The GATE Grind

GATE 2017 CS – Question 39

Engineering Mathematics · Discrete Mathematics: Propositional and First Order Logic · 2 marks · Multiple choice

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

  1. a tautology.
  2. a contradiction.
  3. always TRUE when $p$ is FALSE.
  4. always TRUE when $q$ is TRUE.

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (D) always TRUE when $q$ is TRUE.

Explanation

For $(p \rightarrow q) \rightarrow r$ to be false we need $p \rightarrow q$ true and $r$ false. With $r$ false, $r \rightarrow p$ is true, so $(r \rightarrow p) \rightarrow q$ has the same value as $q$. So it is always true when $q$ is true. It is not a tautology, and when $p$ is false $q$ can still be false.