A tautology is a logical statement that is always true, regardless of the truth values of its constituent propositions. We need to identify which of the given statements is not a tautology.
- \(\sim(p \land q) \lor p\):
- Consider the expression \(\sim(p \land q)\), which is equivalent to \(\sim p \lor \sim q\) (De Morgan's Law).
- Replacing, we get: \((\sim p \lor \sim q) \lor p\).
- This can be simplified further to: \(\sim q \lor \sim p \lor p\), which simplifies to \(\sim q \lor \text{True}\) (since \(p \lor \sim p\) is always true).
- The expression evaluates to True, hence it is a tautology.
- \((p \land q) \rightarrow p\):
- This is equivalent to \(\sim(p \land q) \lor p\) (since \((p \rightarrow q) \equiv (\sim p \lor q)\)).
- From the previous expansion on the first point, this reduces to a tautology.
- \(q \lor \sim(p \land q)\):
- This can also be simplified as: \(q \lor (\sim p \lor \sim q)\)
- This simplifies further to \(q \lor \sim p \lor \sim q\).
- Regardless of \(\sim p\) and \(\sim q\), since there's a disjunction with \(q\), it becomes True, confirming it is a tautology.
- \((\sim p \land q) \cap (\sim p \lor p)\):
- The expression \((\sim p \lor p)\) is a tautology (always true) due to the law of excluded middle.
- However, there's a confusion here, as "\(\cap\)" is unconventional in propositional logic compared to "\(\land\)", assuming it's a logical conjunction.
- This simplifies to \((\sim p \land q) \land \text{True}\), which ultimately depends on the truth values of \(\sim p\) and \(q\).
- Since it does not uniformly evaluate to true across all variable assignments, it is not a tautology.
Thus, the statement that is not a tautology is \((\sim p \land q) \cap (\sim p \lor p)\).