Advertisements
Advertisements
प्रश्न
Without using truth table, show that
~ [(p ∧ q) → ~ q] ≡ p ∧ q
Advertisements
उत्तर
L.H.S.
≡ ~ [(p ∧ q) → ~ q]
≡ (p ∧ q) ∧ ~ (~ q) ....[Negation of implication]
≡ (p ∧ q) ∧ q .....[Negation of a negation]
≡ p ∧ (q ∧ q) ....[Associative law]
≡ p ∧ q .....[Idempotent law]
≡ R.H.S.
APPEARS IN
संबंधित प्रश्न
The negation of p ∧ (q → r) is ______________.
Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p
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.
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) ∧ ∼(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) ∧ 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
~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p
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)
For any two statements p and q, the negation of the expression (p ∧ ∼q) ∧ ∼p is ______
(p ∧ ∼q) ∧ (∼p ∧ q) is a ______.
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 ______.
Which of the following is not a statement?
Negation of the Boolean expression `p Leftrightarrow (q \implies p)` is ______.
∼ ((∼ p) ∧ q) is equal to ______.
Without using truth table prove that
[(p ∧ q ∧ ∼ p) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) ≡ (p ∨ q) ∧ r
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.
The logically equivalent statement of \[\left(\sim p\wedge q\right)\vee\left(\sim p\wedge\sim q\right)\] \[\vee\left(\ p\wedge\sim q\right)\] is
