Question regarding systems of equations

Question

If I have the following system of equations:

$2+x^2-y^2=0$

$x^2-y^2-2=0$

And if I substitute $y$ by a function of $x$ and vice versa I get:

$2+x^2-x^2+2=0$

$y^2-y^2-4=0$

I therefore get:

$4=0, -4=0$ Therefore I don't have any solutions for that system of equations

In theory, am I allowed to get to this conclusion?

Answer 1

1. Let's suppose one and only one condition: that there exists at least one $(x,y)$ in $R^2$ solution of the system of equations. Let $(x,y)$ be that solution.
2. In consequence, the two equations are verified by $(x,y)$.
3. Therefore, [...] 4=0. Since $x$ and $y$ are both in the $(R,+,*)$ ring, at least one of our supposed conditions is false.
4. Conclude.

This is a typical Reductio ad absurdum. As long as you don't violate the Zermelo–Fraenkel set theory and the properties of the objects you're using, there's no reason for your reasoning to be incorrect.