Advertisements
Advertisements
प्रश्न
Let p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). Then, this law is known as ______.
पर्याय
Commutative law
Associative law
De-Morgan's law
Distributive law
Advertisements
उत्तर
Let p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). Then, this law is known as distributive law.
APPEARS IN
संबंधित प्रश्न
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 2 is a rational number then `sqrt2` is irrational number.
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
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 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
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
Without using truth table, show that
(p ∨ q) → r ≡ (p → r) ∧ (q → r)
Using the algebra of statement, prove that (p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q).
(p → q) ∨ p is logically 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: ______
If p ∨ q is true, then the truth value of ∼ p ∧ ∼ q 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.
The simplified form of [(~ p v q) ∧ r] v [(p ∧ ~ q) ∧ r] is ______.
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
