Question

Among the two statements
(S1) : (p ⇒ q)∧(q ∧(~q)) is a contradiction and  
(S2) : (p∧q) v ((~p)∧q) v (p ∧ (~q)) v ((~p) ∧ (~q)) is a tautology 

• A. both are true
• B. both are false
• C. only (S1) is true
• D. only (S2) is true

Hint:

Remember the truth tables for basic logical operations (and, or, not, implication). A contradiction is always false, and a tautology is always true

Answer 1

The Correct Option is A

Let's analyze the logical statements provided one by one:

Statement S1: \((p \Rightarrow q) \land (q \land (\neg q))\) 

• The expression \((p \Rightarrow q)\) is logically equivalent to \((\neg p \lor q)\).
• Now consider the expression \((q \land (\neg q))\). By definition, \((q \land (\neg q))\) is a contradiction because it is impossible for a statement and its negation to both be true. Therefore, this part of the expression is always false.
• Thus, the entire expression \((p \Rightarrow q) \land (q \land (\neg q))\) simplifies to \((\neg p \lor q) \land \text{False}\), which is always false, hence it is a contradiction.

Statement S2: \((p \land q) \lor ((\neg p) \land q) \lor (p \land (\neg q)) \lor ((\neg p) \land (\neg q))\)

• This expression systematically covers all possible truth values for \(p\) and \(q\).
• Evaluate each part:
  • \((p \land q)\) covers the case when both \(p\) and \(q\) are true.
  • p is false and \(q\) is true.
  • \((p \land (\neg q))\) covers the case when \(p\) is true and \(q\) is false.
  • p and \(q\) are false.
• As each of these covers all possible truth conditions for \(p\) and \(q\), at least one part will always be true, making the whole expression a tautology.

Since both statements are logically correct as analyzed, the correct answer is that both are true.