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UPSC 2015 Maths Optional Paper 2 Q5c — Step-by-Step Solution

10 marks · Section B

Boolean algebra · Numerical Analysis · asked 15× in 13 yrs · Read the full method →

Question

Find the principal (or canonical) disjunctive normal form in three variables p,q,rp,q,r for the Boolean expression ((pq)r)((pq)¬r)((p\wedge q)\to r)\vee((p\wedge q)\to\neg r). Is the given Boolean expression a contradiction or a tautology?

Technique

Reduce (Ar)(A¬r)=¬Ar¬r=True(A\to r)\vee(A\to\neg r)=\neg A\vee r\vee\neg r=\text{True} symbolically; principal DNF of a tautology is the disjunction of all minterms.

Solution

Recall. ab¬aba\to b\equiv\neg a\vee b. The expression has 3 variables p,q,rp,q,r giving 23=82^3=8 rows.

Step 1 — Simplify symbolically

Let A=pqA=p\wedge q. Then expression =(Ar)(A¬r)=(¬Ar)(¬A¬r)=¬Ar¬r=¬ATrue=True=(A\to r)\vee(A\to\neg r)=(\neg A\vee r)\vee(\neg A\vee\neg r)=\neg A\vee r\vee\neg r=\neg A\vee\text{True}=\text{True}.

So the expression is a tautology (true for all 8 valuations).

Step 2 — Principal DNF

A tautology’s principal DNF (canonical sum-of-products with all variables) is the disjunction of all 23=82^3=8 minterms.

Minterms (one per row of the truth table):

ppqqrrMinterm
000¬p¬q¬r\neg p\wedge\neg q\wedge\neg r
001¬p¬qr\neg p\wedge\neg q\wedge r
010¬pq¬r\neg p\wedge q\wedge\neg r
011¬pqr\neg p\wedge q\wedge r
100p¬q¬rp\wedge\neg q\wedge\neg r
101p¬qrp\wedge\neg q\wedge r
110pq¬rp\wedge q\wedge\neg r
111pqrp\wedge q\wedge r

Step 3 — Principal DNF (full sum)

Answer

  (all 8 minterms)=(¬p¬q¬r)(¬p¬qr)(¬pq¬r)(¬pqr)(p¬q¬r)(p¬qr)(pq¬r)(pqr).  \boxed{\;\bigvee\text{(all 8 minterms)}=(\neg p\wedge\neg q\wedge\neg r)\vee(\neg p\wedge\neg q\wedge r)\vee(\neg p\wedge q\wedge\neg r)\vee(\neg p\wedge q\wedge r)\vee(p\wedge\neg q\wedge\neg r)\vee(p\wedge\neg q\wedge r)\vee(p\wedge q\wedge\neg r)\vee(p\wedge q\wedge r).\;}
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