# Why does negation reverse order of strength?

1. It is the negation that does the trick. Think of a "condition" as a restriction on the class of things that satisfy it, the stronger the restriction the narrower the class. Normally NC are weaker than SC, but [2.] negation always reverses the order of strength: [End of 2.]
¬{a weaker condition} is always stronger than ¬{a stronger condition}.

I already saw these pictures, but the following (created by me) depict 1: P ⊆ Q ⟺ QC ⊆ PC.
But I still sense that I can delve more into 1: why is 2 true intuitively?

• I'm not sure if I understand exactly what is the question. In case my answer does not answer the question, could you try to be more precise? – user2953 Aug 2 '16 at 6:39
• Because 100-3 is larger than 0+64. – Dan Bron Aug 2 '16 at 11:05
• @Keelan I think that your answer does answer my question helpfully. But before I accept anything, I should try to clarify: If P is strong/weak, why does ¬P become weak/strong? – NNOX Apps Aug 2 '16 at 17:33
• Didn't you already ask this question as part of philosophy.stackexchange.com/q/31652/9166 – user9166 Aug 3 '16 at 2:22

Consider splitting a set in two based, one subset of which the elements satisfy the predicate, and one subset of which the elements don't satisfy the predicate. If the predicate is stronger, the Yes-set will be smaller ("the stronger the restriction, the narrower the class"). Therefore, the No-set will be bigger. And since it works both ways, this means that the negation of the predicate is now weaker.

This does not only work for predicates that become stronger or weaker. If we have a strong predicate, that means the Yes-set is small. Since the No-set is the complement of the Yes-set, it has to be big, and therefore the negation of the original predicate (that is, the predicate of the No-set) has to be weak.

The same, mathematically.

Suppose we have a set X and a predicate P. Then Y = {x ∈ X ∣ P x} and N = {x ∈ X | ¬P x} partition X, that is, Y ∪ N = X and Y ∩ N = ∅.

Since "the stronger the restriction, the narrower the class", Y will be smaller when P is stronger. And since N = X ∖ Y, this means that N will be greater, therefore, ¬P will be weaker. A similar argument can be made when P gets weaker: Y will be greater, so N smaller, therefore, ¬P will be stronger.

Here is a simple illustration:

Condition = being over 2 meters (about 6' 7") tall. Strong condition, few satisfy it.

Condition = not being over 2 meters tall. Weak condition, many satisfy it.

Condition = owning 6 cats. Strong condition, few satisfy it.

Condition = not owning 6 cats. Weak condition, many satisfy it.

However, when the set of things satisfying or not satisfying a given condition is infinite, then this does not work. E.g. when the condition is 'prime number', then its negation isn't weaker or stronger. Both are satisfied by infinitely many numbers.

• Thanks. I understand your examples, but I was asking 'why' rather than 'what'? But I think that your last independent clause in each condition ('few/many satisfy it') somewhat explains. – NNOX Apps Aug 2 '16 at 17:36