Question

Consider the statement patterns
A. \((q\rightarrow p)∨(p\rightarrow q)\)
B. \((\sim p∨\sim q)\leftrightarrow \sim (p∧q)\)
C. \([(p∨q)∧\sim p]∧\sim q\)
D. \((p∧q)∧(\sim p∨\sim q)\)
then statement patterns

• A. A and B are contradictions, C and D are tautologies.
• B. A and B are tautologies, C and D are contradictions.
• C. A, B, C, D all are tautologies.
• D. A, B, C, D all are contradictions.

Hint:

Simplify each pattern using standard laws and decide whether it is always true or always false.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A tautology is true for every truth value assignment and a contradiction is false for every assignment. We simplify each pattern. In pattern B the connective between the two brackets is not shown clearly in print, and we read it so that it is the De Morgan type equivalence (an OR or an equivalence between \(\sim p \vee \sim q\) and the negation of \(p \wedge q\)).

Step 2: Pattern A.
\((q \to p) \vee (p \to q)\). Write \(p \to q \equiv \sim p \vee q\) and \(q \to p \equiv \sim q \vee p\). Then A \(\equiv \sim q \vee p \vee \sim p \vee q\). This contains \(p \vee \sim p\), so it is always true. A is a tautology.

Step 3: Pattern B.
\(\sim p \vee \sim q \equiv \sim(p \wedge q)\), so joining it with \(p \wedge q\) by OR gives \(\sim(p\wedge q) \vee (p \wedge q)\), which is of the form \(r \vee \sim r\). B is a tautology.

Step 4: Pattern C.
\([(p \vee q) \wedge \sim p] \wedge \sim q\). If \(p\) is false then \(p \vee q\) needs \(q\) true, but \(\sim q\) needs \(q\) false. Both cannot hold. C is a contradiction.

Step 5: Pattern D.
\((p \wedge q) \wedge (\sim p \vee \sim q)\). The second bracket is \(\sim(p \wedge q)\), so D is \(r \wedge \sim r\). D is a contradiction.

Final Answer:
A and B are tautologies, and C and D are contradictions, option (B). \[ \boxed{\text{A, B tautologies; C, D contradictions}} \]