Logical Equivalence Laws of Logic Proving logical equivalences using Laws of Logic and Truth Tables

1. Discrete Mathematical Structures
Fundamentals of Logic
BY,
Lakshmi R
Asst. Professor
Dept. of ISE

2. Logical Equivalence
• Two propositions u and v are said to be logically equivalent whenever
u and v have the same truth value.
• Symbol used ⇔
• Eg: (p → q) ⇔ (¬p ∨ q)
Lakshmi R, Asst. Professor, Dept. Of ISE
¬ ∧ ∨ ⊻ → ↔
p q ¬p (p → q) (¬p ∨ q)
0 0 1 1 1
0 1 1 1 1
1 0 0 0 0
1 1 0 1 1

3. Logical Equivalence contd.,
• Eg: [(p ∨ q) ∧ (¬ p ∨ ¬ q) ] ⇔ p ⊻ q
Lakshmi R, Asst. Professor, Dept. Of ISE
¬ ∧ ∨ ⊻ → ↔
p q ¬p ¬q (p ∨ q) (¬p ∨ ¬q) (p ∨ q) ∧ (¬ p ∨ ¬ q) p ⊻ q
0 0 1 1 0 1 0 0
0 1 1 0 1 1 1 1
1 0 0 1 1 1 1 1
1 1 0 0 1 0 0 0

4. Logical Equivalence contd.,
• Eg: [p → (q ∧ r) ] ⇔ [(p → q) ∧ (p → r)]
Lakshmi R, Asst. Professor, Dept. Of ISE
¬ ∧ ∨ ⊻ → ↔
p q r (q ∧ r) (p → q) (p → r) p → (q ∧ r) (p → q) ∧ (p → r)
0 0 0 0 1 1 1 1
0 0 1 0 1 1 1 1
0 1 0 0 1 1 1 1
0 1 1 1 1 1 1 1
1 0 0 0 0 0 0 0
1 0 1 0 0 1 0 0
1 1 0 0 1 0 0 0
1 1 1 1 1 1 1 1

5. Logical Equivalence contd.,
• Eg: [p ∧ (¬ q ∨ r) ] ⇔ [p ∨ (q ∧ ¬r) ]
Lakshmi R, Asst. Professor, Dept. Of ISE
¬ ∧ ∨ ⊻ → ↔
p q r ¬ q (¬ q ∨ r) p ∧ (¬ q ∨ r) ¬r (q ∧ ¬r) p ∨ (q ∧ ¬r)
0 0 0 1 1 0 1 0 0
0 0 1 1 1 0 0 0 0
0 1 0 0 0 0 1 1 1
0 1 1 0 1 0 0 0 0
1 0 0 1 1 1 1 0 1
1 0 1 1 1 1 0 0 1
1 1 0 0 0 0 1 1 1
1 1 1 0 1 1 0 0 1

6. Laws of Logic
Lakshmi R, Asst. Professor, Dept. Of ISE
¬ ∧ ∨ ⊻ → ↔
1. Law of double
negation
¬ ¬ p ⇔ p
2. DeMorgan’s
Laws
i. ¬ (p ∨ q) ⇔ ¬p ∧ ¬q
ii. ¬ (p ∧ q) ⇔ ¬p ∨ ¬q
3. Commutative
Laws
i. (p ∧ q) ⇔ (q ∧ p)
ii. (p ∨ q) ⇔ (q ∨ p)
4. Associative
Laws
i. p ∨ (q ∨ r) ⇔ (p ∨ q ) ∨ r
ii. p ∧ (q ∧ r) ⇔ (p ∧ q ) ∧ r
5. Distributive
Laws
i. p ∨ (q ∧ r) ⇔ (p ∨ q ) ∧ (p ∨ r )
ii. p ∧ (q ∨ r) ⇔ (p ∧ q ) ∨ (p ∧ r )
6. Idempotent
Laws
i. (p ∨ p) ⇔ p
ii. (p ∧ p) ⇔ p
7. Identity Laws
i. (p ∨ F) ⇔ p
ii. (p ∧ T) ⇔ p
8. Inverse Laws
i. (p ∨ ¬p) ⇔ T
ii. (p ∧ ¬p) ⇔ F
9. Domination
Laws
i. (p ∨ T) ⇔ T
ii. (p ∧ F) ⇔ F
10. Absorption
Laws
i. p ∨ (p ∧ q )⇔ p
ii. p ∧ (p ∨ q ) ⇔ p
Important p → q ⇔ ¬ p ∨ q

7. Law of negation for a conditional
• Let x be a specific number. Write the negation of the following conditional.
“if x is an integer, then x is a rational number”
Lakshmi R, Asst. Professor, Dept. Of ISE
¬ ∧ ∨ ⊻ → ↔
Let
p: x is an integer
q: x is a rational number
The given statement can be symbolically
represented as
p → q
We need to find ¬ (p → q)
Important p → q ⇔ ¬ p ∨ q
¬ (p → q) ⇔ ¬ (¬p ∨ q)
⇔ ¬ ¬p ∧ ¬q (using DeMorgan’s Law)
⇔ p ∧ ¬q (using law of double negation)
 ¬ (p → q) is
“x is an integer and x is not a rational number”
Solution
Problem 9:

8. Law of negation for a conditional
• Let x be a specific number. Write the negation of the following conditional.
“if x is not a real number, then it is not a rational number and not an
irrational number”
Lakshmi R, Asst. Professor, Dept. Of ISE
¬ ∧ ∨ ⊻ → ↔
Let
p: x is a real number
q: x is a rational number
r: x is an irrational number
The given statement can be symbolically
represented as
¬p →( ¬q ∧ ¬r)
We need to find ¬ (¬p →( ¬q ∧ ¬r))
¬ (¬p →( ¬q ∧ ¬r)) ⇔ ¬ (¬ ¬ p ∨ (¬q ∧ ¬r))
⇔ ¬ ( p ∨ (¬q ∧ ¬r)) (using law of double
negation)
⇔ ¬p ∧ ¬(¬q ∧ ¬r) (using DeMorgan’s Law)
⇔ ¬ p ∧ (¬ ¬q ∨ ¬ ¬r) (using DeMorgan’s Law)
⇔ ¬ p ∧ (q ∨ r) (law of double negation)
 ¬ (¬p →( ¬q ∧ ¬r)) is
“x is not a real number and it is a rational number or an
irrational number”
Solution

9. 1. Prove the following logical equivalences without using truth tables.
i. p ∨ [p ∧ (p ∨ q) ] ⇔ p
ii. [p ∨ q ∨ (¬ p ∧ ¬ q ∧ r) ] ⇔ (p ∨ q ∨ r)
iii. [(¬ p ∨ ¬ q) → (p ∧ q ∧ r) ] ⇔ p ∧ q
Compulsorily you have to use laws of logic to solve
Lakshmi R, Asst. Professor, Dept. Of ISE
¬ ∧ ∨ ⊻ → ↔

10. Solution i:
To prove that, p ∨ [p ∧ (p ∨ q) ] ⇔ p
p ∨ [p ∧ (p ∨ q) ]
⇔ p ∨ [ p ] ---- Absorption law
⇔ p ---- Idempotent Law
Lakshmi R, Asst. Professor, Dept. Of ISE
¬ ∧ ∨ ⊻ → ↔

11. Solution ii:
To prove that, [p ∨ q ∨ (¬ p ∧ ¬ q ∧ r) ] ⇔ (p ∨ q ∨ r)
[( p ∨ q) ∨ ( ¬ p ∧ ¬ q ∧ r) ]
⇔ [( p ∨ q ) ∨ ( ¬ ( p ∨ q ) ∧ r) ] ----- DeMorgan’s Law – eq(1)
Let P = ( p ∨ q ) an d Q = ¬ ( p ∨ q ) , eq ( 1 ) b ecomes
⇔ [P ∨ ( Q ∧ r ) ]
⇔ [ ( P ∨ Q ) ∧ ( P ∨ r) ] --- Distributive Law - eq(2)
Substituting P = ( p ∨ q ) an d Q = ¬ ( p ∨ q ) in eq ( 2 )
⇔ [ { ( p ∨ q ) ∨ ¬ ( p ∨ q ) } ∧ ( p ∨ q ∨ r) ]
⇔ [ T ∧ ( p ∨ q ∨ r) ] ---- Inverse Law
⇔ ( p ∨ q ∨ r) ---- Identity Law
Lakshmi R, Asst. Professor, Dept. Of ISE
5. Distributive
Laws
8. Inverse Laws
i. (p ∨ ¬p) ⇔ T
ii. (p ∧ ¬p) ⇔ F
i. p ∨ (q ∧ r) ⇔ (p ∨ q ) ∧ (p ∨ r )
ii. p ∧ (q ∨ r) ⇔ (p ∧ q ) ∨ (p ∧ r )
5. Distributive
Laws
7. Identity Laws
i. (p ∨ F) ⇔ p
ii. (p ∧ T) ⇔ p

12. Exercise
iii. [(¬ p ∨ ¬ q) → (p ∧ q ∧ r) ] ⇔ p ∧ q
iv. [(p ∨ q) ∧ (p ∨ ¬ q)] ∨ q ⇔ p ∨q
Lakshmi R, Asst. Professor, Dept. Of ISE

13. Prove that (p → q)∧ [(¬ q ∧ (r ∨ ¬ q)] ⇔ ¬ (q ∨ p)
Solution:
(p → q)∧ [(¬ q ∧ (r ∨ ¬ q)]
⇔ (p → q)∧ [(¬ q ∧ (¬ q ∨ r)] --- Commutative Law
⇔ (p → q)∧ ¬ q -------------------- Absorption Law
⇔ (¬p ∨ q) ∧ ¬ q ------------------ Fact p → q ⇔ ¬ p ∨ q
⇔ ¬ q ∧ (¬p ∨ q) ----------------- Commutative Law
⇔ (¬ q ∧ ¬p) ∨ (¬ q ∧ q) --------- Distributive Law
⇔ (¬ q ∧ ¬p) ∨ F ------------------ Inverse Law
⇔ (¬ q ∧ ¬p) ----------------------- Identity Law
⇔ ¬(q ∨ p) ------------------------- DeMorgan’s Law
Lakshmi R, Asst. Professor, Dept. Of ISE

14. Prove that [¬p ∧ (¬ q ∧ r ) ] ∨ (q ∧ r) ∨ ( p ∧ r) ⇔ r
Solution:
[¬p ∧ (¬ q ∧ r ) ] ∨ (q ∧ r) ∨ ( p ∧ r)
⇔ [¬p ∧ (¬ q ∧ r ) ] ∨ (r ∧ q) ∨ ( r ∧ p) ----- Commutative Law
⇔ [(¬p ∧ ¬ q ) ∧ r ] ∨ (r ∧ q) ∨ ( r ∧ p) ----- Associative Law
⇔ [¬(p ∨ q ) ∧ r ] ∨ (r ∧ q) ∨ ( r ∧ p) ------- DeMorgan’s Law
⇔ [¬(p ∨ q ) ∧ r ] ∨ r ∧ (q ∨ p) -------------- Distributive Law
⇔ [r ∧ ¬(p ∨ q ) ] ∨ [r ∧ (p ∨ q)] ------------- Commutative Law
Let A = ¬(p ∨ q ) and B = (p ∨ q)
⇔ [r ∧ A] ∨ [r ∧ B]
⇔ r ∧ [ A ∨ B ] ------------------------------------ Distributive Law
Substituting A = ¬(p ∨ q ) and B = (p ∨ q) in the above equation
⇔ r ∧ [¬(p ∨ q ) ∨ (p ∨ q) ] ⇔ r ∧ T --------------- Inverse Law
⇔ r -------------- Identity Law
Lakshmi R, Asst. Professor, Dept. Of ISE

15. Prove that
[ ( p ∨ q) ∧ ¬ {¬p ∧ (¬ q ∨ ¬r ) } ] ∨ (¬ p ∧ ¬q ) ∨ (¬ p ∧ ¬r )
is a tautology without using truth tables.
Solution:
To prove that
[ ( p ∨ q) ∧ ¬ {¬p ∧ (¬ q ∨ ¬r ) } ] ∨ (¬ p ∧ ¬q ) ∨ (¬ p ∧ ¬r ) = T
Lakshmi R, Asst. Professor, Dept. Of ISE

16. [ ( p ∨ q ) ∧ ¬ { ¬ p ∧ ( ¬ q ∨ ¬ r ) } ] ∨ ( ¬ p ∧ ¬ q ) ∨ ( ¬ p ∧ ¬ r )
⇔ [ ( p ∨ q ) ∧ ¬ { ¬ p ∧ ¬ ( q ∧ r ) } ] ∨ ( ¬ p ∧ ¬ q ) ∨ ( ¬ p ∧ ¬ r )
- - - - - DeMorgan’s Law
⇔ [ ( p ∨ q ) ∧ { ¬ ¬ p ∨ ¬ ¬ ( q ∧ r ) } ] ∨ ( ¬ p ∧ ¬ q ) ∨ ( ¬ p ∧ ¬ r )
- - - - - DeMorgan’s Law
⇔ [ ( p ∨ q ) ∧ { p ∨ ( q ∧ r ) } ] ∨ ( ¬ p ∧ ¬ q ) ∨ ( ¬ p ∧ ¬ r )
- - - - - Law of Double Negation
⇔ [ p ∨ ( q ∧ ( q ∧ r ) ) ] ∨ ( ¬ p ∧ ¬ q ) ∨ ( ¬ p ∧ ¬ r ) - - - - - - - - - - - - Distributive Law
⇔ [ p ∨ ( ( q ∧ q ) ∧ r ) ] ∨ ( ¬ p ∧ ¬ q ) ∨ ( ¬ p ∧ ¬ r ) - - - - - - - - - - - - - Associative Law
⇔ [ p ∨ ( q ∧ r ) ] ∨ ( ¬ p ∧ ¬ q ) ∨ ( ¬ p ∧ ¬ r ) - - - - - - - - - - - - - Idempotent Law
⇔ [ p ∨ ( q ∧ r ) ] ∨ ¬ ( p ∨ q ) ∨ ¬ ( p ∨ r ) - - - - - - - - - - - - - DeMorgan’s Law
⇔ [ p ∨ ( q ∧ r ) ] ∨ ¬ [ ( p ∨ q ) ∧ ( p ∨ r ) ] - - - - - - - - - - - - - DeMorgan’s Law
⇔ [ p ∨ ( q ∧ r ) ] ∨ ¬ [ ( p ∨ ( q ∧ r ) ] - - - - - - - - - - - - - Distributive Law
Let u = p ∨ ( q ∧ r )
⇔ [ p ∨ ( q ∧ r ) ] ∨ ¬ [ ( p ∨ ( q ∧ r ) ] ⇔ u ∨ ¬ u ⇔ T - - - - - - - - Inverse Law
Lakshmi R, Asst. Professor, Dept. Of ISE

17. Application to switching networks
• We can relate switches and their states with propositions and their truth
values.
• Value 0 ----> when switch is open
• Value 1 ----> when switch is closed
• Laws of logic – simplify the complex switching networks.
Lakshmi R, Asst. Professor, Dept. Of ISE

18. Application to switching networks
1. Simplify the switching network.
Lakshmi R, Asst. Professor, Dept. Of ISE
Solution:
We write the given network as:
[p ∧ (¬ r ∨ q ∨ ¬ q )] ∨ [(r ∨ t V ¬ r) ∧ ¬ q ]
r
t
¬ r
¬ r
q
¬q
p
T1 T2
¬q

19. Lakshmi R, Asst. Professor, Dept. Of ISE
[p ∧ (¬ r ∨ q ∨ ¬ q )] ∨ [(r ∨ t V ¬ r) ∧ ¬ q ]
⇔ [p ∧ (¬ r ∨ T)] ∨ [(t V T) ∧ ¬ q ] --- Inverse Law
⇔ [p ∧ (T)] ∨ [(T) ∧ ¬ q ] --- Domination Law
⇔ [p ∨ ¬ q ] --- Identity Law
p
¬q
T1 T2
The simplified network is

20. Application to switching networks
2. Simplify the switching network.
Lakshmi R, Asst. Professor, Dept. Of ISE
p
q
r
p
t
¬q
p
¬ t
r
T1 T2
Solution:
We write the given network as:
(p ∨ q ∨ r) ∧ (p ∨ t ∨ ¬ q ) ∧ (p ∨ ¬ t V r)

21. Lakshmi R, Asst. Professor, Dept. Of ISE
(p ∨ q ∨ r) ∧ (p ∨ t ∨ ¬ q ) ∧ (p ∨ ¬ t V r)
⇔ [p ∨ ((q ∨ r) ∧( t ∨ ¬ q) )] ∧ (p ∨ ¬ t V r)
Distributive Law
⇔ p ∨ [((q ∨ r) ∧( t ∨ ¬ q)) ∧ (¬ t V r) ]Distributive Law
⇔ p ∨ [((q ∨ r) ∧ (¬ t V r)) ∧ ( t ∨ ¬ q)]Associative law
⇔ p ∨ [((r ∨ q) ∧ (r V ¬ t)) ∧ ( t ∨ ¬ q)]Commutative law
⇔ p ∨ [r ∨ ( q ∧ ¬ t)] ∧ ( t ∨ ¬ q)] Distributive Law

22. Lakshmi R, Asst. Professor, Dept. Of ISE
⇔ p ∨ [r ∨ ( q ∧ ¬ t)] ∧ ( t ∨ ¬ q)]
⇔ p ∨ [( t ∨ ¬ q) ∧ [r ∨ ( q ∧ ¬ t)]] Commutative Law
⇔ p ∨ [[( t ∨ ¬ q) ∧ r] ∨ [( t ∨ ¬ q) ∧ ( q ∧ ¬ t)]] -Distributive Law
⇔ p ∨ [[( t ∨ ¬ q) ∧ r] ∨ [( t ∨ ¬ q) ∧ (¬ t ∧ q)]] -Commutative law
⇔ p ∨ [[( t ∨ ¬ q) ∧ r] ∨ [( t ∨ ¬ q) ∧ ¬ (t ∨ ¬ q)]] -DeMorgan’s Law
⇔ p ∨ [[( t ∨ ¬ q) ∧ r] ∨ [( t ∨ ¬ q) ∧ ¬ (t ∨ ¬ q)]]
⇔ p ∨ [( t ∨ ¬ q) ∧ r] ∨ F] -Inverse Law
⇔ p ∨ [( t ∨ ¬ q) ∧ r] -Identity Law
p
t
¬q
rT1 T2

23. Lakshmi R, Asst. Professor, Dept. Of ISE
p
q
r
p
t
¬q
p
¬ t
r
T1 T2
Simplified circuit is
p
t
¬q
rT1 T2
Given circuit is

24. Exercise
Prove the following logical equivalences without using truth tables.
i. [(¬ p ∨ ¬ q) → (p ∧ q ∧ r) ] ⇔ p ∧ q
ii. [(p ∨ q) ∧ (p ∨ ¬ q)] ∨ q ⇔ p ∨q
iii. p → (q → r) ⇔ (p ∧ q) → r
iv. ¬ [{(p ∨ q) ∧ r} → ¬q] ⇔ ¬ (¬ {(p ∨ q) ∧r} ∨ ¬q) ⇔ q ∧ r

25. Exercise
¬ p
p
q
r r
T1 T2
¬q
r
Simplify the following switching networks
¬ p ¬q r
p r
q rT1 T2
2
1

26. Converse, Inverse, and Contrapositive
Consider a conditional p → q. then,
i. q →p is called the converse of p → q
ii. ¬p → ¬q is called the inverse of p → q
iii. ¬q → ¬p is called the contrapositive of p → q
Lakshmi R, Asst. Professor, Dept. Of ISE
p q ¬ p ¬ q p → q q → p ¬ p → ¬ q ¬ q → ¬ p
0 0 1 1 1 1 1 1
0 1 1 0 1 0 0 1
1 0 0 1 0 1 1 0
1 1 0 0 1 1 1 1
p → q q → p ¬ p → ¬ q ¬ q → ¬ p
1 1 1 1
1 0 0 1
0 1 1 0
1 1 1 1
p → q ⇔ ¬q→¬p
q → p ⇔ ¬p→ ¬q
Conditional ⇔ Contrapositive
Converse ⇔ Inverse

27. Converse, Inverse, and Contrapositive cont.,
State converse, inverse, and contrapositive of the following implication.
“If a triangle is not isosceles, then it is not equilateral”
Solution: The given implication p → q is If a triangle is not isosceles, then it is
not equilateral
Let p: Triangle is not isosceles
q: Triangle is not equilateral
Converse: q →p : If a triangle is not equilateral, then it is not isosceles
Inverse: ¬p → ¬q : If a triangle is isosceles, then it is equilateral.
Contrapositive: ¬q → ¬p: If a triangle is equilateral, then it is isosceles.
Lakshmi R, Asst. Professor, Dept. Of ISE