GATE CS 2017 Set 1 Q.29 (CS Q.29): Propositional and First-Order Logic

Let $p, q$ , and $r$ be propositions and the expression $(p \to q) \to r$ be a contradiction. Then, the expression $(r \lor p) \to q$ is

a tautology.

a contradiction.

always TRUE when $p$ is FALSE.

always TRUE when $q$ is TRUE.

Official answer and solution

Official answer: D
From: Official GATE 2017 answer key

Solution

Determine the truth values of the sub-formulas from the condition that $(p \to q) \to r$ is a contradiction, and evaluate $(r \lor p) \to q$ .

We are given that the expression $(p \to q) \to r$ is a contradiction (it evaluates to FALSE under the considered scenario).
An implication $A \to B$ is FALSE if and only if $A = \text{TRUE}$ and $B = \text{FALSE}$ .

Therefore:

$p \to q = \text{TRUE}$
$r = \text{FALSE}$

Now evaluate the target expression:

E = (r \lor p) \to q

Substituting $r = \text{FALSE}$ :

E = (\text{FALSE} \lor p) \to q = p \to q

Since $p \to q = \text{TRUE}$ , $E$ evaluates to TRUE.

Furthermore, looking at the general properties of an implication $(r \lor p) \to q$ :
Whenever the consequent $q$ is TRUE, the implication $(r \lor p) \to \text{TRUE}$ is identically TRUE regardless of the truth values of $r$ and $p$ .
Therefore, the expression is always TRUE when $q$ is TRUE (Option D).

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2017 Computer Science and Information Technology (CS) Session 1
Question: 29 of 65, printed as CS Q.29
Section: Computer Science and Information Technology

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