Difference between $(\exists z)[z > 0 ∧ z^2 = 2]$ and $(\exists z )[z > 0 \Rightarrow z^2 = 2]$

Question

Existential quantifier confusion: what is the difference between $(\exists z)[z > 0 ∧ z^2 = 2]$ and $(\exists z )[z > 0 \Rightarrow z^2 = 2]$?

What are the differences between those two expressions and the expression $(\exists z > 0)[z^2 = 2]$?

Answer 1

The meaning of the first one is fairly straight forward. The second statement is actually a little werid. It is equivalent to the statement $$ \exists z \Big(\lnot (z>0)\lor (z>0\land z^2=0)\Big) $$

So this says "there is a negative number" or "there is a positive number whose square is 2". If we're working in the universe of the reals, then $-1$ would witness this sentence. Certainly $-1$ doesn't witness the first statement.

The third statement is informal according to the standard rules for constructing sentences. However, it's common and normally interpreted as the first statement.

Answer 2

The usual convention for "composite quantifiers" like $\exists x\in A\, P(x)$ and $\forall x\in A\, P(x)$ is

$$\exists x\in A\, P(x)\iff\exists x\,(x\in A\land P(x))$$ $$\forall x\in A\, P(x)\iff\forall x\,(x\in A\to P(x))$$

This rule extends to $\exists z>0(z^2=2)\iff\exists z\,(z>0\land z^2=2)$.

The proposition $\exists z\,(z>0\to z^2=2)$ is true if there are any $z\le0$, which ignores the $z^2=2$ condition completely. In fact, $\exists z\,(z>0\to z^2=2)$ always trivializes to tautology unless $\forall z,z>0$ in which case the $z>0$ part is redundant and

$$\exists z\,(z>0\to z^2=2)\iff\exists z\,(z^2=2)\iff\exists z\,(z>0\land z^2=2).$$

Similar remarks apply to $\forall x\,(x\in A\land P(x))$, which is trivially false when $A$ is not the universe of all objects.