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: Truth table by substitution
With $p = T, q = F, r = T$: $\sim p = F$, $\sim q = T$, $\sim r = F$.

Step 2: Evaluate
(A): antecedent false, so true. (B): antecedent false, so true. (D): $\sim q \vee r$ is $T$, so the disjunction is true.

Step 3: Check (C)
Left side: $(F \vee F) \wedge F = F$. Right side: $p = T$. A biconditional of F and T is F.

Step 4: Answer
Only (C) is false.

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