Advertisements
Advertisements
Question
Using rules in logic, prove the following:
p ↔ q ≡ ∼(p ∧ ∼q) ∧ ∼(q ∧ ∼p)
Advertisements
Solution
(p ↔ q) ≡ (p → q) ∧ (q → p)
p ↔ q ≡ (∼ p ∨ q) ∧ (∼ q ∨ p)] .......(Conditional Law)
p ↔ q ≡ ∼ (p ∧ ∼ q) ∧ ∼ (q ∧ ∼p)
APPEARS IN
RELATED QUESTIONS
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 :
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.
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)
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
~ [(p ∧ q) → ~ q] ≡ p ∧ q
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 ______
(p → q) ∨ p is logically equivalent to ______
The logically equivalent statement of (p ∨ q) ∧ (p ∨ r) is ______
The statement pattern p ∧ (∼p ∧ q) is ______.
The statement pattern [∼r ∧ (p ∨ q) ∧ (p ∨ q) ∧ (∼p ∧ q)] is equivalent to ______
The negation of the Boolean expression (r ∧ ∼s) ∨ s is equivalent to: ______
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
If p ∨ q is true, then the truth value of ∼ p ∧ ∼ q is ______.
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 ______.
The statement p → (q → p) is equivalent to ______.
