Prove $(\forall x, y \in E)(\exists ! z\in Z)$ s. t. $x + z = y$. I am currently studying Introduction to Hilbert Spaces with Applications, the third edition, by Debnath and Mikusinski. Chapter 1, exercise 1, is as follows:


Prove that for every $x, y \in E$ there exists a unique $z \in E$ such that $x + z = y$.


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.

Chapter 1 already contains a proof that it seems would prove this (modified by me):

Let $x, y \in E$. By (c), there exists a $w \in E$ such that $x + w = y$. If $x + z_x = x$ for some $z_x \in E$, then, by (a) and (b),
$$y + z_x = (x + w) + z_x = (x + z_x) + w = x + w = y.$$
This proves that $z$ exists for every $x, y \in E$.
Now let $z_1, z_2 \in E$ such that $x + z_1 = x$ and $y + z_2 = y$ for all $x, y \in E$.
Therefore, we have that
$$z_1 + z_2 = z_1$$
and
$$z_2 + z_1 = z_2$$
Therefore, we can conclude that $z_1 = z_2$.

I would greatly appreciate it if people would please take the time to review this proof for correctness.
Existence is guaranteed by (c). You just need to prove uniqueness. Suppose there exists $z_1,z_2 \in E$ such that $$x+z_1=y, x+z_2=y.$$ Then $$x+z_1=x+z_2.\tag{1}$$ By (c), corresponding to $x \in E$ and $0 \in E$ (whose existence and uniqueness are proved here), there exists $x' \in E$ such that $x+x'=0.$
Therefore by $(1),$ $$x'+(x+z_1)=x'+(x+z_2)$$ and finally (b) implies $z_1=z_2.$

Your solution actually shows the uniqueness of additive identity.
Let $z, z^\prime, w \in E$.

From (c), it follows that there exists a $w \in E$ such that $z + w = z^\prime$.

Let $x + z = y = x + z^\prime$ and $z + w = z^\prime$.

Then 

$$y = x + z^\prime = x + z + w = y + w.$$

Therefore, we must have that $w = 0$.

And so, we have that

$$z^\prime = z + w = z.$$