Advertisements
Advertisements
Question
Without using truth table, show that
p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Advertisements
Solution
L.H.S.
≡ p ↔ q
≡ (p → q) ∧ (q → p)
≡ (~p ∨ q) ∧ (~q ∨ p)
≡ [~ p ∧ (~ q ∨ p)] ∨ [q ∧ (~ q ∨ p)] ....[Distributive law]
≡ [(~ p ∧ ~ q) ∨ (~ p ∧ p)] ∨ [(q ∧ ~ q) ∨ (q ∧ p)] .....[Distributive Law]
≡ [(~ p ∧ ~ q) ∨ F] ∨ [F ∨ (q ∧ p)] ....[Complement Law]
≡ (~ p ∧ ~ q) ∨ (q ∧ p) ....[Identity Law]
≡ (p ∧ q) ∨ (~ p ∧ ~ q) ....[Commutative Law]
≡ R.H.S.
APPEARS IN
RELATED QUESTIONS
The negation of p ∧ (q → r) is ______________.
If A = {2, 3, 4, 5, 6}, then which of the following is not true?
(A) ∃ x ∈ A such that x + 3 = 8
(B) ∃ x ∈ A such that x + 2 < 5
(C) ∃ x ∈ A such that x + 2 < 9
(D) ∀ x ∈ A such that x + 6 ≥ 9
Using the rules of negation, write the negatlon of the following:
(a) p ∧ (q → r)
(b) ~P ∨ ~q
Rewrite the following statement without using if ...... then.
If a man is a judge then he is honest.
Rewrite the following statement without using if ...... then.
It 2 is a rational number then `sqrt2` is irrational number.
Without using truth table prove that:
∼ [(p ∨ ∼ q) → (p ∧ ∼ q)] ≡ (p ∨ ∼ q) ∧ (∼ p ∨ q)
Using rules in logic, prove the following:
∼p ∧ q ≡ (p ∨ q) ∧ ∼p
Using rules in logic, prove the following:
∼ (p ∨ q) ∨ (∼p ∧ q) ≡ ∼p
Using the rules in logic, write the negation of the following:
(p → q) ∧ r
Let p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). Then, this law is known as ______.
Without using truth table, show that
(p ∨ q) → r ≡ (p → r) ∧ (q → r)
Using the algebra of statement, prove that
[p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p] ≡ p
Using the algebra of statement, prove that (p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q).
The statement pattern p ∧ ( q v ~ p) is equivalent to ______.
(p → q) ∨ p is logically equivalent to ______
The negation of p → (~p ∨ q) is ______
The statement pattern [∼r ∧ (p ∨ q) ∧ (p ∨ q) ∧ (∼p ∧ q)] is equivalent to ______
The logical statement [∼(q ∨ ∼r) ∨ (p ∧ r)] ∧ (q ∨ p) is equivalent to: ______
Without using truth table prove that (p ∧ q) ∨ (∼ p ∧ q) v (p∧ ∼ q) ≡ p ∨ q
Which of the following is not a statement?
Show that the simplified form of (p ∧ q ∧ ∼ r) ∨ (r ∧ p ∧ q) ∨ (∼ p ∨ q) is q ∨ ∼ p.
