## In modern logic, why does "All S is P" contradict "Some S is not P"?

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In modern logic, the existential import is removed from universal statements. So All S is P may still be true if there is no S at all.

Contradictory statements must have opposite truth values.

Why is the contradictory relationship between "All S is P" and "Some S is not P" in modern logic?

There are 2 cases, one where there is at least one object that is S, one where there isn't.

If there isn't an object that is S, then "Some S is not P" must be false, because it implies the existence of at least one object that is S and not P. But in a world without S, "All S is P" doesn't have to be true. It is conceivable that it is false too. Doesn't this mean there is one case where both statements are false?

Why do you think it is possible that "All S is P" can be false if there are no objects which are S? (It isn't.) – Noah Schweber – 2016-04-03T06:51:37.183

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Why in modern logic:

does “All S is P” contradict “Some S is not P”?

Because “All S is P” is ∀x(Sx → Px); negating it, we get: ¬∀x(Sx → Px).

Due to equivalence between ¬∀ and ∃¬, this in turn is equivalent to:

∃x¬(Sx → Px).

Now, in propositional logic, ¬(R → Q) is equivalent to: (R & ¬Q), and thus we finally get:

∃x(Sx & ¬Px).

Thus, negating “All S is P” we have obtained: “Some S is not P”.

What happens if there isn't an object that is S ? Obviously, "Some S is not P" is false, because ∃x(Sx & ¬Px) is false.

But "All S is P" is true, because it is ∀x(Sx → Px) and a conditional with a false antecedent is true.

If there are no Ss, then Sx → Px is true for any possible value of x, and thus ∀x(Sx → Px) is true.

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In a world without S, "All S is P" is true. You must keep in mind how it is formulated in first-order logic:

∀x(Sx→Px)

Since there are no S things, the implication is always (vacuously) true.

So there's no case in which "All S is P" and "Some S is not P" are both false.