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Let p, q, and r be propositions and the expression (p → q) → r be a contradiction. Then the expression (r → p) → q is
- a)A tautology
- b)A contradiction.
- c)Always TRUE when p is FALSE
- d)Always TRUE when q is TRUE
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Let p, q, and r be propositions and the expression (p → q) →...
(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
So, from truth table also, it is clear that (r → p) → q is always true when q is true, and Y is false.
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
So, from truth table also, it is clear that (r → p) → q is always true when q is true, and Y is false.
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