Discrete Mathematical Structures Fundamentals of Logic Principle of Duality

1. PRINCIPLE OF DUALITY
By,
Lakshmi R
Asst. professor,
Dept. of ISE

2. PRINCIPLE OF DUALITY
Duality:
Let s be a statement. If S contains no logical connectives other than ∧ and ∨ ,
then dual of s is denoted sd, is the statement obtained from s by replacing each
occurrence of ∧ and ∨ by ∨ and ∧, respectively, and each occurrence of T and F
by F and T,respectively.
Lakshmi R, Asst. Professor, Dept. Of ISE
(p ∧ q) ∧ (r ∨ s ∧ T) ∨ F
(p q) (r s F) T∨ ∨ ∨ ∧∧
s =
sd =

3. PRINCIPLE OF DUALITY
Let s and t be statements that contain no logical
connectives other than ∧ and ∨. If s ⇔ t, then sd ⇔ td
Lakshmi R, Asst. Professor, Dept. Of ISE

4. Write duals for the following propositions.
1. ( p → q) → r
2. p → ( q → r)
Solution:
Lakshmi R, Asst. Professor, Dept. Of ISE
1.( ( p → q) → r)d
⇔ ( (¬ p ∨ q) → r)d
⇔ (¬ (¬ p ∨ q) ∨ r)d
⇔ ((p ∧ ¬ q ) ∨ r)d
= (p ∨ ¬ q) ∧ r
2.( p → (q → r))d
⇔ (p → (¬ q ∨ r) )d
⇔ (¬ p ∨ (¬ q ∨ r) )d
= ¬ p ∧ (¬ q ∧ r)
Steps:
1. Write the given proposition in
terms of logical ∧ and ∨ by
using known equivalences
2. Replace all ∧ and ∨ by ∨
and ∧
1 . ( ( p → q ) → r ) d = ( p ∨ ¬ q ) ∧ r
2 . ( p → ( q → r ) ) d = ¬ p ∧ ( ¬ q ∧ r )

5. 2.Verify the principle of duality for the following equivalence
1. [ (¬ (p ∧ q) → ¬ p ) ∨ (¬ p ∨ q ) ] ⇔ (¬ p ∨ q )
Solution:
Let s = [ (¬ (p ∧ q) → ¬ p ) ∨ (¬ p ∨ q ) ]
and
t = (¬ p ∨ q )
We need to prove that
sd ⇔ td
Lakshmi R, Asst. Professor, Dept. Of ISE

6. s = [ (¬ (p ∧ q) → ¬ p ) ∨ (¬ p ∨ q ) ]
s = [ (¬ ¬ (p ∧ q) ∨ ¬ p ) ∨ (¬ p ∨ q ) ]
s = [ ((p ∧ q) ∨ ¬ p ) ∨ (¬ p ∨ q ) ]
s = [ (p ∧ q) ∨ ¬ p ∨ ¬ p ∨ q ]
sd = [ (p ∨ q) ∧ ¬ p ∧ ¬ p ∧ q ]
⇔ [(p ∨ q) ∧ ¬ p ∧ q ] ----- Idempotent Law
⇔ [q ∧ (q ∨ p) ∧ ¬ p ] ----- Associative Law
⇔ [q ∧ ¬ p ] ----- Absorption Law
⇔ [¬ p ∧ q] ----- Commutative Law
⇔ td ----- By eq(1)
Lakshmi R, Asst. Professor, Dept. Of ISE
t = (¬ p ∨ q )
td = (¬ p ∧ q ) ----- eq(1)