I am currently studying Introduction to Hilbert Spaces with Applications, the third edition, by Debnath and Mikusinski. Chapter 1, exercise 2.(b), is as follows:
Prove that, for any vectors $x$, $y$, and $z$, we have that $x - (y - z) = x - y + z$.
The definition of vector space is given as follows:
By a vector space we mean a nonempty set $E$ with two operations:
$(x, y) \mapsto x + y$ from $E \times E$ into $E$ called addition.
$(\lambda, x) \mapsto \lambda x$ from $\mathbb{F} \times E$ into $E$ multiplication by scalars,
such that the following conditions are satisfied for all $x, y, z \in E$ and $\alpha, \beta \in \mathbb{F}$:
(a) $x + y = y + x$;
(b) $(x + y) + z = x + (y + z)$
(c) For every $x, y \in E$ there exists a $z \in E$ such that $x + z = y$;
(d) $\alpha (\beta x) = (\alpha \beta) x$;
(e) $(\alpha + \beta)x = \alpha x + \beta x$;
(f) $\alpha(x + y) = \alpha x + \alpha y$
(g) $1x = x$
Elements of $E$ are called vectors. If $\mathbb{F} = \mathbb{R}$, then $E$ is called a real vector space, and if $\mathbb{F} = \mathbb{C}$, $E$ is called a complex vector space.
My proof is as follows:
$$\begin{align} x - (y - z) &= x + (-1)(y - z) \\ &= x + \{(-1)(y) + (-1)[(-1)(z)]\} \ \ \ \text{ (By (f).) } \\ &= x + [ -y + [(-1)(-1)(z)] \ \ \ \text{(By (d).)} \\ &= x + [ -y + (1)(z)] \\ &= x + (-y + z) \\ &= (x - y) + z \ \ \ \text{(By (b).)} \end{align}$$
I would greatly appreciate it if people would please take the time to review this proof for correctness.
Your proof is correct. Note it also works to prove this identity on a field.