The compound propositions (p rightarrow q) rightarrow r and p rightarrow (q righ

Question

The compound propositions (p rightarrow q) rightarrow r and p rightarrow (q rightarrow r) are not logically equivalent because _______. when p, q, and r are all true, (p rightarrow q) rightarrow r is false, but p rightarrow (q rightarrow r) is true when p, q, and r are all false. both (p rightarrow q) rightarrow r and p rightarrow (q rightarrow r) are false when p, q, and r are all false, both (p rightarrow q) rightarrow r and prightarrow (q rightarrow r) are true when p, q, and r are all false. (p rightarrow q) rightarrow ris false, but p rightarrow (q rightarrow r) is true

Explanation / Answer

When all of p, q and r are true, obviously both expression are true.

When p, q and r are all false,

(p -> q) -> r = (F -> F) -> F = T -> F = F

p -> (q -> r) = F -> (F -> F) = F -> T = T

The answer is:

when p,q and r are all false, (p -> q) -> r is false, but p -> (q -> r) is true.

This is given in the last option.

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