Advertisements
Advertisements
Question
Without using truth table, show that
~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p
Advertisements
Solution
L.H.S.
≡ ~r → ~ (p ∧ q)
≡ ~(~ r) ∨ ~ (p ∧ q) ....[p → q ≡ ~ p ∨ q]
≡ r ∨ ~(p ∧ q) ....[Negation of negation]
≡ r ∨ (~p ∨ ~q) ....[De Morgan’s law]
≡ ~p ∨ (~q ∨ r) .....[Commutative and associative law]
≡ ~p ∨ (q → r) ....[p → q ≡ ~ p ∨ q]
≡ (q → r) ∨ ~p ......[Commutative law]
≡ ~[~ (q → r)] ∨ ~ p ......[Negation of negation]
≡ [~ (q → r)] → ~ p .....[p → q ≡ ~ p ∨ q]
= R.H.S.
APPEARS IN
RELATED QUESTIONS
The negation of p ∧ (q → r) is ______________.
Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p
Without using the truth table show that P ↔ q ≡ (p ∧ q) ∨ (~ p ∧ ~ q)
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.
Using rules in logic, prove the following:
p ↔ q ≡ ∼(p ∧ ∼q) ∧ ∼(q ∧ ∼p)
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) ∨ (p ∧ ∼q)
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 ∧ [(~ p ∨ q) ∨ ~ q] ≡ p
Without using truth table, show that
~ [(p ∧ q) → ~ q] ≡ p ∧ q
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 ______.
For any two statements p and q, the negation of the expression (p ∧ ∼q) ∧ ∼p is ______
(p → q) ∨ p is logically equivalent to ______
The logically equivalent statement of (p ∨ q) ∧ (p ∨ r) is ______
The statement pattern [∼r ∧ (p ∨ q) ∧ (p ∨ q) ∧ (∼p ∧ q)] is equivalent to ______
(p ∧ ∼q) ∧ (∼p ∧ q) is a ______.
Which of the following is not a statement?
Without using truth table, prove that:
[p ∧ (q ∨ r)] ∨ [∼r ∧ ∼q ∧ p] ≡ p
The statement p → (q → p) is equivalent to ______.
