CmSc 175 Discrete Mathematics
Lesson 02: Tautologies and Contradictions. Logical Equivalences. De Morgan’s Laws
Solutions to Exercises
1. Tautologies and Contradictions
Exercise:
a) Show that P → Q  ¬ P V Q is a tautology, i.e. P → Q ≡ ¬ P V Q
P
Q
¬P
¬PVQ
P→Q
P→Q¬PVQ
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T
F
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F
F
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F
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F
F
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b) Show that ( P Q)  ( ( P Λ Q ) V ( ¬P Λ ¬Q) ) is a tautology
i.e. P Q ≡ ( P Λ Q ) V ( ¬P Λ ¬Q)
P
Q
P Q
---------------------------T
T
T
T
F
F
F
T
F
F
F
T
P
Q
~P
~Q
PΛQ
¬P Λ ¬Q
( P Λ Q ) V ( ¬P Λ ¬Q)
--------------------------------------------------------------------------------------------------T
T
F
F
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Let A = ( P Q)  ( ( P Λ Q ) V ( ¬P Λ ¬Q) )
P
Q
P Q
( P Λ Q ) V ( ¬P Λ ¬Q)
A
---------------------------------------------------------------------------T
T
T
T
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F
F
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1
c) Show that ( P Q)  ( ( P Λ ¬Q ) V ( ¬P Λ Q) ) is a tautology
i.e. P Q ≡ ( P Λ ¬Q ) V ( ¬P Λ Q)
P
Q
P Q
---------------------------T
T
F
T
F
T
F
T
T
F
F
F
P
Q
~P
~Q
P Λ ~Q
¬P Λ Q
( P Λ ~Q ) V ( ¬P Λ Q)
--------------------------------------------------------------------------------------------------T
T
F
F
F
F
F
T
F
F
T
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F
T
F
T
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F
F
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F
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F
Let A = ( P Q)  ( ( P Λ ¬Q ) V ( ¬P Λ Q) )
P
Q
P Q
( P Λ ¬Q ) V ( ¬P Λ Q)
A
---------------------------------------------------------------------------T
T
F
F
T
T
F
T
T
T
F
T
T
T
T
F
F
F
F
T
2. Logical equivalences
Exercises
Use the equivalence A → B = ~A V B, and the equivalence laws.
1. Show that (A → B) Λ A is equivalent to A Λ B
2. Show that (A → B) Λ B is equivalent to B
3. Show that (A → B) Λ (B → A) is equivalent to A B
4. Show that ~((A → B) Λ (B → A)) is equivalent to A B
5. Show that ¬ ((P → Q) → P ) Λ P is a contradiction
2
1. Show that (A → B) Λ A is equivalent to A Λ B
(A → B) Λ A = (~A V B) Λ A =
Apply distributive law:
(~A Λ A) V (B Λ A) =
Apply complement properties:
F V (B Λ A) =
Apply identity laws:
B Λ A=AVB
2. Show that (A → B) Λ B is equivalent to B
(A → B) Λ B = (~A V B) Λ B =
Apply distributive law:
(~A Λ B) V (B Λ B) =
Apply idempotency (consumption) laws:
(~A Λ B) V B =
Apply absorption laws
B
3. Show that (A → B) Λ (B → A) is equivalent to A B
(A → B) Λ (B → A) = (~A V B) Λ (~B V A) =
Apply distributive law:
(~A Λ ~B) V (~A Λ A) V (B Λ ~B) V (B Λ A) =
Apply complement properties:
(~A Λ ~B) V F
V F
V (B Λ A) =
Apply identity laws:
(~A Λ ~B) V (B Λ A) =
Apply commutative laws:
(B Λ A)
V (~A Λ ~B) =
Apply commutative laws:
(A Λ B) V (~A Λ ~B) = A 
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4. Show that ~((A → B) Λ (B → A)) is equivalent to A B
In problem 3 we showed that (A → B) Λ (B → A) = (A Λ B) V (~A Λ ~B)
~((A → B) Λ (B → A)) = ~( (A Λ B) V (~A Λ ~B) ) =
Apply De Morgan’s laws:
~(A Λ B) Λ ~(~A Λ ~B) =
Apply De Morgan’s laws:
(~A V ~B) Λ (A V B) =
Apply distributive law:
(~A Λ A) V (~A Λ B) V (~B Λ A) V (~B Λ B) =
Apply complement properties:
F V (~A Λ B) V (~B Λ A) V F =
Apply identity laws:
(~A Λ B) V (~B Λ A) =
Apply commutative laws:
(~A Λ B) V ( A Λ ~B) =
Apply commutative laws:
( A Λ ~B) V (~A Λ B) = A B
5. Show that ¬ ((P → Q) → P ) Λ P is a contradiction
¬ ((P → Q) → P ) Λ P = ~((~P V Q) → P ) Λ P =
~(~(~P V Q) V P ) Λ P =
Apply De Morgan’s laws:
~( ( P Λ ~Q) V P) Λ P =
Apply De Morgan’s laws:
~ ( P Λ ~Q) Λ ~P Λ P =
Apply complement properties:
~ ( P Λ ~Q) Λ F =
Apply identity laws:
F
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6. Replace the conditions in the following if statements with equivalent conditions without using the
logical operators || and &&
a) if ((a > 0 && b > 0) || (b > 0))
c = a*b;
Let P = a > 0, Q = b > 0
The expression is: (P Λ Q) V P)
By the absorption law (P Λ Q) V P) = P
Therefore we can replace ((a > 0 && b > 0) || (b > 0)) with b > 0
if (b > 0)
c = a*b;
b) if (( a > 0 || b > 0 ) && (b > 0)) c = a*b;
Let P = a > 0, Q = b > 0
The expression is: (P V Q) Λ Q)
By the absorption law (P V Q) Λ Q) = Q
Therefore we can replace ((a > 0 || b > 0) && (b > 0)) with b > 0
if (b > 0)
c = a*b;
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