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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

Best answer
67 votes
67 votes
Given $(p\to q)\to r$ is false. It is possible only when $r$ is FALSE and $(p\to q)$ is TRUE.
Now even without checking any other option we can directly conclude option $D$ is correct as $(r\to p)\to q$ can be written as $\neg(r\to p)\vee q.$

Since, $r$ is False, $r\to p$ is true and $\neg (r\to p)$ is False. So, it becomes $(\text{False} \vee q).$
which is TRUE whenever $q$ is TRUE

Hence, option (D)
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13 votes
13 votes

D)

Given that ( p → q ) → r  is a contradiction so  r = and ( p → q ) = T 

We have to evaluate the expression ( r → p ) → q  

Since r = F, ( r → p )  = T ( As F → p, is always true )

The final expression is T → q and this is True when q is True, hence option D.  

Answer:

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