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
(S1) : (p ⇒ q)∧(q ∧(~q)) is a contradiction and
(S2) : (p∧q) v ((~p)∧q) v (p ∧ (~q)) v ((~p) ∧ (~q)) is a tautology
Show Hint
Remember the truth tables for basic logical operations (and, or, not, implication). A contradiction is always false, and a tautology is always true
- both are true
- both are false
- only (S1) is true
- only (S2) is true
The Correct Option is A
Solution and Explanation
For \( S_1 : (p \rightarrow q) \land (p \land \sim q) \), the truth table is:
| \( p \) | \( q \) | \( p \rightarrow q \) | \( p \land \sim q \) | \( S_1 \) |
|---|---|---|---|---|
| T | T | T | F | F |
| T | F | F | T | F |
| F | T | T | F | F |
| F | F | T | F | F |
From the truth table, \( S_1 \) is a Contradiction, as \( S_1 \) is always false.
For \( S_2 : (\sim p \land q) \lor (p \land \sim q) \), the truth table is:
| \( p \) | \( q \) | \( p \land q \) | \( \sim p \land q \) | \( p \land \sim q \) | \( (\sim p \land q) \lor (p \land \sim q) \) | \( S_2 \) |
|---|---|---|---|---|---|---|
| T | T | T | F | F | F | F |
| T | F | F | F | T | T | T |
| F | T | F | T | F | T | T |
| F | F | F | F | F | F | F |
From the truth table, \( S_2 \) is a Tautology, as \( S_2 \) is always true.
Hence:
\( S_1 \) is Contradiction, and \( S_2 \) is Tautology.
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