Advertisements
Advertisements
Question
Using the algebra of statement, prove that
(p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q) ≡ (p ∨ ~ q)
Advertisements
Solution
L.H.S.
= (p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q)
≡ (p ∧ q) ∨ [(p ∧ ~ q) ∨ (~ p ∧ ~ q)] ....[Associative Law]
≡ (p ∧ q) ∨ [(~q ∧ p) ∨ (~ q ∧ ~ p)] ....[Commutative Law]
≡ (p ∧ q) ∨ [~q ∧ (p ∨ ~ p)] ....[Distributive Law]
≡ (p ∧ q) ∨ (~q ∧ t) .....[Complement Law]
≡ (p ∧ q) ∨ (~q) .....[Identity Law]
≡ (p ∨ ~ q) ∧ (q ∨ ~q) .....[Distributive Law]
≡ (p ∨ ~ q) ∧ t ....[Complement Law]
≡ p ∨ ~ q .....[Identity 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.
Rewrite the following statement without using if ...... then.
It f(2) = 0 then f(x) is divisible by (x – 2).
Without using truth table prove that:
(p ∨ q) ∧ (p ∨ ∼ q) ≡ p
Without using truth table prove that:
(p ∧ q) ∨ (∼ p ∧ q) ∨ (p ∧ ∼ q) ≡ p ∨ q
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 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 ≡ (p ∧ q) ∨ (~p ∧ ~q)
Without using truth table, show that
~ [(p ∧ q) → ~ q] ≡ p ∧ q
Without using truth table, show that
(p ∨ q) → r ≡ (p → r) ∧ (q → r)
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 ______
(p ∧ ∼q) ∧ (∼p ∧ q) is a ______.
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
∼ ((∼ p) ∧ q) is equal to ______.
Without using truth table, prove that:
[p ∧ (q ∨ r)] ∨ [∼r ∧ ∼q ∧ p] ≡ p
The statement p → (q → p) is equivalent to ______.
Show that the simplified form of (p ∧ q ∧ ∼ r) ∨ (r ∧ p ∧ q) ∨ (∼ p ∨ q) is q ∨ ∼ p.
