I was reading about quantifiers from this book.
I decided to jot down all implications due to different orders of quantifiers.
While talking about the orders of the quantifiers the author states following:
1) ∀x∀y P(x, y) ≡ ∀y∀x P(x, y)
2) ∃x∃y P(x,y) ≡ ∃y∃x P(x,y)
3) ∃y∀x P(x, y) ≢ ∀x∃y P(x, y)
However,
4) ∃y∀x P(x, y) → ∀x∃y P(x, y)
But,
5) ∀x∃y P(x, y) ↛ ∃y∀x P(x, y)
that is, not necessarily
6) ∀x∃y P(x, y) → ∃y∀x P(x, y)
However the book does not say anything about other orders.
Assuming all above are correct (if not please correct), I googled for more, but no use. So I was just guessing if we cannot infer anything firm about any other orders, thats why it is not given.
Or just guessing if we can follow simple rule: The implication between two orders of quantifiers is correct if premise contain more number universal quantifiers than conclusion, that is following correct:
7) ∀x∃y P(x, y) → ∃y∃x P(x, y) (one universal quantifiers in premise none in conclusion)
8) ∀x∀y P(x, y) → ∃y∃x P(x, y) (two universal quantifiers in premise none in conclusion)
9) ∀x∀y P(x, y) → ∀y∃x P(x, y) (two universal quantifiers in premise one in conclusion)
10) ∀x∀y P(x, y) → ∃y∀x P(x, y) (two universal quantifiers in premise one in conclusion)
Are above incorrect? If incorrect, are there others? Or the ones given in the book only possible? Can these implications rule be extended to 3 or more quantifier's order?
All of these are almost correct, as in each one you are essentially going from $\forall x$ P(x) to $\exists x P(x)$. This fails if the universe is empty. You also use the fact (quoted in the book) that you can commute like quantifiers in some of your examples.
With three or more, the simple count fails. You cannot go from $\forall x \forall y \exists z$ to $\exists x \exists y \forall z$ or any permutation. The first only puts a condition on one $z$, while the second requires it for all $z$.