Home
Class 12
MATHS
STATEMENT-1 : ~(phArrq)=~phArrq=phArr~q ...

STATEMENT-1 : `~(phArrq)=~phArrq=phArr~q`
and
STATEMENT-2 : `(phArrq)hArrr=p hArr(q hArrr)`

A

Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1

B

Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1

C

Statement-1 is True, Statement-2 is False

D

Statement-1 is False, Statement-2 is True

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • MATHEMATICAL REASONING

    AAKASH INSTITUTE|Exercise Assignment (SECTION-C) (Linked Comprehension Type Questions) (Comprehension-I)|3 Videos

  • LIMITS AND DERIVATIVES

    AAKASH INSTITUTE|Exercise Section - j|3 Videos

  • MATRICES

    AAKASH INSTITUTE|Exercise Assignment (Section - J) Aakash Challengers Questions|3 Videos

Similar Questions

Explore conceptually related problems

STATEMENT-1 : (phArrq)=~ p hArr q and STATEMENT-2 : (phArrq) =~p hArr ~q

STATEMENT-1 : ~(prArrq)=p^^~q and STATEMENT-2 : p rArr q = ~p vvq

STATEMENT-1 : (p^^~q)^^(~p^^q) and STATEMENT-2 : (pvv~q) vv (~p ^^q)

STATEMENT-1 : (p rArr q) hArr (~q rArr~p) and STATEMENT-2 : p rArr q is logically equivalent to ~q rArr ~p

If p, q, r be anY three. statements STATEMENT-1 : pvv(q^^r)hArr(pvvq)^^(pvvr) and STATEMENT-2 : pvv(q^^r)=(pvvq)^^(pvvr)

Let p, q and r be three statements. Consider two compound statements S_(1):(prArr q)rArr r -= p rArr (q rArr r) S_(2):(p hArr q)hArrr -=phArr (q hArr r) State in order, whether S_(1), S_(2) are true of false. (where, T represents true F represents false)

STATEMENT-1 : p hArr ~q is true, when pis false and q is true. and STATEMENT-2 : p rArr ~q is true, when pis true and q is false, and -q ⇒ p is true, when q is false and p is true.

STATEMENT-1 : The inverse of (p^^~q) rArrr " is "~pvvqrArr ~r and STATEMENT-2 : ~(p^^q)=~pvv~q

STATEMENT-1 : The converse of p rArr q " is " q rArr p and STATEMENT-2 : The inverse of p rArr q " is " ~p rArr ~q

STATEMENT-1 : (p rArr ~q) ^^ (~q rArr q) is a contradication and STATEMENT-2 : The inverse of p rArr ~p " is " ~p rArr p