\((p \Rightarrow q) \Leftrightarrow(\sim q \Rightarrow \sim p)\) is a (a) contradiction (b) tautology (c) both tautology \& contradiction (d) None of above

First, let's create a truth table for all possible combinations of p and q: | p | q | |---|---| | T | T | | T | F | | F | T | | F | F |

Next, calculate the values of the implications \((p \Rightarrow q)\) and \((\sim q \Rightarrow \sim p)\) for each combination of p and q: | p | q | p => q | ~q | ~p | ~q => ~p | |---|---|--------|----|----|----------| | T | T | T | F | F | T | | T | F | F | T | F | F | | F | T | T | F | T | T | | F | F | T | T | T | T | Note that \((\sim q \Rightarrow \sim p)\) is the contrapositive of \((p \Rightarrow q)\).

To check if the statement is a tautology (always true), contradiction (always false), both, or neither, compare the values of the implications \((p \Rightarrow q)\) and \((\sim q \Rightarrow \sim p)\): | p | q | p => q | ~q => ~p | (p => q) <=> (~q => ~p) | |---|---|--------|----------|-------------------------| | T | T | T | T | T | | T | F | F | F | T | | F | T | T | T | T | | F | F | T | T | T | Since \((p \Rightarrow q) \Leftrightarrow (\sim q \Rightarrow \sim p)\) is true in all possible cases, it is a: (a) contradiction: no, it's not always false (b) tautology: yes, it's always true (c) both tautology & contradiction: no, it can't be both at the same time (d) None of above: no, it fits the definition of a tautology So, the correct answer is (b) tautology.