Question

The following statement (p → q) → [(~p→q)→q] is:

a tautology
equivalent to ~p → q
equivalent to p → ~q
a fallacy

Answer 1

Answer:

♠ The correct option is A.

Explanation:

(p → q) → ((~p → q) → q)

= (p → q) → ((p ∨ q) → q)

= (p → q) → ((~p ∧ ~q) ∨ q)

= (p → q) → ((~p ∨ q) ∧ (~q ∨ q))

= (p → q) → (~p ∨ q)

= (p → q) → (p → q)

= T

Answer 2

Explanation:

One approach is just to start writing a truth table - after all, there are only four cases. And this can be reduced: If q is true, the statement is true (this is two cases). If p is false, the statement is true because

p∧(p→q)

is false. The only case that's left is when p is true and q is false, and I'll leave it to you to verify that the proposition is true in this case too.

option A is your right answer