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Official PYQ · 1995

GATE CS 1995 Q2.19 — Propositional and First Order Logic

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If the proposition ¬p→q\lnot p \to q is true, then the truth value of the proposition ¬p∨(p→q)\lnot p \lor \left ( p \to q \right ), where ¬\lnot is negation, ∨\lor is inclusive OR and →\to is implication, is

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Answer: D

We are given that the proposition ¬p→q\lnot p \to q is True.

First, let us simplify the given premise:

¬p→q≡¬(¬p)∨q≡p∨q\lnot p \to q \equiv \lnot(\lnot p) \lor q \equiv p \lor q

So, we have:

p∨q=Truep \lor q = \text{True}

Now, let us simplify the target proposition, say EE:

E=¬p∨(p→q)E = \lnot p \lor (p \to q)

Using the implication identity p→q≡¬p∨qp \to q \equiv \lnot p \lor q:

E≡¬p∨(¬p∨q)E \equiv \lnot p \lor (\lnot p \lor q)

Using the associative and idempotent laws of disjunction:

E≡(¬p∨¬p)∨q≡¬p∨qE \equiv (\lnot p \lor \lnot p) \lor q \equiv \lnot p \lor q

To determine if the truth value of EE is uniquely fixed when p∨qp \lor q is True, let us test different truth assignments for pp and qq:

  1. Case 1: Let p=Truep = \text{True} and q=Trueq = \text{True}.
    • Premise: p∨q=True∨True=Truep \lor q = \text{True} \lor \text{True} = \text{True} (Satisfied)
    • Target: E=¬p∨q=False∨True=TrueE = \lnot p \lor q = \text{False} \lor \text{True} = \text{True}
  2. Case 2: Let p=Truep = \text{True} and q=Falseq = \text{False}.
    • Premise: p∨q=True∨False=Truep \lor q = \text{True} \lor \text{False} = \text{True} (Satisfied)
    • Target: E=¬p∨q=False∨False=FalseE = \lnot p \lor q = \text{False} \lor \text{False} = \text{False}

Since the target proposition EE can be either True or False depending on the specific truth values of pp and qq, its truth value cannot be determined.

Therefore, the correct option is D.

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