Advertisements
Advertisements
Question
Using the algebra of statement, prove that (p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q).
Advertisements
Solution
L.H.S. = (p ∨ q) ∧ (~ p ∨ ~ q)
≡ [(p ∨ q)] ∧ [(p ∨ q) ∧ ~ q] ...[Distributive law]
≡ [(p ∧ ~ p) ∨ (q ∧ ~ p)] ∨ [(p ∧ ~ q) ∨ (q ∧ ~ q)] ...[Distributive law]
≡ [F ∨ (q ∧ ~ p)] ∨ [(p ∧ ~ q) ∨ F] ...[Complement law]
≡ (q ∧ ~ p) ∨ (p ∧ ~ q) ...[Identity law]
≡ (p ∧ ~ q) ∨ (~ p ∧ q) ...[Commutative law]
≡ R.H.S.
Notes
The question is modified.
APPEARS IN
RELATED QUESTIONS
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)
Write the Truth Value of the Negation of the Following Statement :
The Sun sets in the East.
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 f(2) = 0 then f(x) is divisible by (x – 2).
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 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)
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
~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p
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 ______
(p ∧ ∼q) ∧ (∼p ∧ q) is a ______.
The negation of the Boolean expression (r ∧ ∼s) ∨ s is equivalent to: ______
Which of the following is not a statement?
Negation of the Boolean expression `p Leftrightarrow (q \implies p)` is ______.
Without using truth table, prove that:
[p ∧ (q ∨ r)] ∨ [∼r ∧ ∼q ∧ p] ≡ p
Without using truth table, prove that : [(p ∨ q) ∧ ∼p] →q is a tautology.
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.
