# Thread: Easy doubt on second-order logic

1. ## Easy doubt on second-order logic

Why does the negation of the operator "For all" results in the operator "There exists", and vice-versa?

2. ## Re: Easy doubt on second-order logic

This happens in first-order logic as well. Basically, it's common sense. You can prove that $M\models\neg\forall x\,P(x)$ iff $M\models\exists x\,\neg P(x)$, but this still relies on the feature of informal, meta-level quantifiers that if it is not the case that a property holds for all elements, then it does not hold for some element, i.e., its negation holds for that element.