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

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 $(\exists z > 0)[z^2 = 2]$?

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.

• Is there a particular value of z, where each may differ? Because I cannot still see, where they can differ. They seem equivalent. Or is there another example which might make more sense? Commented Jun 17, 2015 at 1:35
• Does my edit answer your questions? Commented Jun 17, 2015 at 1:38
• Thank you Zach. It makes makes more sense now. Commented Jun 17, 2015 at 1:55
• For the expression ∃z(¬(z>0)∨(z>0∧z² = 2)) , could it just be (∃z) (¬(z>0)∨(z² = 2)) using the rules of contra-positive? Commented Jun 17, 2015 at 2:18
• Yeah, you can distribute $\lor$ over $\land$ to get that simplification. That works fine. Commented Jun 17, 2015 at 2:25

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.

• "∃z(z > 0 → z² = 2), always trivializes to tautology, unless ∀z, z > 0, ", do you mean that it will be tautology for all z less than or equal to zero? Commented Jun 17, 2015 at 2:15
• @user3330840 I mean that if the universe of discourse contains any nonpositive numbers, then $\exists z\,(z>0\to z^2=2)$ is equivalent to truth, and if it doesn't, then it is pointless to even put $z>0$ in the expression (it is just a long-winded way of saying $\exists z\,(z^2=2)$). Using Zach's decomposition, it could also be written $\lnot\forall z\,(z>0)\lor\exists z\,(z^2=2)$ or $\lnot\forall z\,(z>0)\lor\exists z(z>0\land z^2=2)$. Commented Jun 17, 2015 at 2:20