# 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.

Related Precalculus Mathematics Homework Help News on Phys.org
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!