1.5.2.
Contradiction
Definition: Contradiction.
Any statement that is false regardless of the truth values of
the constituent parts is called a contradiction or
contradictory statement.
Examples:
Complete the truth table for the statement
~ ( P ∧ Q ) ⇔ (Q ∧ P )
P
Q
~
T
T
F
∧
T
T
F
T
F
T
F
F
Step:
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⇔
F
∧
T
F
F
F
T
F
F
F
T
F
F
F
2
1
4*
3
(P
Logic
Q)
(Q
P)
43
Exercises:
•
Complete the truth table for the statement
~ ( P ∨ Q ) ∧ P to show it is a contradiction.
P
Q
~(P
∨
Q)
∧
P
Step:
•
Complete the truth table for the statement
( P ∧ Q ) ∧ ~ Q to show it is a contradiction.
P
Q
(P
∧
Q)
∧
~Q)
Step:
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1.5.2.1
Quick Method for Showing a
Contradiction
The quick method for determining if a compound statement
is a tautology can be used similarly for showing a
contradiction.
The quick method relies on the fact that if a truth value of
“T” can occur under the main connective (for some
combination of truth values for the components), then the
statement is not a contradiction. If this truth value is not
possible, then we have a contradiction.
Therefore, to determine whether a statement is a
contradiction, we place a “T” under the main connective
and work backwards.
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Example:
•
Use the “quick” method for the statement
~ ( P ∨ Q ) ∧ P to determine if it is a contradiction.
~
Step:
(P
2
∨
1
Q)
1.Place “T” under main
∧
3*
P
T
connective
2. For “T” to occur
T
T
under the main
connective, ~ must be
“T” and P must be “T”
F
3. For “T” to occur
under ~, P ∨ Q must be
“F”.
4. For “F” to occur
F
F
under P ∨ Q , P must be
“F” and Q must be “F”
P cannot be both “T” and “F”, thus ~ ( P ∨ Q ) ∧ P can only
ever be false and is a contradiction.
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Exercise:
•
Use the “quick” method for the statement
( P ∧ Q ) ∧ ~ Q to determine if it is a contradiction.
(P
∧
Q)
∧
~Q
Step:
1.
2
3
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1.5.3.
Contingent
Definition: Contingent.
Any statement that is neither a tautology nor a
contradiction is called a contingent or intermediate
statement.
Examples:
Complete the truth table for the statement Q ∨ (Q ⇒ P )
Q
T
T
∨
T
T
F
T
T
F
T
F
F
F
F
T
T
2*
1
Step:
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⇒
P
Q
Logic
(Q
P)
T
48
Exercises:
•
Complete the truth table for the statement
( p ∨ r ) ⇒ ( p ∧ q ) to show it is contingent.
p
q
r
(p
∨
r)
⇒
(p
∧
q)
Step:
•
Complete the truth table for the statement
~ (( p ∧ ~ q ) ∨ r ) ⇔ (r ⇒ q ) to show it is contingent.
p
q
r
~( (p ∧
~
q) ∨
r) ⇔ (r ⇒ q)
Step:
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49
1.6.
Logical Equivalence
Definition: Logical Equivalence.
Two statements are logically equivalent if, and only if, they
have identical truth values for each possible substitution of
statements for their statements variables.
The logical equivalence of two statements P and Q is
denoted P ≡ Q .
If two statements P and Q are logically equivalent then
P ⇔ Q is a tautology
1.6.1.
Determining Logical Equivalence.
To determine if two statements P and Q are logically
equivalent, construct a full truth table for each statement. If
their truth values at the main connective are identical, the
statements are equivalent.
Alternatively show P ⇔ Q is a tautology and hence
conclude P ≡ Q .
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50
Examples:
•
Determine if the following statements are logically
equivalent. P : p ⇒ q,
Q :~ p ∨ q
⇒
p
q
T
T
T
F
∨
T
T
F
F
F
F
F
T
T
T
T
F
F
T
T
T
1*
1
2*
p
Step:
q
~p
q
Since the main connectives * are identical, the statements P
and Q are equivalent. Thus P ≡ Q
•
i.e. p ⇒ q ≡ ~ p ∨ q
Determine if the following statements are logically
equivalent. P :~ ( p ∧ q ),
Q :~ p ∧ ~ q
p
q
~(
T
T
F
∧
T
T
F
T
F
T
F
F
Step:
~q
F
∧
F
F
F
F
T
T
F
T
F
F
T
F
T
T
T
2*
1
1
2*
1
p
q)
~p
F
Since the main connectives * are not identical, the
statements P and Q are not equivalent.
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Exercises:
•
Determine if the following statements are logically
equivalent. P :~ ( p ∨ q ),
p
q
~(
Q :~ p ∧ ~ q
∨
p
q)
∧
~p
~q
Step:
•
Determine if ~ ( p ∧ q ) ⇔ ~ p ∨ ~ q is a tautology, and
hence if ~ ( p ∧ q ) ≡~ p ∨ ~ q .
p
q
~(
p
∧
q)
⇔
~p
∨
~q
Step:
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1.6.2.
Substitution
There are two different types of substitution into
statements.
Rule of Substitution: If in a tautology all occurrences of a
variable are replaced by a statement, the result is still a
tautology.
Examples:
•
We know P ∨ ~ P is a tautology.
Thus, by the rule of substitution, so too are:
∗
Q ∨ ~ Q , by letting Q = P .
∗
(( p ∧ q ) ⇒ r )∨ ~ (( p ∧ q ) ⇒ r ) , by letting
( p ∧ q) ⇒ r = P .
Note: We have simply replaced every occurrence of P in
the tautology P ∨ ~ P , by some other statement.
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Rule of Substitution of Equivalence: If in a tautology we
replace any part of a statement by a statement equivalent to
that part, the result is still a tautology.
Example:
•
Determine if P ⇒ (~ Q ∨ P ) is a tautology.
We know: P ⇒ (Q ⇒ P ) is a tautology and
( P ⇒ Q ) ≡~ P ∨ Q
By the rule of substitution (Q ⇒ P ) ≡ ~ Q ∨ P
Thus, by the rule of substitution of equivalence,
P ⇒ (Q ⇒ P ) ≡ P ⇒ (~ Q ∨ P ) , and hence
P ⇒ (~ Q ∨ P ) is also a tautology.
Exercise:
•
~ T ∨ (~ S ∨ T ) a tautology? Yes.
We know ( P ⇒ Q ) ≡~ P ∨ Q . So, ( S ⇒ T ) ≡~ S ∨ T and
T ⇒ (~ S ∨ T ) ≡ ~ T ∨ (~ S ∨ T ) (by RoS).
Hence, ~ T ∨ (~ S ∨ T ) ≡ T ⇒ ( S ⇒ T ) (by SoE).
P ⇒ (Q ⇒ P ) is a known tautology, thus (by (SoE)
T ⇒ ( S ⇒ T ) is a tautology, and since
~ T ∨ (~ S ∨ T ) ≡ T ⇒ ( S ⇒ T ) , ~ T ∨ (~ S ∨ T ) is a
tautology.
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1.6.3.
Laws
The following logical equivalences hold:
1.
Commutative Laws:
•
•
•
( P ∨ Q ) ≡ (Q ∨ P )
( P ∧ Q ) ≡ (Q ∧ P )
( P ⇔ Q ) ≡ (Q ⇔ P )
2.
Associative Laws:
•
•
•
(( P ∨ Q ) ∨ R ) ≡ (P ∨ (Q ∨ R ))
(( P ∧ Q ) ∧ R ) ≡ (P ∧ (Q ∧ R ))
(( P ⇔ Q ) ⇔ R ) ≡ (P ⇔ (Q ⇔ R ))
3.
•
•
Distributive Laws:
(P ∨ (Q ∧ R) ) ≡ (( P ∨ Q) ∧ ( P ∨ R) )
(P ∧ (Q ∨ R) ) ≡ (( P ∧ Q) ∨ ( P ∧ R) )
4.
•
Double Negation (Involution) Law:
~~ P ≡ P
5.
•
•
De Morgan’s Laws:
~ ( P ∨ Q ) ≡ (~ P ∧ ~ Q )
~ ( P ∧ Q ) ≡ (~ P ∨ ~ Q )
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6.
Implication Laws:
•
( P ⇒ Q ) ≡ (~ P ∨ Q )
•
( P ⇔ Q ) ≡ (( P ⇒ Q ) ∧ (Q ⇒ P ) ) ( Biconditional)
Identity Laws:
7.
•
•
(P ∨ F ) ≡ P
(P ∧ T ) ≡ P
Negation (Complement) Laws:
8.
•
•
( P∨ ~ P ) ≡ T
( P∧ ~ P) ≡ F
9.
•
•
Dominance Laws:
(P ∨ T ) ≡ T
(P ∧ F ) ≡ F
Idempotent Laws:
10.
•
•
( Implication )
( P ∨ P) ≡ P
( P ∧ P) ≡ P
11. Absorption Laws:
•
P ∧ (P ∨ Q) ≡ P
•
P ∨ (P ∧ Q) ≡ P
Property of Implication:
12.
•
•
(P ⇒ (Q ∧ R) ) ≡ (( P ⇒ Q) ∧ ( P ⇒ R) )
(( P ∨ Q) ⇒ R ) ≡ (( P ⇒ R) ∧ (Q ⇒ R) )
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Example:
Prove the first of De Morgan’s Laws using truth tables.
P
Q
~(
T
T
F
∨
T
T
F
F
F
T
F
F
Step:
~Q
F
∧
F
T
F
F
T
F
T
T
F
F
T
F
T
T
T
2*
1
1
2*
1
P
Q)
~P
F
Since the main connectives are identical, the statements are
equivalent, and first of De Morgan’s Laws is true.
Exercise:
Prove the second of De Morgan’s Laws using truth tables.
P
Q
~(
P
∧
Q)
~P
∨
~Q
Step:
Since the main connectives are identical, the statements are
equivalent, and second of De Morgan’s Laws is true.
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57
Example:
Using logically equivalent statements, without the direct
use of truth tables, show: ~ (~ p ∧ q ) ∧ ( p ∨ q ) ≡ p
~ (~ p ∧ q ) ∧ ( p ∨ q ) ≡ (~ (~ p ) ∨ ~ q ) ∧ ( p ∨ q )
(De Morgan )
≡ ( p ∨ ~ q ) ∧ ( p ∨ q ) (Double Negation)
(Distributivity)
≡ p ∨ (~ q ∧ q )
(Commutativity)
≡ p ∨ (q ∧ ~ q )
(Negation)
≡ p∨F
(Identity)
≡ p
Exercises:
Using logically equivalent statements, without the direct
use of truth tables, show:
•
~ ( p ⇔ q ) ≡ ( p ∧ ~ q ) ∨ (q ∧ ~ p )
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58
•
( p ⇒ q ) ≡ (~ q ⇒~ p )
•
p ⇒ (q ∧ r ) ≡ ( p ⇒ q ) ∧ ( p ⇒ r ) , without using the
property of implication
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