Question

If the truth value of the compound statement
\([(p∨q)∧(q\rightarrow r)∧(\sim r)]\rightarrow (p∧q)\) is false, then the truth values of \(p\rightarrow q\) and \(q\rightarrow p\) are respectively...

• A. \((F,T)\)
• B. \((T,F)\)
• C. \((T,T)\)
• D. \((F,F)\)

Hint:

The whole statement is false only when the bracket is true and p and q is false; solve for r, then q, then p.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The statement has the form \(A\rightarrow B\). It is false only when \(A\) is true and \(B\) is false. Here \(A=(p\vee q)\wedge(q\rightarrow r)\wedge(\sim r)\) and \(B=p\wedge q\).

Step 2: Make A true:
All three parts of \(A\) must be true. From \(\sim r\) true we get \(r=F\). Then \(q\rightarrow r\) is true with \(r=F\) only when \(q=F\). Then \(p\vee q\) is true with \(q=F\) only when \(p=T\).

Step 3: Check B:
With \(p=T\) and \(q=F\) we get \(p\wedge q=F\), as needed.

Step 4: Find the required values:
\(p\rightarrow q=T\rightarrow F=F\). \(q\rightarrow p=F\rightarrow T=T\). So the pair is \((F,T)\).

Step 5: Why the other options are wrong:
Since \(q=F\), the statement \(q\rightarrow p\) is true, which rules out \((T,F)\) and \((F,F)\). Since \(p=T\) and \(q=F\), the statement \(p\rightarrow q\) is false, which rules out \((T,T)\).

Final Answer:
The truth values are \((F,T)\), which is option (A). \[ \boxed{\text{Option A: }(F,T)} \]