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## Let p, q, and r be the propositions and the expression (p -> q) -> r be a contradiction. Then, the expression (r -> p)-> q is : -gate computer science 2017

A) a tautology

B) a contradiction

C) always TRUE when p is FALSE

D) always TRUE when q is TRUE

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## Answers

**D) always TRUE when q is TRUE**

Explanation:

The expression (r → p) → q is always true when q is true, regardless of the value of the expression (p → q) → r.

Alternate Solution

Option (A): If (p → q) → r is false, then (p → q) is true and r is false.

The possible cases are

(i) p is true, q is true, r is false

(ii) p is false, q is true, r is false

(iii) p is false, q is false, r is false

For case (iii), (r → p) → q is false . It is not a tautology . Option (A) is not true.

Option (B): For case (i) and case (ii), (r → p) → q is true.

It is not a contradiction , option (B) is not true.

Option (C): For case (iii), p is false and (r → p) → q is also false

option (C) is not true.

Option (D): Only for case (i) and case (ii), q is true For both cases, (r →p) →q is true. Option (D) is true

This explanation is contributed by

__Deepak Raj__deepuckraj

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