Proof verification: show that $x - a = b$ can be rewritten as $x = b + a$. I am beginning an introductory college math course to catch up from my bad high school education. This is one my first proofs.

Prove that $x - a = b$ can be rewritten as $x = b + a$.

We have been given the properties of the operations of the set of the real numbers (not sure how to latex that).
My proof is this:
$x - a - (-a) = b - (-a)$
$x - 0 = b + a$
$x = b + a$
I'm not completely sure this is correct.
I have another, more important and general doubt. In the proof I use the fact that adding something to both sides of an equation does not change the equation. Do I need to prove this, since no proof has been given in this course, if I want to use it? We are proving very intuitively obvious theorems, so I'm not sure what other intuitively obvious theorems I can use without proving first!
Here's an attempt:
Theorem: adding $x \in R$ to both sizes of an equation does not change the equation.
If $a, b$ are real numbers and $a = b$, then $a$ and $b$ are the same. $a + x = b + x$ can then be rewritten as $a + x = a + x$, since a = b. Both sides are the same, so $a + x = b + x$.
Here I'm not sure how to say that a bunch of operations in the real numbers is a real number. This also should work for all operations we haven't mentioned yet: if $a^{2/3} = b^{4/7}$, then $a^{2/3} + c = b^{4/7} + c$. I'm not sure if this complicates things, but I have no idea how to say this either way.
Let's also ignore for a second that this is an introductory course. Would I need to prove this if I was asked to prove the theorem in an exam?
 A: I encourage you to ask yourself the following questions:


*

*What do you mean for "="?

*What is $+$? It's first of all a $\textbf{function}$
$$+:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$$
So for any couple $(a,b)$ there is one and only one value $a+b$.
The fact that $+$ send a couple of real numbers to a real number (as it happens for example for $\mathbb{Z}$ or $\mathbb{Q}$) is related to a not so trivial question: what is a real number?
A: 
$x - a - (-a) = b - (-a)$

What is "$-$"? Well, we do learn it to be an operation, and indeed you can define subtraction of reals because $\mathbb R$ is additive group, and then by definition it means that for all real $x$ there exists a real denoted by $-x$ with property $x+(-x) = -x+x = 0$. This allows us to define subtraction:
$$x-y:=x+(-y).$$
However, there are subtleties here, the first one being that additive inverse is unique (which can be easily proved from group axioms) and therefore the above will really be well defined operation.
But, what properties of subtraction do you know? For example, it is neither associative nor commutative and, because of that, ill-suited for proofs like these where you need to carefully pay attention to axioms.
More importantly, it is a mystery why you would choose to subtract $-a$ instead of just adding $a$. Note that $x-a-(-a) = x-a+a$.
Furthermore, what does $x-a-(-a)$ really mean? It should actually be $(x-a)-(-a)$ and after that one should invoke associativity to simplify.
With the above comments in mind, what I would do is the following:
\begin{align}
x-a = b &\implies x+(-a) = b\\
&\implies (x+(-a))+a = b+a\\
&\implies x+(-a+a) = b+a\\
&\implies x+0 = b+a\\
&\implies x = b+a\\
\end{align}
You should be able to recognize what axiom was used in each line, except maybe the second line. So, what is going on there?
Well, reals come equipped with binary operation $+\colon\mathbb R\times\mathbb R\to \mathbb R$, either by construction or from axiomatic definition. Binary operation is by definition a function, so, for any real number $a$, we can define a new function $f_a\colon\mathbb R\to\mathbb R$ with formula $f_a(x) = x + a$. It is important that $f_a$ is function, since all functions must satisfy $$x = y\implies f(x) = f(y)$$
and letting $f = f_a$, we get
$$x = y \implies x + a = y + a,$$
which is what we did in the above second line.
A: Here is a proof with all steps spelled out:
$$
\begin{array}{rcll}
x-a &=& b & \qquad\text{hypothesis} \\
(x-a)+a &=& b+a & \qquad\text{add $a$ to both sides} \\
(x+(-a))+a &=& b+a & \qquad\text{definiton of subtraction} \\
x+((-a)+a) &=& b+a& \qquad\text{associativity of addition} \\
x+0 &=& b+a & \qquad\text{definition of additive inverse} \\
x &=& b+a & \qquad\text{definition of additive neutral element} \\
\end{array}
$$
A: To your second question.
When you have two terms $t_1$ and $t_2$ which are equal
$$ t_1 = t_2 $$
then since a function $f$ is well-defined we also have 
$$f(t_1) = f(t_2)$$
or in other words
$$t_1 = t_2 \quad\Rightarrow \quad f(t_1) = f(t_2)$$
Your particular function is $f:\mathbb{R} \to \mathbb{R},\; x \mapsto x+a$. Since this function has an inverse function which is $f^{-1}: \mathbb{R}\to \mathbb{R},\; x \mapsto x-a$. We conclude:
$$t_1 = t_2 \quad\Rightarrow \quad f(t_1) = f(t_2)
\quad\Rightarrow \quad f^{-1}(f(t_1)) = f^{-1}(f(t_2))
\quad\Rightarrow \quad t_1 = t_2.
$$
In a more compact notation this means
$$t_1 = t_2 \quad\Leftrightarrow \quad f(t_1) = f(t_2)$$
which should vanish your concerns.
A: As you say, $$x-a=b\implies (x-a)+a=b+a$$ because both members represent the same values before and after addition (actually, $\iff$ holds.)
Then by properties of the addition,
$$(x-a)+a=x+(-a+a)=x+0=x.$$
