Advertisements
Advertisements
प्रश्न
Using the rules of negation, write the negatlon of the following:
(a) p ∧ (q → r)
(b) ~P ∨ ~q
Advertisements
उत्तर
(a) p ∧ (q → r)
~[p ∧ ~ (q → r)]
≡ ~ p ∨ ~ (q → r )
≡ ~ p ∨ ~ (q ∧ ~ r )
(b) ~P ∨ ~q
~[~p ∨ ~q]
≡ ~(~p) ∧ ~(~q)
≡ P ∧ q
APPEARS IN
संबंधित प्रश्न
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
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 f(2) = 0 then f(x) is divisible by (x – 2).
Without using truth table prove that:
(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)
Using the rules in logic, write the negation of the following:
(∼p ∧ q) ∨ (p ∧ ∼q)
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
~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p
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)
(p → q) ∨ p is logically equivalent to ______
The logically equivalent statement of (p ∨ q) ∧ (p ∨ r) is ______
The negation of p → (~p ∨ q) is ______
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
Negation of the Boolean expression `p Leftrightarrow (q \implies p)` is ______.
The statement p → (q → p) is equivalent to ______.
