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48 votes
48 votes

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

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

3 votes
3 votes

If a given compound proposition is a contradiction then it is FALSE for every possible combination of the truth value of  it’s atomic propositions

1 votes
1 votes
Given, ((p → q) → r) $\equiv$ F  => (p → q) $\equiv$ T and r $\equiv$ F  (T: true ; F: false)

((r → p) → q) $\equiv$ ((F → p) → q) $\equiv$ (T → q) $\equiv$ q

$\therefore$ The expression ((r → p) → q) $\equiv$ q is a contingency. It is always true when q is true.
0 votes
0 votes

Check below 2 approaches for solving this question.

First approach is just for understanding and longer method.

0 votes
0 votes
Given, (p → q) → r == F

We can conclude that from here that is false and p and q can be either true or false.

NOW,

(r → p) → q can be expressed as (F → p) → q == T → q == q.

So the truth value will depend on Q.

It will be true when Q= true.

Option d is correct!
Answer:

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