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Logic
Shuang (Sean) Luan
Department of Computer Science
University of New Mexico
Albuquerque, NM 87131
E-mail: [email protected]
1
What is Logic?
Definition 1 Logic is the discipline that studies the method of reasoning.
The discipline of logic aims to abstract our thought process and rigorously formalize the rules of
inferences. In this course, we study logic to help us form valid arguments and construct correct proofs.
However, it would be a mistake to try to convert everything into logic. Instead the correct approach
is to think whether the argument we make are logical.
The development of proof techniques proceeds the discipline of logic. Proof techniques were developed by ancient Greek mathematicians, and can be traced to as early as the Thales Theorem by
Thales, who lives 624BC - 546BC. Logic was first studied by Aristotle (384-322BC), and was formalized in the 19th century by George Boole and Augustus De Morgan, who presented the systematic
mathematical treatment of logic.
2
Proposition
At the center of logic is the concept of proposition.
Definition 2 A proposition is a declarative sentence that is either true or false but not both. The
truth value of a proposition is denoted by true (T) or false (F). Notationwise, we usually use lower
case letters p, q, r, ... to represent a proposition.
The following are propositions:
• “The earth is round.”
• “Today’s temperature is 70◦ F.”
• “2+3 = 4”
The following are not propositions:
• “Do you speak Spanish?”
• “3 − x = 5”
1
In most cases, we use proposition, statement and claim interchangeably. In other words, all of
them refer to a declarative sentence that is either true or false but not both.
An argument is a sentence aiming to persuade someone to accept a particular conclusion.
A proof is a sequence of valid arguments that establish the truth of a proposition.
An axiom or postulate is a proposition that is accepted to be true without proof.
A theorem is a proposition that has been proven to be true. Theorem is usually reserved for
important result.
A lemma is a “baby theorem” or “helping theorem”.
A corollary is a proposition that follows with little or no proof from a theorem.
A scientific law is a proposition based on experimental observations. As a result, a scientific law
is only applicable under certain conditions.
3
Logic Connectors
If we compare logic to algebra, then propositions are the counterparts of numbers. What is the
counterpart of algebraic operators such as addition, subtraction, multiplication, etc? The answer is
logic connectors. Yes, we can also perform computation on propositions!
How are the logic connectors defined? How is an algebraic operation defined? Observe that an
algebraic operation is actually a function. Take the addition for natural numbers for example. Addition
between two natural numbers is actually a function from the Cartesian product N × N to N, where
the ordered pair (a, b) is mapped to a + b by the addition function. To describe the addition function,
one could certainly build a table that describes all possible input and output to the addition function.
From our experiences with cardinality, we know that this will be an infinite table with countably
number of entries.
Input
(1, 1)
(1, 2)
(2, 1)
(1, 3)
(2, 2)
(3, 1)
···
Output
2
3
3
4
4
4
···
From the above discussion, a logic operator can also be defined just like the addition. It will be
a function that can operate on one or more propositions. When defining a logic operator, we can use
a table that specifies the outcome of the operation for each possible input values. Such a table that
defines a logic operator is called a truth table. However, since proposition only has two possible values,
i.e., T (true) or F (false), the truth table for logic connectors are usually not very big.
3.1
Negation
Definition 3 The negation of a proposition p is “not p”, and is denoted by ¬p. The truth table for
negation is:
p
T
F
¬p
F
T
2
3.2
Conjunction
Definition 4 The conjunction of two propositions p and q is “p and q”, and is denoted by p ∧ p. The
truth table for conjunction is:
p
T
T
F
F
3.3
q
T
F
T
F
p∧q
T
F
F
F
Disjunction
Definition 5 The disjunction of two propositions p and q is “p or q”, and is denoted by p ∨ q. The
truth table for disjunction is:
p
T
T
F
F
3.4
q
T
F
T
F
p∨q
T
T
T
F
Exclusive Or
Definition 6 The exclusive or of two propositions p and q is “p or q but not both”, and is denoted
by p ⊕ p. The truth table for exclusive or is:
p
T
T
F
F
q
T
F
T
F
p⊕q
F
T
T
F
Example 1 Calculate the truth table of the following logic expression: (p ∧ (¬q)) ∨ (q ∧ (¬p)).
Solution:
p
T
T
F
F
q
T
F
T
F
¬p
F
F
T
T
¬q
F
T
F
T
p ∧ ¬q
F
T
F
F
q ∧ ¬p
F
F
T
F
(p ∧ (¬q)) ∨ (q ∧ (¬p))
F
T
T
F
2
Observe that p⊕q and (p ∧ (¬q))∨(q ∧ (¬p)) have the same logical values, thus they are essentially
the same logical functions. Note that this actually makes sense. In order to have (p ∧ (¬q))∨(q ∧ (¬p))
evaluate to T, either p ∧ ¬q or q ∧ ¬p is true. In either situation, p and q are going to have different
logical values, which is exactly the definition of exclusive or.
Definition 7 Two logic expressions are said to be logically equivalent if they have the same truth
table.
3
3.5
Implication
Definition 8 The implication of two propositions p and q is “if p then q”, and is denoted by p → q.
In an implication, the proposition p is called the premise and the proposition q is called the conclusion.
The truth table for implication is:
p
T
T
F
F
q
T
F
T
F
p→q
T
F
T
T
One confusing aspect of p → q is when p is false, the implication is always true. What does this
mean? This means that if you start out with a premise or an assumption that is wrong, then you may
end up with anything. For example, the statement “if you are Shaq O’Neil, then I am Michael Jordan”
is in fact a logically correct statement. Note that in this case, p =“You are Shaq O’Neil” and q =“I
am Michael Jordan”. Since you are not Shaq O’Neil, p = F . From the truth table, the implication is
always true.
The implication is one of the more difficult logic connectors. Other ways for expressing p → q in
English are:
• “p implies q.”
• “p is sufficient for q.”
• “q unless ¬p.”
• “p only if q.”
• “A necessary condition for p is q.”
The best ways to appreciate this is by some real world examples. For example, consider the
following situations. p =“Your cumulative is over 90%”, and q =“You will get a final letter grade of
A”. Consider the following arguments:
• “If your cumulative is over 90%, then you will get a final letter grade A.”
• “Your cumulative is 90% implies you will get a final letter grade A.”
• “A cumulative of 90% is sufficient for a final letter grade A.”
• “Your final letter grade is A unless your cumulative is not over 90%”.
Comment: this is to say that the only way your final grade is not A is that you didn’t get a
cumulative of over 90%.
• “A cumulative of over 90% only if you get a final letter grade A.”
Comment: this is to say that if your final letter grade is not A, then your cumulative is not over
90%.
• “A necessary condition for a cumulative of over 90% is a final letter grade of A.”
Comment: this is to say that a final letter grade of other than “A” is sufficient that your
cumulative is less than 90%”.
4
Example 2 Calculate the truth table of q ∨ (¬p).
Solution:
p
T
T
F
F
q
T
F
T
F
¬p
F
F
T
T
q ∨ (¬p)
T
F
T
T
Observe that q ∨ (¬p) is logically equivalent to p → q. This can also serves an explanation of “q
unless ¬p”, because the only way for q ∨ ¬p to be false is when q if false and p is true.
2
Definition 9 The contrapositive of p → q is (¬q) → (¬p).
Example 3 Calculate the truth table of (¬q) → (¬p).
Solution:
p
T
T
F
F
q
T
F
T
F
¬q
F
T
F
T
¬p
F
F
T
T
(¬q) → (¬p)
T
F
T
T
Observe that (¬q) → (¬p) is logically equivalent to p → q. In other words the contrapositive of an
implication is logically equivalent to the implication.
2
Definition 10 The converse of p → q is q → p.
Example 4 Calculate the truth table of q → p.
Solution:
p
T
T
F
F
q
T
F
T
F
q→p
T
T
F
T
Observe that q → p is not logically equivalent to p → q. This makes sense. Recall the example
p =“a cumulative of ≥ 90%”, and q =“a final letter grade of A”. The implication p → q is “if your
cumulative is ≥ 90%, then your final letter grade is A”. The converse of this is “If your final letter
grade is A, then your cumulative is ≥ 90%”, which is obviously not true, because the professor may
curve and even if your cumulative is not ≥ 90%, your final letter grade may still be A.
2
Definition 11 The inverse of p → q is (¬p) → (¬q).
Example 5 Calculate the truth table of (¬q) → (¬p).
5
Solution:
p
T
T
F
F
q
T
F
T
F
¬p
F
F
T
T
¬q
F
T
F
T
(¬p) → (¬q)
T
T
F
T
Observe that (¬p) → (¬q) is not logically equivalent to p → q, but is logically equivalent to the
converse, i.e., q → p.
Recall the example p =“a cumulative of ≥ 90%”, and q =“a final letter grade of A”. The implication
p → q is “if your cumulative is ≥ 90%, then your final letter grade is A”. The inverse of this is “If your
cumulative is < 90%, then your final letter grade is not A”, which is obviously not true, because the
professor may curve and even if your cumulative is not ≥ 90%, your final letter grade may still be A.
2
3.6
Biconditional
Definition 12 The biconditional statement of two propositions p and q is “p if and only if q”, and
is denoted by p ↔ q. The truth table of biconditional is:
p
T
T
F
F
q
T
F
T
F
p↔q
T
F
F
T
Example 6 Calculate the truth table of (p → q) ∧ (q → p).
Solution:
p
T
T
F
F
q
T
F
T
F
p→q
T
F
T
T
q→p
T
T
F
T
(p → q) ∧ (q → p)
T
F
F
T
(p → q) ∧ (q → p) is logically equivalent to the biconditional statement p ↔ q.
3.7
2
Precedence of Logic Connectors
Similar to the algebraic operations, the logic connectors also have precedence or order of operations.
The precedence follows the order ¬, ∧, ∨, →, ↔.
Example 7 ¬p ∨ q → r ∧ s ↔ t should be evaluated as (((¬p) ∨ q) → (r ∧ s)) ↔ t
4
Propositional Equivalence
Using the logic connectors, we can form compound propositions. In this section, we present some
special compound propositions and discuss propositional equivalence.
6
4.1
Tautology, Contradiction, and Contingency
Definition 13 A compound proposition that is true for all possible values of its propositional variables
is called a tautology.
Example 8 Determine if p ∧ (p → q) → q is a tautology.
Solution:
p
T
T
F
F
q
T
F
T
F
p→q
T
F
T
T
p ∧ (p → q)
T
F
F
F
p ∧ (p → q) → q
T
T
T
T
Since its values is T for all possible values of p and q, p ∧ (p → q) → q is a tautology.
2
Definition 14 A compound proposition that is false for all possible values of its propositional variables
is called a contradiction.
Example 9 Determine if p ∧ ¬p is a contradiction.
Solution:
p
T
F
¬q
F
T
p ∧ ¬p
F
F
Since its values is F for all possible values of p, p ∧ ¬p is a contradiction.
2
Definition 15 A compound proposition that can be either true or false is called a contingency.
4.2
Logic Equivalence
Definition 16 Two compound propositions p and q are called logically equivalent if p ↔ q is a tautology, and is denoted by p ≡ q.
Example 10 Determine if the two propositions p and ¬p → F are logically equivalent.
Solution:
p
T
F
¬p
F
T
¬p → F
T
F
p ↔ ¬p → F
T
T
Thus, p ≡ ¬p → F
2
Observe that Example 10 is a way of explaining proof by contradiction. Suppose we want to show
a statement p is true. We will assume p is wrong, i.e., ¬p and show that this assumption will lead to
a contradiction.
Example 11 Determine if the two propositions p → q and (¬q ∧ p) → F are logically equivalent.
7
Solution:
p
T
T
F
F
q
T
F
T
F
p→q
T
F
T
T
¬q
F
T
F
T
¬q ∧ p
F
T
F
F
(¬q ∧ p) → F
T
F
T
T
p → q ↔ (¬q ∧ p) → F
T
T
T
T
Since p → q ↔ (¬q ∧ p) → F is a tautology, the two statement are logically equivalent, i.e.,
p → q ≡ (¬q ∧ p) → F
2
Example 11 is another way to explain the principle behind proof by contradiction. Recall the
framework for proof by contradiction. To prove a statement in the form p → q, we will start by
assuming the conclusion q is wrong, i.e., ¬q is correct. We will show that this, i.e., ¬q together with
the original premise p will lead to a contradiction. Expressed logically, this is exactly (p ∧ ¬q) → F .
Example 12 (DeMorgan’s Law) Show that the following are logically equivalent:
(a) ¬(p ∧ q) and ¬p ∨ ¬q
(b) ¬(p ∨ q) and ¬p ∧ ¬q
Proof:
p
T
T
F
F
q
T
F
T
F
¬p
F
F
T
T
¬q
F
T
F
T
p ∧ q)
T
F
F
F
¬(p ∧ q)
F
T
T
T
¬p ∨ ¬q
F
T
T
T
¬(p ∧ q) ↔ ¬p ∨ ¬q
T
T
T
T
Since ¬(p ∧ q) ↔ ¬p ∨ ¬q is a tautology, ¬(p ∧ q) ≡ ¬p ∨ ¬q.
Similarly, we can show that ¬(p ∨ q) ≡ ¬p ∧ ¬q.
4.3
2
Common Logic Equivalence
There are many logic equivalences. The Table below shows some of the most important ones.
Identity Laws
Domination Laws
Idempotent Laws
Double Negation Law
Commutative Laws
Associative Laws
Distributive Laws
Absorption Laws
Negation Laws
p ∧ T ≡ p, p ∨ F ≡ p
p ∨ T ≡ T, p ∧ F ≡ F
p ∨ p ≡ p, p ∧ p ≡ p
¬¬p ≡ p
p ∧ q ≡ q ∧ p, p ∨ q ≡ q ∨ p.
(p ∧ q) ∧ r ≡ p ∧ (q ∧ r), (p ∨ q) ∨ r ≡ p ∨ (q ∨ r).
p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r), p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r).
p ∨ (p ∧ q) ≡ p, p ∧ (p ∨ q) ≡ p
p ∨ ¬p ≡ T , p ∧ ¬p ≡ F .
The Table below shows some important logic equivalence involving implications. The right column
shows the laws used to derive the logic equivalence in the left column.
8
p → q ≡ ¬p ∨ q
p → q ≡ ¬q → ¬p
p ∨ q ≡ ¬p → q
p ∧ q ≡ ¬(p → ¬q)
¬(p → q) ≡ p ∧ ¬q
(p → q) ∧ (p → r) ≡ p → (q ∧ r)
(p → r) ∧ (q → r) ≡ (p ∨ q) → r
(p → q) ∨ (p → r) ≡ p → (q ∨ r)
(p → r) ∨ (q → r) ≡ (p ∧ q) → r
Truth Table
Contrapositive
Double Negation Law
DeMorgan’s Law
DeMorgan’s Law
Distributive Law
Distributive Law
Distributive Law
Distributive Law
The Table below shows some important logic equivalence involving biconditional. The right column
shows the laws used to derive the logic equivalence in the left column.
p ↔ q ≡ (p → q) ∧ (q → p)
p ↔ q ≡ ¬p ↔ ¬q
p ↔ q ≡ (p ∧ q) ∨ (¬p ∧ ¬q)
¬(p ↔ q) ≡ p ↔ ¬q
5
Truth
Truth
Truth
Truth
Table
Table
Table
Table
Logic and Digital Circuit
One may not realize that the logic we have learnt so far is enough for us to design a digital computer.
To demonstrate this, we will show the design of a full adder.
5.1
1 Bit Adder
Digital computers use binary representation of numbers. Let’s consider a 1 bit full adder. Figure 1
shows the function of a 1-bit adder. A 1-bit adder is capable of adding two binary bits input a and b
and calculate their sum s and carry c. Since the input and output are bits, i.e., 0 and 1, we can also
view them as logic variables with F = 0 and T = 1. Thus the function of a 1-bit adder:
a
0
0
1
1
b
0
1
0
1
s
0
1
1
0
c
0
0
0
1
can be viewed as a truth table describing s and c:
a
F
F
T
T
b
F
T
F
T
s
F
T
T
F
c
F
F
F
T
Observe that logically, s = a ⊕ b and c = a ∧ b.
9
Figure 1: Illustrating the function of a 1-bit adder.
Figure 2: The symbols of some common logic gates.
5.2
Logic Gates
Logic gates are essentially symbols for representing logical connectors. In the next section we will show
the connection between logic gates and physical devices. Figure 2 shows the symbols of the 3 most
fundamental logic gates, the NOT gate, the AND gate, and the OR gate, since all the logic connectors
can be represented by combining these three type of logic gates. For example, a⊕b = (a∧¬b)∨(¬a∧b).
Figure 3 shows how a 1-bit adder can be realized using logic gates.
Most logic design uses NAND gate (see Figure 4) because it is a universal gate and can be used
to implement any other logic gates. To see this observe that:
• ¬p = ¬p ∧ p
• p ∧ q = ¬ (¬ (p ∧ q)) = ¬ (¬ (p ∧ q) ∧ ¬ (p ∧ q))
• p ∨ q = ¬ (¬p ∧ ¬q) = ¬ (¬ (p ∧ p) ∧ ¬ (q ∧ q))
We can implement the 1-bit adder using NAND gate logic. Observe that, a⊕b = (a∧¬b)∨(¬a∧b) =
¬¬(a ∧ ¬b) ∨ (¬a ∧ b) = ¬ (¬(a ∧ ¬b) ∨ ¬(¬a ∧ b)). Further a ∧ ¬(a ∧ b) = a ∧ (¬a ∨ ¬b) = (a ∧ ¬a) ∨
(a ∧ ¬b) = a ∧ ¬b. Thus a ⊕ b = ¬ (¬(a ∧ ¬b) ∨ ¬(¬a ∧ b)) = ¬ (¬(a ∧ ¬(a ∧ b)) ∨ ¬(¬(a ∧ b) ∧ b))
Figure 5 shows the implementation of a 1-bit adder using NAND gates. Observe that compared to
10
Figure 3: The symbols of some common logic gates.
Figure 4: The symbols for NAND gate.
the implementation in Figure 3, the NAND gate implementation is actually more efficient because it
uses 5 logic gates instead of 6.
5.3
1-bit Full Adder and Ripple Adder
A goal for designing a 1-bit adder is to use it as a building block for a multiple-bit full adder. In order
to do that, we need a 1-bit full adder, which is also capable of handling carry in and carry out. From
this regards, the 1-bit adder that we presented in the previous sections is usually referred to as a half
adder. Figure 6 shows a 1-bit full adder. It function is described in the truth table: Like the 1-bit half
adder, the 1-bit full adder can also be implemented using NAND gate logic. We will not go into the
details here.
a
0
0
0
0
1
1
1
1
b
0
0
1
1
0
0
1
1
cin
0
1
0
1
0
1
0
1
s
0
1
1
0
1
0
0
1
cout
0
0
0
1
0
1
1
1
With the 1-bit full adder, one can built an adder for arbitrary number of bits. The easiest way to
Figure 5: The NAND gate implementation of a 1-bit adder.
11
Figure 6: Illustrating the function of a 1-bit full adder.
Figure 7: Illustrating a ripple adder.
realize this is to connect a lot of 1-bit full adders sequentially, connect the carry out of a less significant
bit to the carry in to the adder for a more significant bit. Figure 7 shows the schematic overview of a
ripple adder.
5.4
Transistor
Interested readers may wonder how the logic gates implementation can be translated to actual physical
circuits. In this section, we will show how a NAND gate can be implemented using transistors. The
underlying principle also extends to the more modern CMOS technology.
A transistor is a semiconductor device invented in 1947 by William Shockley, John Bardeen, and
Walter Brattain at AT&T’s Bell Labs, who later shared the 1956 Nobel Prize in Physics “for their
researches on semiconductors and their discovery of the transistor effect”.
Generally speaking a transistor has 3 terminals, referred to as base, collector, and emitter (see
Figure 8 for illustrations). An important property of a transistor is that the base can act as a switch.
When the voltage on base is high, the switch is open and current can flow from collector to the emitter.
When the voltage on base is low or close to 0, the switch is turned off and no current can flow from
the collector to the emitter.
Using the transistor as a switch, we can build NAND gate. Figure 9 shows the implementation of
a NAND gate using transistor to transistor logic (TTL). The input of the NAND gate is a and b. The
12
Figure 8: Illustrating a ripple adder.
Figure 9: Illustrating a ripple adder.
output is c. To understand how this NAND gate work, we first remark that physically, we use a high
voltage to represent a logic 1 or T, and a low voltage to represent a logic 0 or F. Observe that when a
and b are both 1, i.e., high voltage, then both switches are turned on, essentially making c grounded,
i.e., 0. If either a and b are 0, i.e., a low voltage, then there is no current flowing from the VCC to the
ground. This keeps c at high voltage, i.e. 1.
With the NAND implemented physically, it is not hard to imagine that we can build an entire
computer. However, all of these starts out with the logic operations that we have discussed so far.
6
Propositional Functions and Logic Quantifiers
Definition 17 A predicate is a function such that when applied to the individual becomes a proposition.
Predicates are also referred to as propositional functions.
Example 13 P (x) : x ≥ 3 is a predicate. It becomes a proposition when applied to individual values
of x. For example, P (4) is “4 ≥ 3”, which is a proposition with logic value T.
We can also make predicates a proposition using universal and existential quantifiers.
Definition 18 The universal quantification of a predicate P (x) is the proposition: “for every x, P (x)
is true”, and is denoted by ∀xP (x).
Definition 19 The existential quantification of a predicate P (x) is the proposition: “there exists an
x, P (x) is true”, and is denoted by ∃xP (x).
Example 14 Let P (x) : x ≥ 3, then:
13
• ∀P (x) is “for all x, x ≥ 3”. ∀P (x) is false.
• ∃P (x) is “there exists x, x ≥ 3”. ∃P (x) is true.
Propositions involving predicates and quantifiers are logically equivalent if and only if they have
the same truth value no matter which predicates are substituted into these statements and which
domain of discourse is used for the variables in these propositional functions.
Theorem 1 ∀xP (x) ≡ ¬(∃xP (x)) and ∃xP (x) ≡ ¬(∀xP (x)).
7
Rules of Inference
We study logic to help us construct arguments. Proofs are essentially valid arguments that establish the
truth of a mathematical proposition. In this section we present the rules of inference for constructing
valid arguments.
7.1
Argument
Definition 20 An argument is a sequence of statements that end with a conclusion. All proceeding
statements before the conclusion are called premises. An argument is valid if it is impossible for all
the premises to be true and the conclusion is false.
Example 15 The following argument is valid:
“If you pass discrete math, you can take algorithms.”
“You have passed discrete math.”
“You can take algorithms.”
Observe that if we let p = “you pass discrete math”, q = “you can take algorithms”, then the
argument is essentially (p → q) ∧ p → q. In fact all arguments are essentially implications. The
conjunction of the premises of the argument forms the premise of the implication. The conclusion forms
the conclusion of the implication. In other words, consider an argument with statements p1 , p2 , ..., pn , q.
Expressing the argument logically, the argument is the implication p1 ∧ p2 ∧ ... ∧ pn → q.
Recall that if the argument is valid, it is impossible for the conclusion r to be false when all the
premises p1 , p2 , ...pn are true. On the other hand, observe that for the implication, p1 ∧ p2 ∧ ... ∧ pn → q
if any of the p1 , p2 , ...pn is false, their conjunction p1 ∧ p2 ∧ ... ∧ pn is false, resulting the implication
to be true. Thus, the implication correspond to a valid argument is in fact a tautology because the
implication will always evaluate to true. This is actually one way for verifying if an argument is valid.
7.2
Logical Form
The logical form of an English sentence or collection of sentences is the abstraction in logic expressions.
The logic expression (p → q) ∧ p → q is the logic form of the sentences from Example 15. The logic
form is important because it allows us to analyze the statements from a pure logic perspective.
Example 16 Construct the logic form of the following sentences:
14
If the professor travels to a conference, then the lecture will be replaced with an exam.
The discrete math lecture is replaced with an exam.
Therefore the professor is traveling to a conference.
Solution: Let p = “the professor travels to a conference”, and q = “lecture is replaced with an exam”.
Then the logic form is (p → q) ∧ q → p.
2
Observe that since discrete math has midterm and final exams. Even if the professor is not traveling,
some lectures will be replaced with the exams. Thus the argument is obviously false. What’s interesting
here is that the proposition (p → q) ∧ q → p is not a tautology, therefore the argument is not valid.
Thus even if we have no idea what the proposition is about, if we have its correct logic abstraction,
then we can analyze whether the argument is valid or not.
7.3
Argument Form
An argument form is the logic abstraction of an argument. Because of the particular structure of an
argument, which consists of a list of premises and followed by a conclusion, we will write the argument
form in a manner, where all the proposition involved are listed in a column, whose last entry is the
conclusion. To distinguish conclusion from the premises, we include ∴ in front of the conclusion. The
general argument form is the following:
∴
p1
p2
...
pn
q
For Example 16, its argument form is:
∴
p→q
q
p
In most situations, it is rather straightforward to obtain the logic form of an argument. The things
can be a little bit tricky when quantifiers are involves.
Example 17 Construct the logic form of the following sentences:
All whales are mammals.
Mammals are warm blooded.
Therefore all whales are warm blooded.
Solution: Let P (x) : “x is a whale.”, and Q(x) : “x is a mammal”. Then the first sentence “All
whales are mammals” is ∀x(P (x) → Q(x)).
Similarly, let R(x) : “x is warm blooded”. Then the second sentence is ∀x(Q(x) → R(x)). The
third sentence is ∀x(P (x) → R(x)).
The argument form is:
15
∴
∀x(P (x) → Q(x))
∀x(Q(x) → R(x))
∀x(P (x) → R(x))
2
Example 18 Construct the logic form of the following argument:
All computer science students are good at math.
I am not good at math.
Therefore I am not a computer science student.
Solution: Let P (x) : “x is computer science student.”, and Q(x) : “x is good at math”. Then the
first sentence “All computer science students are smart” is ∀x(P (x) → Q(x)).
So, what is the logic abstraction of “I am not good at math”? Observe that this is essentially saying
that when applying the function Q to me, the result is a F, i.e., ¬Q(I). Similarly, the conclusion “I
am not a computer science student” is ¬P (I).
The argument form is:
∴
∀x(P (x) → Q(x))
¬Q(I).
¬P (I).
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7.4
Rules of Inference
From the previous discussion, it is clear that we can determine if an argument is valid by examining its
argument form. If the argument form is a tautology, then the argument is valid. It is worthwhile noting
that a valid argument doesn’t necessarily imply the argument is correct. For example, in Example
18, the argument is valid. However, the argument is not correct because the assumption that “all
computer science students are good at math” is not correct.
Even though it is possible to determine if an argument form is valid by writing out its truth tables,
this can be cumbersome for argument with a lot of premises. In such situations, it is very helpful to
come up with a set of “short cuts” that can help us simplify the argument. These shortcuts are in fact
“building blocks” for constructing a valid argument and are called Rules of Inferences.
The Table below shows some of the most well-known rules of inference:
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Name
Tautology
Modus Ponens
p ∧ (p → q) → q
Modus Tollens
¬q ∧ (p → q) → ¬p
Hypothetical Syllogism
(p → q) ∧ (q → r) → (p → r)
Disjunctive Syllogism
(p ∨ q) ∧ ¬p → q
Addition
p→p∨q
Simplification
p∧q →p
Conjunction
p∧q →p∧q
Resolution
(p ∨ q) ∧ (¬p ∨ r) → (q ∨ r)
Argument Form
p
p→q
∴ q
¬q
p→q
∴ ¬p
p→q
q→r
∴ p→r
p∨q
¬p
∴ q
p
∴ p∨q
p∧q
∴ p
p
q
∴ p∧q
p∨q
¬p ∨ r
∴ q∨r
Example 19 Determine if the following argument form is valid:
∴
¬p ∧ q
r→p
¬r → s
s→t
t
Solution:
∴
¬p ∧ q
¬p
r→p
¬r
¬r → s
s
s→t
t
Simplification
Modus Tollens
Modus Ponens
Modus Ponens
2
Example 20 Determine if the following argument form is valid:
17
∴
p→q
¬p → r
r→s
¬q → s
Solution:
∴
p→q
¬q → ¬p
¬p → r
¬q → r
r→s
¬q → s
Logic Equivalence
Hypothetical Syllogism
Hypothetical Syllogism
2
Example 21 Determine if the following argument form is valid:
∴
(p ∧ q) ∨ r
r→s
p∨s
Solution:
∴
(p ∧ q) ∨ r
¬(p ∧ q) → r
r→s
¬(p ∧ q) → s
(p ∧ q) ∨ s
(p ∨ s) ∧ (q ∨ s)
p∨s
Logic Equivalence
Hypothetical Syllogism
Logic Equivalence
Logic Equivalence
Simplification
2
7.5
Rules of Inference for Quantifiers
The table below shows some useful rules of inference for quantifiers.
18
Name
Universal Instantiation
Universal Generalization
Existential Instantiation
Existential Generalization
Argument Form
∴
∀xP (x)
P (c), c is a particular member of the domain of P .
∴
P (c) for arbitrary c in the domain of P
∀xP (x)
∴
∃xP (x)
P (c), c is a particular member of the domain of P .
∴
P (c), c is a particular member of the domain of P .
∃xP (x)
Example 22 Determine if the following argument is valid:
All computer science students are smart.
Sean is a computer science student.
Therefore Sean is smart.
Solution: Let P (x) : “x is computer science student.”, and Q(x) : “x is smart”. Then the first
sentence “All computer science students are smart” is ∀xP (x) → Q(x).
The second sentence “Sean is a computer science student” is P (sean).
The third sentence “Sean is smart” is Q(Sean).
The argument form is:
∴
∀x(P (x) → Q(x))
P (Sean).
Q(Sean).
This is a valid argument.
∴
∀x(P (x) → Q(x))
P (Sean) → Q(Sean)
P (Sean)
Q(Sean).
Universal Instantiation
Modus Ponens
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8
Logic and Proof
We study logic to help us construct valid proofs. In this section, we will show a few examples illustrating how logic can be used to construct proofs. However, as the readers will soon find out, writing
mathematical proofs in logic can result in lengthy arguments, and not practical. As we have mentioned
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before, when proving a mathematical problem, we should use the knowledge gained in this section and
think whether our argument is logical rather than trying to convert everything to logic.
8.1
Direct Proof
Example 23 Prove if n is odd, then n2 is odd.
Proof: Let P (n) = “n is odd”, and Q(n) = “n2 is odd”. Then the theorem is the proposition
∀n(P (n) → Q(n)). Our strategy is to prove that for arbitrary c, P (c) → Q(c). Then we use universal
generalization:
∴
P (c) → Q(c) for arbitrary c
∀n(P (n) → Q(n))
to conclude that ∀n(P (n) → Q(n)) is true.
P (c) = “c is odd.”
For any x, if x is odd, then x = 2k + 1 for some integer k.
If c is odd, then c = 2k + 1 for some integer k.
c = 2k + 1
2
2
2
2
c = (2k + 1) = 4k + 4k + 1 = 2(2k + 2k) + 1
For any x, if x = 2m + 1 for some integer m, then x is odd.
If c2 = 2m + 1 for some integer m, then n2 is odd.
Q(c) = “c2 is odd.”
Premise
Definition
Universal Instantiation
Modus Ponens
Premise/Algebra
Definition
Universal Instantiation
Modus Ponens
Thus P (c) → Q(c) is true for arbitrary c. Now from universal generalization, we can conclude that
∀n(P (n) → Q(n)).
2
8.2
Proof by Contrapositive
Example 24 If 3n + 2 is odd, then n is odd.
Proof: Let P (n) = “3n + 2 is odd” and Q(n) = “n is odd”. Then we are trying to prove ∀n(P (n) →
Q(n)).
We will use the same strategy and show that P (c) → Q(c) for arbitrary c, and use universal
generalization to conclude that ∀n(P (n) → Q(n)) is true. However instead of using direct proof to
show that P (c) → Q(c), we will prove its equivalent contrapositive, i.e., ¬Q(c) → ¬P (c)
¬Q(c) = “c is even” for some arbitrary c.
For any x, if x is even, then x = 2k for some integer k.
If c is even, then c = 2k for some integer k.
c = 2k.
3c + 2 = 3 · (2k) + 2 = 6k + 2 = 2(3k + 1)
For any x, if x = 2m for some integer m, then x is even.
If 3c + 2 = 2m for some integer m, then 3c + 2 is even.
¬P = “3c + 2 is even.”
Premise
Definition
Universal Instantiation
Modus Ponens
Premise/Algebra
Definition
Universal Instantiation
Modus Ponens
Thus we know that ¬Q(c) → ¬P (c) for arbitrary c, which is equivalent to P (c) → Q(c). Using
universal generalization, we can conclude ∀n(P (n) → Q(n)).
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2
8.3
Proof by Counter Example
Example 25 Show that the proposition “the square of any odd number is even” is wrong.
Proof: Let P (n) = “n is odd” and Q(n) = “n2 is even”. Then we are trying to show that the
argument ∀n(P (n) → Q(n)) is false. In other words, we are trying to prove that ¬∀n(P (n) → Q(n)).
This is logically equivalent to ∃n¬(P (n) → Q(n)). In other words, if we can find a particular n such
that P (n) → Q(n) is wrong, then we are done. This is what we called a counter example. In particular
consider n = 3, then n= 9 is not even.
2
8.4
Proof by Contradiction
Example 26
√
2 is irrational.
√
Proof: Let p = “ 2 is irrational”. We will use proof by contradiction. Recall that the tautology for
√
proof by contradiction is ¬p → F . We will show that ¬p = “ 2 is rational” will lead to a contradiction.
√
2 is rational.
√If x is irrational, then√x = for relative prime integers a and b.
2 is irrational, then√ 2 = for relative prime integers a and b.
2 = ab for relatively prime integers a and b.
a and b are relatively prime.
a2 = 2b2
If x = 2k for some integer k, then x is even.
If a2 iseven, then a is even.
a2 is even
2
If x is even, then x is even.
If a2 is even, then a is even.
a is even
If x is even, then x = 2k for some integer k.
If a is even, then a = 2k for some integer k
a = 2k for some integer k
a2 = 4k 2
2
2
b2 = a2 = 4k2 = 2k 2
b is even
“a and b are relative prime” and “a, b are both even”.
a
b
a
b
Premise
Premise/Definition
Universal Instantiation
Modus Ponens
Simplification
Premise/Algebra
Definition
Universal Instantiation
Modus Ponens
Theorem
Universal Instantiation.
Modus Ponens
Definition
Universal Instantiation
Modus Ponens
Algebra
Algebra
Similar to a is even.
Conjunction
To complete the proof, we need to invoke the following arguments.
¬p → F
¬¬p ∧ F
¬F
¬¬p
Premise
Logic Equivalence
Premise
Premise
Disjunctive Syllogism
2
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8.5
Proof by Induction
Let the domain D = {n0 , n0 + 1, ...}. Let P (n) = be a predicate with variable n. Then the induction
theorem says ∀nP (n) ≡ P (n0 ) ∧ (∀(P (n) → P (n + 1))). To put this in plain English, to show that
P (n) is true for all integers n ≥ n0 , we only need to show that “P (n0 ) is true” and “for all n, if P (n)
is true then O(n + 1) is true”. The first part “P (n0 ) is true” is called the basis of induction. The
second part “for all n, if P (n) is true then P (n + 1) is true” is called the induction step, where the
premise “P (n) is true” is called the induction hypothesis. To prove the induction step, we need to
use show that “for arbitrary k, if P (k) is true then P (k + 1) s true”. Then use universal generalization
to claim the correctness of the induction step.
The strong induction theorem says ∀nP (n) ≡ P (n0 ) ∧ (∀P (n0 ) ∧ P (n0 + 1) ∧ ... ∧ P (k)) → P (k + 1)).
The difference between strong induction and weak induction is the premise of the induction step. In
the strong induction theorem, the premise is P (n0 ) ∧ P (n0 + 1) ∧ ... ∧ P (k).
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