I've been taught that the "subtraction operation" is not accurate/forward stable as the relative error can be arbitrary large.
I tried to prove it formally but I end up with a contradiction.
What I'm given
In this question we will consider the stability of some elementary operations using the “axioms” of computer arithmetic. Thus we assume that
(P1): For all $x \epsilon \mathbb{R}$, there exists $\varepsilon$ with |$\varepsilon$| < $\varepsilon_m$ and $fl(x) \epsilon \mathbb{F}$ such that $fl(x) = x(1 + \varepsilon)$
(P2): For every elementary operation $∗$ and all $x, y \epsilon \mathbb{F}$, there exists $\varepsilon$ with $|\varepsilon|$ < $\varepsilon_m$ such that $x ∗' y = (x ∗ y)(1 + ε)$.
(Note $∗$ denotes the exact operation whereas $∗'$ represents the "computer arithmetic operation")
My attempt
We have $$f(x)=x_1-x_2$$ and
$$g(x)=x_1 -' x_2=x_1(1+\varepsilon_1)(1+\varepsilon_3)-x_2(1+\varepsilon_2)(1+\varepsilon_3)$$ with $|\varepsilon_1|,|\varepsilon_2|,|\varepsilon_3|$ $\lt$ $\varepsilon_m$
hence
$$\frac{|g(x)-f(x)|}{|f(x)|}=\frac{|x_1(\varepsilon_1+\varepsilon_3+\varepsilon_1\varepsilon_3)-x_2(\varepsilon_2+\varepsilon_3+\varepsilon_2\varepsilon_3)|}{|x_1-x_2|}$$ And thus
$$\frac{|g(x)-f(x)|}{|f(x)|}< \frac{|(x_1-x_2)(2\varepsilon_m+\varepsilon_m^2)|}{|x_1-x_2|}<(2\varepsilon_m+\varepsilon_m^2)$$
which would show that subtraction is accurate.
There is obviously some mistake in my reasoning and I would be very grateful if someone could help me on that.