I attempted a proof before reading the solution Apostol provides. I don't think it is valid but I am trying to determine why

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> Theorem 2, Possibility of Subtraction: Given $a \text{ and } b$, there is exactly one x such that $ a + x = b $ This is denoted by $b - a$. $\ 0-a$ is written as $-a$ and called the negative of a.
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> **Proof from Apostol** - Given $a$ and $b$, choose $y$ so that $a + y = 0$ and let $x = y + b$. Then $a + x = a + (y + b) = (a + y) + b = 0 + b = b$. Therefore there is at least one x such that $a + x = b$. But by Theorem 1.1 there is at most one such x. Hence there is exactly one.
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I will also use theorem 1, previously proved

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> **Theorem 1**
> $ \text{if } a + b = a + c \text{ then } b = c $
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**My attempt**

$$ \begin{align}
\text{Given } a, b \text{ let } &\quad a + x = b &\quad \text{ start }\tag{1}\label{eq1}\\
\text{there is exactly one } x \text{ such that } &\quad a + x = b &\quad\text{from theorem 1.1, show in 2.1}\tag{2}\label{eq2} \\
\text{ proof of (2) } \\
\text{ consider more than one x that satisfies (2) } &\quad a + x = b,\; a + x' = b &\quad\text{ start }\tag{2.1}\label{eq2.1}\\
\text{ then } &\quad a + x = a + x' &\quad\text{ }\tag{2.2}\label{eq2.2}\\
\text{ therefore x and x' are the same and (2) is proven } &\quad x = x' &\quad\text{ direct result of theorem 1.1 }\tag{2.3}\label{eq2.3}\\
\end{align}$$

**Is this valid? If not, why?**

My thoughts - one way that I could see it being invalid is that in (2.1) I break $a + x = b$ into two cases, x and x'. The form of the resulting cases is same as the original, so then, using my logic, you would have to break each case into two more cases on to infinity.

Appreciate the input!