UPSC 2021 Maths Optional Paper 2 Q5c-ii — Step-by-Step Solution

5 marks · Section B

Boolean algebra · Numerical Analysis · asked 15× in 14 yrs · Read the full method →

Question

Obtain the principal conjunctive normal form of $(\neg P\to R)\wedge(Q\leftrightarrow P)$ .

Truth table; conjunction of maxterms at rows where expression = 0.

$\neg P\to R\equiv P\vee R$ (since $\neg(\neg P)\vee R=P\vee R$ ).

$Q\leftrightarrow P\equiv(Q\wedge P)\vee(\neg Q\wedge\neg P)$ .

So expression $=(P\vee R)\wedge((Q\wedge P)\vee(\neg Q\wedge\neg P))$ .

Truth table:

$P$	$Q$	$R$	$P\vee R$	$Q\leftrightarrow P$	Expression
0	0	0	0	1	0
0	0	1	1	1	1
0	1	0	0	0	0
0	1	1	1	0	0
1	0	0	1	0	0
1	0	1	1	0	0
1	1	0	1	1	1
1	1	1	1	1	1

Rows: $(P,Q,R)\in\{(0,0,0),(0,1,0),(0,1,1),(1,0,0),(1,0,1)\}$ .

For row $(P,Q,R)$ , the maxterm is the disjunction of literals, with $0\to$ unprimed and $1\to$ primed.

$(0,0,0)$ : $P\vee Q\vee R$ .
$(0,1,0)$ : $P\vee\neg Q\vee R$ .
$(0,1,1)$ : $P\vee\neg Q\vee\neg R$ .
$(1,0,0)$ : $\neg P\vee Q\vee R$ .
$(1,0,1)$ : $\neg P\vee Q\vee\neg R$ .

\boxed{\;(P\vee Q\vee R)\wedge(P\vee\neg Q\vee R)\wedge(P\vee\neg Q\vee\neg R)\wedge(\neg P\vee Q\vee R)\wedge(\neg P\vee Q\vee\neg R).\;}