I have the following statement: "Every number besides 0 has a unique multiplicative inverse."
I am curious as to why the following does not work:
$$\forall x \in \mathbb{R_{ \neq 0}} \exists y,z \in \mathbb{R_{\neq 0}} : ((xy = 1) \cap (xz =1)) \iff (y =z)$$
From my understanding, the following is correct, but am told that there needs to be universal quantifier for z instead. Why is this the case?
$$\forall x \in \mathbb{R_{ \neq 0}} \exists y,z \in \mathbb{R_{\neq 0}} : (xy =1 ) \land (xz = 1) \implies (y = z) $$
I'm assuming the following doesn't hold because (y = z) can be false, while the other predicate is true:
$$\forall x \in \mathbb{R_{ \neq 0}} \exists y,z \in \mathbb{R_{\neq 0}} : (y = z) \implies (xy =1 ) \land (xz = 1) $$
$\exists y\exists z~(\varphi(y)\land\varphi(z)\leftrightarrow y=z)$ can be satisfied by the existence of two distinct entities where at least one does not satisfy the predicate $\varphi$. If there is some $y,z$ where $y\neq z$ and at least $\lnot\varphi(y)$ or $\lnot\varphi(z)$, then these values do satisfy $\varphi(y)\land\varphi(z)\leftrightarrow y=z$.
In short, the fact that $(2\cdot 3=1\land 2\cdot 4=1)\leftrightarrow 3=4$ is valued as true does not suggest that either $3$ or $4$ are the multiplicative inverse of $2$.
You wish to state "For every (non-zero real) $x$, there is some $y$ that is its multiplicative inverse, and there is no other."
$$\forall x\in\Bbb R_{\neq 0}~\exists y\in\Bbb R_{\neq 0}~\forall z\in\Bbb R_{\neq 0}~\big(xy=1\land(xz=1\to z=y)\big)$$
Note: the parenthesis placement is important.