# Finding fault in proof [contains logical quantifiers]

## Homework Statement

Consider the following incorrect theorem: $∃x∈ℝ ∀y∈ℝ (xy^2 = y-x)$

[Translation (not part of the original problem statement): There is at least an $x∈ℝ$ such that, for every $y∈ℝ$, $(xy^2 = y-x)$.]

What's wrong with the following proof?

Let $x = y(y^2+1)$, then
$y-x=y-y/(y^2+1)=y^3/(y^2+1)=y/(y^2+1) * y^2=xy^2$​

## Homework Equations

$1. (xy^2 = y-x)$

$2. x = y(y^2+1)$

$3. y-x=y-y/(y^2+1)=y^3/(y^2+1)=y/(y^2+1) * y^2=xy^2$

## The Attempt at a Solution

Since the first equation is to be proven and the third equation seem to be correct, i think that the problem lies in the second.
I have transformed the theorem as follow:
$[∃x∈ℝ ∀y∈ℝ (xy^2 = y-x)] = [∃x(x∈ℝ∧∀y(y∈ℝ→(xy^2=y-x))]$
From this, i thought that since one of the things to prove is that there is at least an actual x that is true for all y, the substitution done in equation 2 is not correct (since x is substituted not with an actual value but a free variable).
But i'm not sure if this is really the reason for why the proof is incorrect. Any help will be appreciated.

What's wrong with the following proof?

Let $x = y(y^2+1)$

This is already wrong. x is a single number and can't depend on y

• iopz

This is already wrong. x is a single number and can't depend on y
I see, thank you!