Every non-zero real number has a reciprocal (or multiplicative inverse).
The above statement can be expressed as $$ \forall x \in \mathbb{R} \big( (x \neq 0) \rightarrow \exists y \in \mathbb{R} (xy = 1) \big). $$ Am I right?
Now my question is, is this statement logically equivalent to the following? $$ \forall x \in \mathbb{R} \exists y \in \mathbb{R} \big( (x \neq 0) \rightarrow (xy=1) \big). $$ If so, how? If not, why not?
In general, let $P(x)$ and $Q(x, y)$ be any propositional functions of one and two variables, respectively. Then does the following logical equivalence hold? $$ \forall x \big( P(x) \rightarrow \exists y Q(x, y) \big) \equiv \forall x \exists y \big( P(x) \rightarrow Q(x, y) \big), \tag{1} $$ where the domains of discourse for $x$ and for $y$ on the two sides of (1) are the same.
no they are not equivalent. In the first statement i can replace $x\ne 0$ by $x+y\ne 0$ or $y\ne 0$ by universal specification.
Since the scope of $y$ is only the second part of the formula.