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Mathematical Logic
Lecture 3
Propositional Logic
Continuing
Marie Duží
1
Normal forms of formulas of
propositional logic
• To each formula of propositional logic there is just one truth-value
function, i.e., mapping {p, q, r…}  {0, 1} (the truth-value table).
• On the other hand, to each such function there are infinitely many
formulas, which are mutually equivalent.
• Definition: Formulas A, B are equivalent, notation A  B, if A, B have
exactly the same models, i.e., A and B express the same truth-value
function. In other words, A entails B, and vice versa:
A  B iff A |= B a B |= A.
• Example: p  q  p  q  (p  q) 
(p  q)  (p  p)  (p  q)  (p  p)  ...
• Note: Do not confuse the equivalence of formulas A  B (‘ ‘ is a sign
of meta-language!) with a formula of the form A  B.
However, it holds that A  B if and only if the formula |= A  B (is a
tautology).
• Example: (p  q)  [(p  q)  (q  p)]
iff
•
|= [(p  q)  ((p  q)  (q  p))]
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Normal forms of formulas
(p  q)  [(p  q)  (q  p)] 
[(p  q)  (q  p)] 
[(p  q)  (p  q)]  ….
• It is useful to determine a normal (standardized) form
of a formula: to select one (or two) determined forms out
of these infinitely many equivalent forms.
• The class of equivalent formulas is then represented
by the chosen formula in the normal form.
• In the above example the second and third formulas are
in a normal form.
•
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Normal forms (of PL formulas): definition
Literal is a proposition symbol or its negation:
Ex.: p, q, r, ...
Elementary conjunction (EC) is a conjunction of literals.
Ex.: p  q, r  r, ...
Elementary disjunction (ED) is a disjunction of literals.
Ex.: p  q, r  r, ...
Complete elementary conjunction (CEC) of a given set S of elementary propositional symbols is an
elementary conjunction in which each symbol (element of S) occurs just once:
Ex.: p  q
Complete elementary disjunction (CED) of a given set S of elementary propositional symbols is an
elementary disjunction in which each symbol (element of S) occurs just once:
Ex.: p  q
Disjunctive normal form (DNF) of a formula F is a formula F’ such that F’ is equivalent to F and F’ has the
form of a disjunction of elementary conjunctions.
Example: DNF(p  p): (p  p)  (p  p), p  p
Conjunctive normal form (CNF) of a formula F is a formula F’ such that F’ is equivalent to F and F’ has the
form of a conjunction of elementary disjunctions.
CNF(p  p): (p  p)  (p  p)
Complete disjunctive normal form (CDNF) of a formula F is a formula F’ equivalent to F such that F’ has
the form of a disjunction of complete elementary conjunctions. UDNF(p  q): (p  q)  (p  q)
Complete conjunctive normal form (CCNF) of a formula F is a formula F’ equivalent to F such that F’ has
the form of a conjunction of complete elementary disjunctions. UKNF(p  q): (p  q)  (q  p)
CDNF and CCNF are called canonic (or standard) forms of a formula.
4
Normal forms of PL formulas
How to find canonic forms (i.e., CDNF, CCNF) of a
formula?
CDNF: disjunction = 1, if at least one CEC = 1, i.e.,
all the literals in this CEC = 1.
CCNF: conjunction = 0, if at least one CED = 0, i.e.,
all the literals in this CED = 0.
Therefore: CDNF (CCNF) can be constructed from
the truth-table in such a way, that we take into
account only rows with the value 1 (0) and ensure
“the correct value of literals” – 1 (0).
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CDNF, CCNF  the truth-table
method
Find the CDNF, CCNF
of the formula:
(pq)
p q (pq) UEK
UED
pq
1 1
0
1 0
1
CDNF: pq
0 1
CCNF:
0 0
(pq)(pq)(pq)
0
pq
0
pq
Mathematical Logic
pq
6
CDNF, CCNF  by equivalent
transforming method:
p  q  p  q  (p  q) CCNF 
[p  (q  q]  [q  (p  p] 
 p  q  p  q  p  q CDNF
Note: In the second line we make use of the fact that
a disjunction of any formula A with a contradiction F
is equivalent to A: A  F  A
In the third line we make use of distributive laws
Statement: Each formula which is not a contradiction
has a CDNF, and
each formula which is not a tautology has a CCNF
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Reverse task: given a CDNF, CCNF, find a simpler
original formula
Example
• An alchemist is imprisoned. He received 5 messages with the following content:
– p:
– q:
– r:
You will succeed in converting lead to gold.
On April 1st your brother-in-law will become an attorney.
After April 1st you may expect pleading.
1st message says:
pqr
2nd message says:
p  q  r
3rd message says:
p  q  r
4th message says:
p  q  r
5th message: At least one of the previous messages is true.
– Question: What did the poor alchemist learn?
•
Solution: (p  q  r)  (p  q  r)  (p  q  r)  (p  q  r).
Hence we need to find a formula F which has the above CDNF.
By means of distribute laws we obtain:
(p  q  r)  (p  q  r)  (p  q  r)  (p  q  r) 
(p  q)  (r  r)  (p  q)  (r  r)  (p  q)  (p  q)  (p  q)
•
Answer: You will succeed in converting lead to gold if and only if on April 1st your
brother-in-law becomes an attorney.
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Question: how many binary truth-value functions
(hence how many binary connectives) are there?
X Y 0 1 2 3 4 5 6 7 8 9 1 1 1 1 1 1
0 1 2 3 4 5
1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

  
 
Which is the minimal number of truthfunctional connectives?
•
According to the Statement on normal forms (see slide 7)
the following connectives suffice: , , 
(funkcionally complete system)
The following systems of truth-functional connectives are
functionally complete:
1. {, , },
2. {, } or {, },
3. {, },
4. {} or {}.
Hence in order to express any truth-value function (and thus
any PL formula) just one connective suffices!
Either Scheffer’s NAND  or Pierce’s NOR 
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Proof on the minimum of connectives
•
•
•
•
The system {, , } is sufficient according to the
Statement on normal forms
Transition to the system {, } or {, }:
A  B  (A  B),
A  B  (A  B)
Transition to the system {, }:
A  B  A  B,
A  B  (A  B)
Transition to the system {} or {}:
A  AA, AB  (AB)(AB), ( is NAND),
A  AA, AB  (AB)(AB), ( is NOR).
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A method of creating DNF, CNF:
semantic tableau
• Disjunctive tableau: a tree where branching means
disjunction and leaves are conjunctions of literals
• Conjunctive tableau: a tree where branching means
conjunction and leaves are disjunctions of literals
• |= A (tautology)  all the leaves of the conjunctive
tableau have to be closed, i.e., contain an opposite pair of
literals: p, p, because (p  p) – tautology
• A |= (contradiction)  all the leaves of the disjunctive
tableau have to be closed, i.e., contain an opposite pair of
literals: p, p, because (p  p) – contradiction
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Way of creating a tableau
1. A formula is transformed in such an equivalent way
that the negation is moved inside (to elementary
symbols).
2. Formulas of an implicative or equivalence form are
transformed by means of:
(p  q)  (p  q),
(p  q)  (p  q)  (q  p)  (p  q)  (p  q)
3. Construct the tableau by applying a distributive law
where-ever necessary.
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Applying a distributive law:
disjunctive tableau
•
•
•
(Branching means disjunction, comma conjunction.)
Repeat: (from left to right) branch the formula into a lefthand part and a right-hand part whenever you find a
disjunction.
The left-hand branch contains the left disjunct + the rest
of formula where instead of conjunction there is a
comma.
The right-hand part contains the right disjunct + the rest
of formula where instead of conjunction there is a
comma.
Until: there is no disjunction in any branch and the tree
contains in its leaves only literals separated by commas,
i.e., elementary conjunctions.
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Applying a distributive law:
conjunctive tableau
•
•
•
•
(dual to disjunctive)
(Branching means conjunction, comma disjunction.)
Repeat: (from left to right) branch the formula into a
left-hand part and a right-hand part whenever you find
a conjunction.
The left-hand branch contains the left conjunct + the
rest of formula where instead of conjunction there is a
comma.
The right-hand part contains the right conjunct + the
rest of formula where instead of disjunction there is a
comma.
Until: there is no conjunction in any branch and the
tree contains in its leaves only literals separated by
commas, i.e., elementary disjunctions.
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Prove that a formula is a
a) tautology, b) contradiction
a) Proof that a formula F is a contradiction:
Construct the disjunctive tableau.
If all the branches are closed, i.e., each leave of the tree
contains a couple of opposite literals (e.g. p, p, which
means p  p), the formula F is a contradiction.
b) Proof that a formula F is a tautology:
Construct the conjunctive tableau.
If all the branches are closed, i.e., each leave of the tree
contains a couple of opposite literals (e.g. p, p, which
means p  p), the formula F is a tautology.
Example (next slide): proof of a contradiction.
16
(p  q  r)  (s  q)  (t  r)  p  s  t
p,(s  q),(t  r),p,s,t
+
q,s,(t  r),p,s,t
+
q,(s  q),(t  r),p,s,t
r,(s  q),(t  r),p,s,t
q,q,(t  r),p,s,t
+
r,s,(t  r),p,s,t
+
r,q,t,p,s,t
+
Mathematical Logic
r,q,(t  r),p,s,t
r,q,r,p,s,t
+
17
Find consequences of a formula
•
•
•
•
Complete a disjunctive semantic tableau which
is not closed in each branch by such a set of
literals that closes all the branches.
Negation of the conjunction of added literals is
entailed.
For instance, if you add p, q, then you added
the formula p  q, hence p  q, i.e. p  q
is entailed.
By a dual way you may use a conjunctive
normal form.
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Complete normal forms
Looking at a tableau, you can easily see why
• a tautology does not have a complete conjunctive normal
form (CCNF) and why
• a contradiction does not have a complete disjunctive
normal form (CDNF):
If a formula is a tautoloy, then all the branches of a
conjunctive tableau are closed, i.e., in each ED there is a
couple of opposite literals p, p, which means p  p.
but then the elementary disjunction is not complete.
Hence the CCNF of the formula does not exist.
• Similarly for CDNF.
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CDNF of a tautology with
one (T1), two (T2), three (T3) variables
T1
p  p
T2
(p  q)  (p  q)  (p  q)  (p  q)
T3
(p  q  r)  (p  q  r)  (p  q  r) 
(p  q  r)  (p  q  r)  (p  q  r) 
(p  q  r)  (p  q  r)
For each valuation v there has to be at least one
CEC true.
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If a formula is not a tautology, at least one of these
CECs has to be missing in the respective
CDNF.
Therefore, if we construe a CDNF from a truth
table, we take into account only those rows,
where the function has the value 1 (value
False – 0 does not matter, because A  F 
A) and construe CEC in such a way that the
CEC is true.
Similarly for CCNF – take into account 0-s.
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Prove that the following formula F is a tautology:
F: [(p  (q  r))  (s  q)  (t  r)]  (p  (s  t))
•
Indirect proof.
By means of the disjunctive normal form prove
that the formula F is a contradiction:
(p  (q  r))  (s  q)  (t  r)  p  s  t 
(p  q  r)  (s  q)  (t  r)  p  s  t
Construing a disjunctive tableau (branching –
disjunction, comma – conjunction):
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(p  q  r)  (s  q)  (t  r)  p  s  t
p,(s  q),(t  r),p,s,t
+
q,s,(t  r),p,s,t
+
q,(s  q),(t  r),p,s,t
r,(s  q),(t  r),p,s,t
q,q,(t  r),p,s,t
+
r,s,(t  r),p,s,t
+
r,q,t,p,s,t
+
Mathematical Logic
r,q,(t  r),p,s,t
r,q,r,p,s,t
+
23
What is logically entailed by the formula G ?
G:
(p  (q  r))  (s  q)  (t  r)
Solution: When you make a semantic tableau of G, you
will see that the following literals are missing in all the
branches: p, s, t.
Hence the tableau of the formula (G  p  s  t) is
closed. The formula (G  p  s  t) is a contradiction.
Hence:
|= G  (p  s  t).
G |= (p  s  t),
G |= (p  s  t),
G |= (p  (s  t))
No wonder, because the formula F is a tautology !
Hence:
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Prove that the formula F is a tautology:
[(p  (q  r))  (s  q)  (t  r)]  (p  (s  t))
• Direct proof:
by means if the conjunctive normal form
prove that |= F:
• branching – conjunction, comma – disjunction
F  (p  q  r)  (s  q)  (t  r)  p  s  t
And so on,
similarly (in a dual way) to the indirect proof.
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Rezolution method in
propositional logic





From a practical point of view a semantic tableau is not
favourable
If you want to prove that
P1,...,Pn |= Z, it is equivalent to proving
tautology |= (P1 ...  Pn)  Z,
i.e., P1 ...  Pn  Z is a contradiction.
But, to prove it by the tableau method, you need the
disjunctive normal form.
Which means that you have to transform the formula
that is (almost) in the conjunctive form into the
disjunctive form, which yields a lot of applications of
the distributive law.
Using a tableau often yields a lot distributive steps
It is easier to directly prove the contradiction, i.e., that
the formula P1 ...  Pn  Z is not satisfiable.
Resolution rule in the propositional logic
Let l be a literal. From (A  l)  (B  l) infer (A  B). Notation:
(A  l)  (B  l)
–––––––––––––––
(A  B)
resolvent
This rule does not lead to an equivalent formula. But it preserves satisfiability.
Proof: Let formula (A  l)  (B  l) be non-contradictory. hence there is a
valuation v in which it is true.
But then in this v both the disjuncts (so called clauses) are true:
(A  l) and (B  l).
Let v(l) = 0. Then w(A) = 1 and w(A  B) = 1.
Let v(l) = 1. Then w(l) = 0 and w(B) = 1, a thus w(A  B) = 1.
In both cases the formula A  B is true in the model v of the opriginal
formula, hence it is not a contradiction.
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Proof by resolution in PL
If you realise that the above proof is valid for any
valuation v, then you see that the rule is
truth preserving:
(A  l)  (B  l) |= (A  B).
 Thus you obtained also a method of finding
logical consequences of a formula.
 Moreover:
(A  l)  (B  l)  (A  l)  (B  l)  (A  B)
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28
Clausal form
 The conjunctive normal form is called clausal form.
Particular conjuncts (i.e., elementary disjunctions) are
called clauses.
 Example: Transition of a formula into the clausal form:
[((p  q)  (r  q)  r)  p]

((p  q)  (r  q)  r)  p

(p  q)  (r  q)  r  p
 The formula consists of four clauses.
The proof of contradiction:
 (p  q)  (r  q)  (p  r)  r  p  p
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29
Proof by resolution
• R(F) – conjunctive extension of a formula F by all its resolvents
• R0(F) = F, Ri(F) = R(Ri-1(F)) – rezolution closure of F.
It holds that
Ri(F)  F
• Proof: A is a contradiction (not satisfiable): there is an n such that Rn(A)
contains the empty clause:
• Proof: A is a tautology: A is a contradiction
• Proof: a set of formulas {A1,…,An} is not satisfiable. We prove that the
formula A1 ...  An is a contradiction
• Infer logical consequences of {A1,…,An}: infer all the resolvents
• Proof of validity: A1,…,An |= Z
– Direct: by step-by-step inferring resolvents from A1,…,An you obtain Z
– Indirect: prove that A1 ... An  Z is a tautology, i.e.,
A1 ... An  Z is a contradiction, i.e., the set {A1,..., An,,Z} is not
satisfiable
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Examples
•
•
Particular clauses are numbered. gradually infer
resolvents (that logically follow from the clauses).
When performing an indirect proof, you have to infer
p  p, the empty clause
Prove that the following formula is a contradiction:
(q  p)  (p  r)  (q  r)  p
1. q  p
2. p  r
3. q  r
4. p
5. p  r
rezolution: 1, 3
6. p
rezolution: 2, 5
7.
4, 6
Question: What follows from the formula
(q  p)  (p  r)  (q  r) ?
Answer:
the formula p
31
Examples
•
1.
2.
3.
4.
5.
6.
Direct proof:
p  q  r, s  q, t  r |= p  (s  t)
p  q  r
s  q
t  r
p  s  r
rezolution 1, 2
p  s  t
rezolution 3, 4
p  s  t  p  (s  t) QED
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32
Examples
• Indirect proof:
p  q  r, s  q, t  r |= p  (s  t)
1.
p  q  r
2.
s  q
3.
t  r
4.
p  s  r
rezolution 1, 2
5.
p  s  t
rezolution 3, 4
6.
p
nega7.
s
ted
8.
t
conclusion
9.
p  s  t, p, s, t |= s  t, s, t |= t, t |=
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Logic programming
Strategies of generating resolvents:
• Breadth-first (pursue all closest paths first)
– Resolution closure – generates all resolvents
– It can be rather inefficient due to combinatorial
explosion of resolvents
• Depth-first (top down: pursue first path to
exhaustion before looking an next path)
– It is very simple and efficient to implement, but it is not
complete; it can lead to some very odd infinite loops
though there is a solution in a neighbouring branch.
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Example: incompleteness of the
depth-first strategy
•
•
1.
2.
3.
4.
5.
6.
d  e  (b  a)  (e  b)  (d  a) |= a (?)
Logic program
(depth-first, top down strategy; goal driven):
D.
fact
E.
fact
A : B.
rule (A, if B)
B : E.
rule (B, if E)
A : D.
rule (A, if D)
?A
query, goal
– the program tries to meet the goal A by searching top down; the
clause 3 meets the goal  generates a new goal B. Succeeds
with the clause 4  a new goal E – met by the clause 2. – Yes
; - or generates backtracking: again ?A – clause 5 – goal D –
clause 1. - Yes
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Example: incompleteness of the
depth-first strategy
A small adjustment of the program:
1.
D.
2.
E : B.
3.
A : B.
4.
B : E.
5.
A : D.
6.
?A
A
3.
B
4.
E
2.
B
4. ....
5.
D
1.
etc.
The solution is not found, although it exists
36
Propositional logic:
summary of typical problems
1.
2.
3.
4.
•
A proof of a tautology / contradiction
Prove that a set of formulas is not satisfiable
(Direct / indirect) proof of validity of an argument
Inferring consequences of assumptions
We can solve by:
–
–
–
–
A semantic method, i.e., by a truth-table or indirectly
By means of equivalent transformations – using basic rules of
inference (modus ponens, transposition, disjunction elimination)
By means of normal forms: conjunctive and disjunctive, i.e., by a
semantic tableau
Resolution method: clausal form (conjunctive normal form)
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