Advertisements
Advertisements
Question
The negation of p ∧ (q → r) is ______________.
Options
p ∨ ( ~q ∨ r )
~p ∧ ( q → r )
~p ∧ ( ~q → ~r )
~p ∨ ( q ∧ ~r )
Advertisements
Solution
~ [P ∧ (q → r)
=~[( P)] ∨ [~ (q → r)] ...(By De Morgan's law)
=~[( P)] ∨ [~ (~q ∨ r )] ...(By Conditional Law)
=~[( P)] ∨ [( q ∧ ~r )] ...(By De Morgan's law)
~ [ P ∧ ( q → r )] = ~ P ∨ ( q ∧ ~r )
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)
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 ∧ ∼ q) ≡ p ∨ q
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 ∨ 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 ∧ [(~ p ∨ q) ∨ ~ q] ≡ p
Without using truth table, show that
~ [(p ∧ q) → ~ q] ≡ p ∧ q
Without using truth table, show that
~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p
Without using truth table, show that
(p ∨ q) → r ≡ (p → r) ∧ (q → r)
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)
The statement pattern p ∧ ( q v ~ p) is equivalent to ______.
(p → q) ∨ p is logically equivalent to ______
The statement pattern p ∧ (∼p ∧ q) is ______.
The statement pattern [∼r ∧ (p ∨ q) ∧ (p ∨ q) ∧ (∼p ∧ q)] is equivalent to ______
(p ∧ ∼q) ∧ (∼p ∧ q) is a ______.
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 ______.
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 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
