Question

Let p, q, and r be propositions and the expression (p → q) → r be a contradiction. Then the expression (r → p) → q is
1. A tautology
2. A contradiction.
3. Always TRUE when p is FALSE
4. Always TRUE when q is TRUE

Answer 1

Correct Answer - Option 4 : Always TRUE when q is TRUE

(p → q) → r is a contradiction which is possible only when r is false and (p → q) is true.

Now, from here we can clearly say that option 4 is correct as (r → p) → q means ¬ (r → p) ∨ q.

Since r is false, (r → p) is true and ¬ (r → p) becomes false.

So, it becomes (false ∨ q). Now it totally depends on q. Whenever q is true, this value will always be true.

Alternate Method:

Let X ≡ p → q, Y ≡ (p → q) → r ≡ X → r,

Z ≡ r → p and W ≡ (r → p) → q ≡ Z → p

Using Truth table

p	q	r	X	Y	Z	W
0	0	0	1	0	1	0
0	0	1	1	1	0	1
0	1	0	1	0	1	1
0	1	1	1	1	0	1
1	0	0	0	1	1	0
1	0	1	0	1	1	0
1	1	0	1	0	1	1
1	1	1	1	1	1	1

 

So, from truth table also, it is clear that (r → p) → q is always true when q is true, and Y is false.