5.7.
∴ V (β → γ) = 1.
Thus V is a valuation. But α 6∈ Σ+ . ∴ V (α) = 0.
But Σ ⊆ Σ+ . ∴ V (Σ) = 1. ∴ Σ 2 α.
a
Exercise
1. Prove that every consistent set of TFL formulas is satisfiable.
2. Prove that every satisfiable set of TFL formulas is consistent.
Metatheory of PC
5.4
45
Decidability of PC
Definition 5.3. An effective procedure begins with any one of a specified set
of inputs, and terminates, after a finite number of steps, with exactly one of a
specified set of outputs.
Definition 5.4. A set A is decidable iff there is an effective procedure that will
prove, for any x, either x ∈ A or x 6∈ A. Otherwise, A is undecidable.
It follows from Definition 5.4 that a set Σ of TFL formulas is decidable iff there
is some effective procedure that will take any TFL formula α as input, and
whose only outputs are ‘in Σ’ and ‘not in Σ’.
Our main concern here is the set of PC theorems. Theorems 5.3 and 5.8
together imply that this set corresponds exactly to the set of valid TFL formulas:
for if ` α, then Theorem 5.3 implies that α is valid (i.e. α); and if α is valid,
then Theorem 5.8 implies that ` α. Thus, if there is an effective procedure for
determining whether or not an arbitrary TFL formula is valid, the decidability
of PC will follow immediately.
Recall that a formula α is valid iff V (α) = 1 for any valuation V (Definition
4.3). Obviously we cannot check every valuation, since there are infinitely many
of them. But if two valuations do not differ in the truth values they assign to
the atomic sentences in α, then they will also agree on the value they assign to
α itself. The truth values of atomic sentences that do not occur in α have no
relevance to the truth value of α. (See question 3 in the next Exercise.)
For example, let α be p1 → p2 . Suppose that V (p1 ) = V 0 (p1 ) and V (p2 ) =
0
V (p2 ). Then V (p1 ) ≤ V (p2 ) iff V 0 (p1 ) ≤ V 0 (p2 ), and so by the definition of a
valuation given on page 40, V (α) must be the same as V 0 (α).
If a formula α contains n atomic sentences, then there will be exactly 2n
distinct valuations of α: that is, valuations that do not agree on every atomic
sentence in α.20 And since any TFL formula contains only finitely many propositional letters, that means there are only finitely many valuations that need to
be considered. This yields an effective procedure for testing the validity of a
formula α that contains n distinct propositional letters:
[1] Let i = 2n .
[2] Calculate the value of α on valuation Vi .
[3] If Vi (α) = 0, stop: α is not valid.
[4] If i = 0, stop: α is valid.
[5] Let i = i − 1 and go to step 2.
This procedure will halt after a finite number of steps, yielding either the answer
‘α is valid’ or ‘α is not valid’. Thus we have established:
Theorem 5.9. The set of PC theorems is decidable.
20 These
2.4.
collections of distinct valuations correspond to the valuations discussed in section
46
Chapter 5
However, not every set of TFL sentences is decidable. TFL contains denumerably many sentences. But it can be demonstrated that any denumerable
set has uncountably many subsets.21 Thus, the set of subsets of TFL is uncountable. But any effective procedure must be defined by a finite set of rules,
each of which is finite in length. Hence there are only denumerably many effective procedures; and so there are more subsets of TFL than there are effective
procedures. Thus, some subsets of TFL must be undecidable.
Exercise
1. Prove that the set {α ∈ TFL | ` ⊥ → α} is decidable.
2. Prove that the set {α ∈ TFL | ` α → ⊥} is decidable.
3. Suppose α ∈ TFL, and let Σ be the set of all atomic sentences that occur
in α. Let V and V 0 be valuations where V (pi ) = V 0 (pi ) for each pi ∈ Σ.
Prove that V (α) = V 0 (α). (Hint: Use induction on the length of α.)
5.5
Compactness of PC
Theorem 5.10 (Compactness). Σ ⊆ TFL is satisfiable iff every finite subset
of Σ is satisfiable.
Proof. The ‘only if’ direction is trivial. For the ‘if’ direction, assume that Σ is
not satisfiable. ∴ Σ ⊥ by Lemma 4.8. ∴ Σ ` ⊥ by Theorem 5.8. ∴ ∆ ` ⊥
for some finite ∆ ⊆ Σ by Lemma 4.3. ∴ ∆ ⊥ by Theorem 5.3. ∴ ∆ is not
satisfiable by Lemma 4.8. ∴ Σ has a finite subset that is not satisfiable.
a
21 We call a set denumerable when it is in one-to-one correspondence with the set of natural
numbers N = {0, 1, 2 . . . }. We call a set countable when it is either finite or denumerable.
The null set is considered to be finite.