The negation of (p q) q is

Correct Answer is : (¬p∨q)∨¬q( p q) q(¬p∨q)∨¬q

Correct Answer is : (¬p∨q)∨¬q( p q) q(¬p∨q)∨¬q

Test Index

KEAM 2011 Math Paper

Question : 112 of 120

Marks: +1, -0

(p \lor \lnot q) \land q

Solution:

X = (p \lor \lnot q) \land q

p	q	$\lnot p$	$\lnot q$	$p \lor \lnot q$	X	$(\lnot p \lor q)$	A	$(p \land \lnot q)$	B	$(\lnot p \land q)$	C	D	$\lnot p \lor \lnot q$	E	$\lnot X$
F	F	T	T	T	F	T	T	F	F	F	T	T	T	T	T
F	T	T	F	F	F	T	F	F	T	T	T	F	T	F	T
T	F	F	T	T	F	F	F	F	F	F	T	T	T	T	T
T	T	F	F	T	T	T	F	F	T	F	F	F	F	F	F

Where

A = (\lnot p \lor q) \land \lnot q

B = (p \land \lnot q) \lor q

C = (\lnot p \land q) \lor \lnot q

D = (p \land \lnot q) \lor \lnot q

E = (\lnot p \lor \lnot q) \land \lnot q

Here,

\lnot X \equiv C

i.e.,

\lnot[(p \lor \lnot q) \land q] \equiv (\lnot p \land q) \lor \lnot q

Alternate Method

\lnot[(p \lor \lnot q) \land q]

= \lnot(p \lor \lnot q) \lor \lnot q

( ∵ By D'e Morgan Law)

= (\lnot p \land \lnot(\lnot q)) \lor \lnot q

( ∵ By D'e Morgan Law)

= (\lnot p \land q) \lor \lnot q

( ∵ By Negation)

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p	q	$\lnot p$	$\lnot q$	$p \lor \lnot q$	X	$(\lnot p \lor q)$	A	$(p \land \lnot q)$	B	$(\lnot p \land q)$	C	D	$\lnot p \lor \lnot q$	E	$\lnot X$
F	F	T	T	T	F	T	T	F	F	F	T	T	T	T	T
F	T	T	F	F	F	T	F	F	T	T	T	F	T	F	T
T	F	F	T	T	F	F	F	F	F	F	T	T	T	T	T
T	T	F	F	T	T	T	F	F	T	F	F	F	F	F	F

p	q	$\lnot p$	$\lnot q$	$p \lor \lnot q$	X	$(\lnot p \lor q)$	A	$(p \land \lnot q)$	B	$(\lnot p \land q)$	C	D	$\lnot p \lor \lnot q$	E	$\lnot X$
F	F	T	T	T	F	T	T	F	F	F	T	T	T	T	T
F	T	T	F	F	F	T	F	F	T	T	T	F	T	F	T
T	F	F	T	T	F	F	F	F	F	F	T	T	T	T	T
T	T	F	F	T	T	T	F	F	T	F	F	F	F	F	F

p	q	$\lnot p$	$\lnot q$	$p \lor \lnot q$	X	$(\lnot p \lor q)$	A	$(p \land \lnot q)$	B	$(\lnot p \land q)$	C	D	$\lnot p \lor \lnot q$	E	$\lnot X$
F	F	T	T	T	F	T	T	F	F	F	T	T	T	T	T
F	T	T	F	F	F	T	F	F	T	T	T	F	T	F	T
T	F	F	T	T	F	F	F	F	F	F	T	T	T	T	T
T	T	F	F	T	T	T	F	F	T	F	F	F	F	F	F