← 2015 Paper 2
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,r for the Boolean expression ((p∧q)→r)∨((p∧q)→¬r). Is the given Boolean expression a contradiction or a tautology?
Technique
Reduce (A→r)∨(A→¬r)=¬A∨r∨¬r=True symbolically; principal DNF of a tautology is the disjunction of all minterms.
Solution
Recall. a→b≡¬a∨b. The expression has 3 variables p,q,r giving 23=8 rows.
Step 1 — Simplify symbolically
Let A=p∧q. Then expression =(A→r)∨(A→¬r)=(¬A∨r)∨(¬A∨¬r)=¬A∨r∨¬r=¬A∨True=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=8 minterms.
Minterms (one per row of the truth table):
| p | q | r | Minterm |
|---|
| 0 | 0 | 0 | ¬p∧¬q∧¬r |
| 0 | 0 | 1 | ¬p∧¬q∧r |
| 0 | 1 | 0 | ¬p∧q∧¬r |
| 0 | 1 | 1 | ¬p∧q∧r |
| 1 | 0 | 0 | p∧¬q∧¬r |
| 1 | 0 | 1 | p∧¬q∧r |
| 1 | 1 | 0 | p∧q∧¬r |
| 1 | 1 | 1 | p∧q∧r |
Step 3 — Principal DNF (full sum)
Answer
⋁(all 8 minterms)=(¬p∧¬q∧¬r)∨(¬p∧¬q∧r)∨(¬p∧q∧¬r)∨(¬p∧q∧r)∨(p∧¬q∧¬r)∨(p∧¬q∧r)∨(p∧q∧¬r)∨(p∧q∧r).