Advertisements
Advertisements
Question
Using the algebra of statement, prove that
[p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p] ≡ p
Advertisements
Solution
L.H.S.
= [p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p]
≡ [p ∧ (q ∨ r)] ∨ [(~ r ∧ ~ q)∧ p] ...[Associative Law]
≡ [p ∧ (q ∨ r)] ∨ [(~q ∧ ~r) ∧ p] ....[Commutative Law]
≡ [p ∧ (q ∨ r)] ∨ [~ (q ∨ r) ∧ p] ....[De Morgan’s Law]
≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] .....[Commutative Law]
≡ p ∧ [(q ∨ r) ∨ ~(q ∨ r)] ....[Distributive Law]
≡ p ∧ t ......[Complement Law]
≡ p .....[Identity Law]
= R.H.S.
APPEARS IN
RELATED QUESTIONS
The negation of p ∧ (q → r) is ______________.
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
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.
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) ∧ ∼(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
Using the rules in logic, write the negation of the following:
(∼p ∧ q) ∨ (p ∧ ∼q)
The statement pattern p ∧ ( q v ~ p) is equivalent to ______.
(p → q) ∨ p is logically equivalent to ______
The logically equivalent statement of (p ∨ q) ∧ (p ∨ r) is ______
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
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.
