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

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2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Jun 17, 2015 at 1:35
  • $\begingroup$ Does my edit answer your questions? $\endgroup$
    – Zach Stone
    Commented Jun 17, 2015 at 1:38
  • $\begingroup$ Thank you Zach. It makes makes more sense now. $\endgroup$ Commented Jun 17, 2015 at 1:55
  • $\begingroup$ 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? $\endgroup$ Commented Jun 17, 2015 at 2:18
  • $\begingroup$ Yeah, you can distribute $\lor$ over $\land$ to get that simplification. That works fine. $\endgroup$
    – Zach Stone
    Commented Jun 17, 2015 at 2:25
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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.

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  • $\begingroup$ "∃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? $\endgroup$ Commented Jun 17, 2015 at 2:15
  • $\begingroup$ @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 $\lnot\forall z\,(z>0)\lor\exists z(z>0\land z^2=2)$. $\endgroup$ Commented Jun 17, 2015 at 2:20

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