Advertisements
Advertisements
Question
Without using truth table, show that
~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p
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
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.
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
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)
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 ∧ 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)
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 ______
The logically equivalent statement of (p ∨ q) ∧ (p ∨ r) is ______
The negation of p → (~p ∨ q) 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?
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] →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
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.
