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

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 Apr 3 '16 at 6:51

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.

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.