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
Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p
Rewrite the following statement without using if ...... then.
It 2 is a rational number then `sqrt2` is irrational number.
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) ∧ (q ∨ ∼r)
Using the rules in logic, write the negation of the following:
(p → q) ∧ r
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
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 statement pattern p ∧ (∼p ∧ q) is ______.
(p ∧ ∼q) ∧ (∼p ∧ q) is a ______.
The negation of the Boolean expression (r ∧ ∼s) ∨ s is equivalent to: ______
The logical statement [∼(q ∨ ∼r) ∨ (p ∧ r)] ∧ (q ∨ p) is equivalent to: ______
If p ∨ q is true, then the truth value of ∼ p ∧ ∼ q is ______.
Without using truth table, prove that:
[p ∧ (q ∨ r)] ∨ [∼r ∧ ∼q ∧ p] ≡ p
