# If “All S is P” is true, does it contradict “No non-S is non-P”?

I have a problem I encountered in a logic textbook that I cannot figure out after multiple tries.

Say we assume that "All S is P" is true.

Does this allow us to conclude the truth value of "No Non-S is Non-P", where non-X is the complementary class of X.

The textbook answer is that the truth value cannot be determined. However, I seem to be able to prove the statement is false. This is how I do it:

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.

Yet, when I draw it on a Venn diagram for "All S is P", there is a case where P refers to the collection of ALL objects, which means that Non-P does not exist, hence all Non-S must be P. This admits a rare case where the statement holds, hence the statement's truth value is undetermined.

Both lines of reasoning seem correct, yet contradictory. What went wrong?

• Yes the two propositions are contradictory in Aristotelian ogic. Your reasoning is not even close to WHY the two propositions are contradictory. You are accidentally correct. You must understand there are different types of logic with different rules. So in Mathematical logic this would not be a question at all. It would never be asked. Rules of inference in Aristotelian logic would show that the 2 propositions you state are indeed contradictory. That is, both propositions can't be true at the same time both can't be false. If 1 is true the other must be false & vice versa. – Logikal Sep 6 at 7:31
• In Aristotelian logic you can use inference rules such as obversion, conversion, etc to prove that "No non-s is non-p" is IDENTICAL (not a logical equivalence) to the O type of proposition: Some s are not p. The square of Opposition shows that A type propositions are contradictory to O type propositions. Your reasoning should have been close to the subject material of deductive reasoning--not your own invention. Perhaps you are confusing logic as all logics are the same thing. Perhaps you thought logic is discrete mathematics or something. There is more subject matter to logic than math. – Logikal Sep 6 at 7:39

You are right from the "traditional" point of view.

The issue is with the existential import of categorical propositions:

If a statement includes a term such that the statement is false if the term has no instances, then the statement is said to have existential import with respect to that term. It is ambiguous whether or not a universal statement of the form "All A is B" is to be considered as true, false, or even meaningless if there are no As.

In traditional syllogism the inference from "All S are P" to "Some S are P" (subalternation) is licensed by the assumption that there are Ss (and so also Ps).

In modern logic, while it is (usually) correct that "For all x Px" implies "Some x is Px", the modern translation of the categorical proposition is: "For all x (if Sx, then Px)", that is true also when there are no Ss, and thus we cannot correctly infer: "There are some x (Sx and Px)".

Formally:

∀x(Sx → Px)

is equivalent to:

¬∃x(Sx & ¬Px)

which in turn is:

∀x(¬Px → ¬Sx) [steps 1-3].

Now we have subalternation [step 4]:

∃x(¬Px & ¬Sx)

which is not correct from the modern point of view.

"if P refers to the collection of ALL objects, that means that non-P does not exist";

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].

But this does not contradict the fact that also "There are some non-Water_living that are non-Fishes" is true.

This is the key to the textbook answer:

A valid argument is one that form true premises infers true conclusion.

For modern logic, subalternation is not valid; but this does not means that the conclusion is always false.

• Thank you very much for the answer. I understand now that the assumption of existential import must hold for sub alternation to hold because no conclusion can imply more than the premises. – KohLP Mar 31 '16 at 14:39
• However, I am unable to understand how that eliminates the above contradiction. Please correct me if I am wrong (I know I am somewhere), but it seems to me that the existential import is assumed in both cases. Existential import must be necessary for the soundness of steps 1-6. Existential import is also key for us to say that P exists, only then can P refer to all objects. How can a common assumption be a source of contradiction? – KohLP Mar 31 '16 at 14:50
• Your statement #3 does not have existential import in the all-is-P case. (You state it as true even when there are no instances of non-P.) – Jeff Y Mar 31 '16 at 19:25