PROVING LOGICAL EQUIVALENCES
1. Show that
¬
(p
⟶
q) and p
∧
¬
q are logically equivalent.
¬
(p
⟶
q) =
¬
(
¬
p
∨
q) by Conditional law
=
¬
(
¬
p)
∧
¬
q by the second De Morgan law
= p
∧
¬
q by the double negation law
2. Prove that
¬
(p
∨
(
¬
p
∧
q)) and
¬
p
∧
¬
q.
¬
(p
∨
(
¬
p
∧
q)) =
¬
p
∧
¬
(
¬
p
∧
q) by the second De Morgan law
=
¬
p
∧
[
¬
(
¬
p )
∨
¬
q ] by the first De Morgan law
=
¬
p
∧
(p
∨
¬
q) by the double negation law
= (
¬
p
∧
p )
∨
(
¬
p
∧
¬
q) by the second distributive law
= F
∨
(
¬
p
∧
¬
q) because
¬
p
∧
p = F
= (
¬
p
∧
¬
q)
∨
F by the commutative law for disjunction
=
¬
p
∧
¬
q by the identity law for F
3. Show that (p
∧
q)
⟶
(p
∨
q) is a tautology.
(p
∧
q)
⟶
(p
∨
q) =
¬
(p
∧
q)
∨
(p
∨
q) by Conditional law
= (
¬
p
∨
¬
q)
∨
(p
∨
q) by the first De Morgan law
= (
¬
p
∨
p)
∨
(
¬
q
∨
q) by the associative and commutative laws for
disjunction
= T
∨
T by negation law
= Tby disjunction law
4. Show that (p
∨
q)
∧
(
¬
p
∨
r)
⟶
(q
∨
r) is a tautology.
(p
∨
q)
∧
(
¬
p
∨
r)
⟶
(q
∨
r) = (p
∨
q)
∧
¬
(
¬
p
∨
r)
∨
(q
∨
r) by Conditional law
= (p
∨
q)
∧
(
¬
(
¬
p)
∧
¬
r)
∨
(q
∨
r) by De Morgan law
= (p
∨
q)
∧
(p
∧
¬
r)
∨
(q
∨
r) by Double negation law
= (p
∧
p)
∨
(q
∨
q)
∧
(
¬
r
∨
r) by Basic laws simplification
= p
∨
q
∧
T by Negation law
= T
∧
T by ā law
= Tby Conjunction law
5. Show that (p
⟶
q)
∧
(q
⟶
r)
⟶
(p
⟶
r) is a tautology.
(p
⟶
q)
∧
(q
⟶
r)
⟶
(p
⟶
r)
=
¬
[(
¬
p
∨
q)
∧
(
¬
q
∨
r)]
∨
(
¬
p
∨
r) by Conditional law
= [
¬
(
¬
p
∨
q)
∨
¬
(
¬
q
∨
r)]
∨
(
¬
p
∨
r) by De Morgan law
= (
¬
(
¬
p)
∧
¬
q))
∨
(
¬
(
¬
q)
∧
¬
r))
∨
(
¬
p
∨
r) by De Morgan law
= (p
∧
¬
q)
∨
(q
∧
¬
r)
∨
(
¬
p
∨
r) by Double negation law
= (p
∨
¬
p)
∧
(
¬
q
∨
q)
∧
(
¬
r
∨
r) by Basic laws simplification
= T
∧
T
∧
T by Negation law
= Tby Conjunction law
6. Prove that (p
⟶
q)
∧
(q
⟶
r)
⟶
(p
⟶
r) = T
(p
⟶
q)
∧
(q
⟶
r)
⟶
(p
⟶
r)
= (
¬
p
∨
q)
∧
¬
(
¬
q
∨
r)
∨
(
¬
p
∨
r) by Conditional law
= (
¬
p
∨
q)
∧
(
¬
(
¬
q)
∧
¬
r)]
∨
(
¬
p
∨
r) by De Morgan law
= (
¬
p
∨
q)
∧
(q
∧
¬
r)]
∨
(
¬
p
∨
r) by Double negation law
= (
¬
p
∨
¬
p)
∨
(q
∧
q)
∧
(
¬
r
∨
r) by Basic laws simplification
=
¬
p
∨
q
∧
T by Idempotent and Negation law
= T
∧
T by ā law
= Tby Conjunction law