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Official PYQ · 2000

GATE CS 2000 Q2.7 — Propositional and First Order Logic

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Let a,b,c,da, b, c, d be propositions. Assume that the equivalence a⇔(b∨¬b)a ⇔ ( b \vee \neg b) and b⇔cb ⇔c hold. Then the truth-value of the formula (a∧b)→(a∧c)∨d(a ∧ b) → (a ∧ c) ∨ d is always

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Answer: A

We are given two equivalences:

  1. a  ⟺  (b∨¬b)a \iff (b \lor \neg b)
    Since b∨¬bb \lor \neg b is a tautology (always True), the proposition aa must always be True (a≡Truea \equiv \text{True}).
  2. b  ⟺  cb \iff c
    This means bb and cc always have the same truth value (b≡cb \equiv c).

Now, let us evaluate the given formula:

ϕ=(a∧b)→(a∧c)∨d\phi = (a \land b) \to (a \land c) \lor d

Substituting a≡Truea \equiv \text{True} into the formula:

ϕ≡(True∧b)→(True∧c)∨d\phi \equiv (\text{True} \land b) \to (\text{True} \land c) \lor d ϕ≡b→(c∨d)\phi \equiv b \to (c \lor d)

Since b≡cb \equiv c, we can substitute cc with bb:

ϕ≡b→(b∨d)\phi \equiv b \to (b \lor d)

Using the implication identity P→Q≡¬P∨QP \to Q \equiv \neg P \lor Q:

ϕ≡¬b∨(b∨d)\phi \equiv \neg b \lor (b \lor d) ϕ≡(¬b∨b)∨d\phi \equiv (\neg b \lor b) \lor d

Since ¬b∨b≡True\neg b \lor b \equiv \text{True}:

ϕ≡True∨d≡True\phi \equiv \text{True} \lor d \equiv \text{True}

Thus, the truth-value of the formula is always True.

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