Question

If p, q, r are simple propositions with truth values T, F, T respectively, then which of the following is not a true statement?

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

Hint:

Substitute p = T, q = F, r = T into each compound statement.

Answer 1

The Correct Option is C

Step 1: Understand the task
We have \(p = T\), \(q = F\), \(r = T\). A statement is "not true" when its truth value is False.

Step 2: Option (A)
\(q \wedge (p \to q) = F \wedge (T \to F) = F \wedge F = F\). Then \(F \to p\) is True. So (A) is true.

Step 3: Options (B) and (D)
(B): \(p \wedge q = F\), so \(F \to (\ldots)\) is True. (D): \((p \wedge q) \vee (\sim q \vee r) = F \vee (T \vee T) = T\). Both are true.

Step 4: Option (C)
\(\sim p \vee q = F \vee F = F\). Then \(F \wedge \sim r = F \wedge F = F\). The biconditional \(F \leftrightarrow p\) is \(F \leftrightarrow T = F\). So (C) is false, which means it is not a true statement.

Final Answer:
Option (C) evaluates to False. This is option (C). \[ \boxed{\text{(C) }[(\sim p\vee q)\wedge\sim r]\leftrightarrow p} \]