Question

Which of the following logical statements is a tautology?

• A. \([(p\rightarrow q)∧\sim q]\rightarrow \sim p\)
• B. \((p\rightarrow q)∧(p∧\sim q)\)
• C. \([(p∨q)∧\sim p]∧\sim q\)
• D. \([(p∨\sim q)∨(\sim p∧q)]∧r\)

Hint:

A tautology is true for every combination of truth values; test modus tollens.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A tautology is true in every row of its truth table.

Step 2: Check Option (A):
\([(p\rightarrow q)\wedge\sim q]\rightarrow\sim p\) is the law of contrapositive reasoning (modus tollens). If \(p\rightarrow q\) holds and \(q\) is false, then \(p\) must be false. The only way for the whole implication to fail would need the antecedent true and \(\sim p\) false, i.e. \(p\) true, \(q\) false, and \(p\rightarrow q\) true, which is impossible. So (A) is a tautology.

Step 3: Check Option (B):
\((p\rightarrow q)\wedge(p\wedge\sim q)\): the second part makes \(p\rightarrow q\) false, so the conjunction is always false. It is a contradiction.

Step 4: Check Options (C) and (D):
(C) \([(p\vee q)\wedge\sim p]\wedge\sim q\): needs \(p\) false, \(q\) false, and \(p\vee q\) true, impossible. It is a contradiction.
(D) \([(p\vee\sim q)\vee(\sim p\wedge q)]\wedge r\): false whenever \(r\) is false, so it is not a tautology.

Final Answer:
Only option (A) is a tautology. \[ \boxed{\text{(A) } [(p\rightarrow q)\wedge\sim q]\rightarrow\sim p} \]