FastAPI高速API自動化

**Answer** The two statements are equivalent. The implication “\(p\) is true only if \(q\) is true” is the same as “\(p\) is false or \(q\) is true”. So \[ p \text{ only if } q \;\;\Longleftrightarrow\;\; \neg p \lor q . \] --- **Proof** We prove the equivalence by showing that each side implies the other. --- ### 1. \(p \text{ only if } q \;\Longrightarrow\; \neg p \lor q\) Assume \(p \text{ only if } q\). By definition of “only if”, this means \[ p \;\Longrightarrow\; q . \] Now consider the truth of \(p\). *If \(p\) is true*, the implication forces \(q\) to be true. *If \(p\) is false*, the implication is automatically satisfied, and \(\neg p\) is true. Thus in every case either \(\neg p\) is true or \(q\) is true; i.e. \(\neg p \lor q\) holds. --- ### 2. \(\neg p \lor q \;\Longrightarrow\; p \text{ only if } q\) Assume \(\neg p \lor q\). We must show that \(p \;\Longrightarrow\; q\). Take an arbitrary assignment of truth values to \(p\) and \(q\). *If \(\neg p\) is true*, then \(p\) is false, so the implication \(p \;\Longrightarrow\; q\) is true (a false antecedent makes any implication true). *If \(q\) is true*, the implication is again true regardless of the value of \(p\). Since \(\neg p \lor q\) guarantees that at least one of these two situations occurs, the implication \(p \;\Longrightarrow\; q\) holds in all possible assignments. Hence \(p \text{ only if } q\) is true. --- ### 3. Conclusion We have shown \[ p \text{ only if } q \;\Longrightarrow\; \neg p \lor q \quad\text{and}\quad \neg p \lor q \;\Longrightarrow\; p \text{ only if } q . \] Therefore the two statements are logically equivalent: \[ p \text{ only if } q \;\;\Longleftrightarrow\;\; \neg p \lor q . \] --- **Remark** The equivalence can also be verified by a truth‑table. The table for \(p \text{ only if } q\) (i.e. \(p \rightarrow q\)) and for \(\neg p \lor q\) is identical, confirming the equivalence.