A given text states, “Every real number except zero has a multiplicative inverse" (where mul- tiplicative inverse of a real number x is a real number y such that xy = 1).
It offers the following translation:
$$\forall x((x\neq 0) \rightarrow \exists y(xy = 1)).$$
I personally translated the statement as:
$$\forall x \exists y((x\neq 0)\rightarrow (xy = 1)).$$
Are these two statements logically equivalent? My reasoning being, for every real number x, there exists a real number y, such that if x does not equal zero, then the product of x and y equals 1.
$\endgroup$ 13 Answers
$\begingroup$Yes, it is a general principle that if $y$ does not appear in $\varphi$, then the following are equivalent.
- $\varphi \rightarrow \exists y(\psi)$
- $\exists y(\varphi \rightarrow \psi)$
Yes :
$∀x((x \ne 0 ) → ∃y(xy = 1 ))$
and
$∀x∃y((x \ne 0) → (xy = 1))$
are logically equivalent, because :
$\vdash \exists y (\alpha \rightarrow \beta) \leftrightarrow (\alpha \rightarrow \exists y \beta) \quad $ if $y$ is not free in $\alpha$.
In your case, $\alpha$ is $(x \ne 0 )$ and $y$ is not free in it.
$\endgroup$ $\begingroup$Yes, it is correct. This principle is known as "null quantification rule". In this case, the left-hand $x$ is independent of domain $y$. So, we can place $y$ domain on left hand side. Follow the below link to get more details about null quantification rule.
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