# 如果“All S is P”為真，那麽它是否與“No non-S is non P”相矛盾？

1. If "All S is P" is true, it is also true that S refers to a collection of objects that is smaller than or equal to the collection of objects referred to by P.
2. If S<=P, then No Non-P can be S.
3. All Non-P must therefore be Non-S.
4. By subalternation, since All Non-P is Non-S, there must be some Non-P that is Non-S.
5. If there is some Non-P that is Non-S, then there is some Non-S that is Non-P.
6. If there is some Non-S is Non-P, then the statement "No Non-S is Non-P" must necessarily be false, because it is contradictory with the former statement, which has been arrived at via valid inferences from true premises and must therefore be true.

## 最佳答案

∀x（Sx→Px）

¬∃x(Sx & ¬Px)

∀x（¬Px→¬Sx） [步驟1-3]。

∃x(¬Px & ¬Sx)

“如果 P 指的是所有對象的集合，則表示非P 不存在”;

thus, ∃x(¬Px & ¬Sx) is false: there are no non-Ps, while ∀x(¬Px → ¬Sx) is vacuously true.

This means that, for modern logic, the inference from: ∀x(¬Px → ¬Sx) to ∃x(¬Px & ¬Sx) is not valid.

This does not mean that ∃x(¬Px & ¬Sx) is always false: if S stay for "Fishes" and P for "Water_living", we have that "All Fishes are Water_living" is true, and thus also "All non-Water_living" are non-Fishes" [steps 1-3].