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) ∧ ~ q] ...[Distributive law]
≡ [(p ∧ ~ p) ∨ (q ∧ ~ p)] ∨ [(p ∧ ~ q) ∨ (q ∧ ~ q)] ...[Distributive law]
≡ [F ∨ (q ∧ ~ p)] ∨ [(p ∧ ~ q) ∨ F] ...[Complement law]
≡ (q ∧ ~ p) ∨ (p ∧ ~ q) ...[Identity law]
≡ (p ∧ ~ q) ∨ (~ p ∧ q) ...[Commutative law]
≡ R.H.S.
Notes
The question is modified.
APPEARS IN
RELATED QUESTIONS
Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p
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
Write the Truth Value of the Negation of the Following Statement :
The Sun sets in the East.
Write the truth value of the negation of the following statement :
cos2 θ + sin2 θ = 1, for all θ ∈ R
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
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) ∧ (q ∨ ∼r)
Using the rules in logic, write the negation of the following:
p ∧ (q ∨ r)
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)
The statement pattern p ∧ ( q v ~ p) is equivalent to ______.
The negation of p → (~p ∨ q) is ______
(p ∧ ∼q) ∧ (∼p ∧ q) is a ______.
If p ∨ q is true, then the truth value of ∼ p ∧ ∼ q is ______.
Which of the following is not a statement?
Without using truth table, prove that:
[p ∧ (q ∨ r)] ∨ [∼r ∧ ∼q ∧ p] ≡ p
Without using truth table, prove that : [(p ∨ q) ∧ ∼p] →q is a tautology.
The simplified form of [(~ p v q) ∧ r] v [(p ∧ ~ q) ∧ r] is ______.
Without using truth table prove that
[(p ∧ q ∧ ∼ p) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) ≡ (p ∨ q) ∧ r
Show that the simplified form of (p ∧ q ∧ ∼ r) ∨ (r ∧ p ∧ q) ∨ (∼ p ∨ q) is q ∨ ∼ p.
