Advertisements
Advertisements
Question
Rewrite the following statement without using if ...... then.
It f(2) = 0 then f(x) is divisible by (x – 2).
Advertisements
Solution
Since p → q ≡ ∼ p ∨ q, the given statement can be written as:
f(2) `≠` 0 or f(x) is divisible by (x – 2).
APPEARS IN
RELATED QUESTIONS
Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p
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.
Without using truth table prove that:
(p ∨ q) ∧ (p ∨ ∼ q) ≡ p
Using rules in logic, prove the following:
∼p ∧ q ≡ (p ∨ 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) ∧ r
Using the rules in logic, write the negation of the following:
(∼p ∧ q) ∨ (p ∧ ∼q)
Let p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). Then, this law is known as ______.
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 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 ______
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
Negation of the Boolean expression `p Leftrightarrow (q \implies p)` is ______.
∼ ((∼ p) ∧ q) is equal to ______.
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.
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
Show that the simplified form of (p ∧ q ∧ ∼ r) ∨ (r ∧ p ∧ q) ∨ (∼ p ∨ q) is q ∨ ∼ p.
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
