I am currently studying Introduction to Hilbert Spaces with Applications, the third edition, by Debnath and Mikusinski. Chapter 1, exercise 2.(a), is as follows:
Prove that, for any vectors $x$ and $y$ we have that $x + (y - x) = y$.
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 proceeds as follows:
$$\begin{align} x + (y - x) &= x + (y + (-1)x) \\ &= (x + y) + (-1)x \\ &= (y + x) + (-1)x \\ &= y + (x + (-1)x) \\ &= y + x(1 - 1) \\ &= y + x(0) \end{align}$$
Since we haven't yet proven that $x(0) = 0$ (that is a later exercise), I'm unsure of how to proceed. I would greatly appreciate it if people would please take the time to review my proof.
From Page $3$ of the book:
So $$\begin{array} &x+(y-x)&=(x+y)-x\qquad &\text{by (b)}\\ &=(y+x)-x\qquad &\text{by (a)}\\ &=y+(x-x)\qquad &\text{by (b)}\\ &=y+0 \qquad &\text{by (1)}\\ &=y. \qquad &\text{by (2)}\end{array}$$