Prove the uniqueness of subtraction I have to prove this proposition: Given $m,n \in\mathbb Z$, there exists one and only one $x \in\mathbb Z$ such that $m + x = n$. So, just to be sure: I am given an equation and asked to first prove the existence of solution x and then its uniqueness, right?
From what I understand, "uniqueness" simply means that if two solutions fit in the equation, they must be equal. I can only use the axioms for addition and multiplication and derived propositions.
Proof of existence:
\begin{align*}
m + x &= n\\
(-m) + (m + x) &= (-m) + n\\
(-m + m) + x &= (-m) + n\\
0 + x &= (-m) + n\\
x &= (-m) + n\\
\end{align*}
Hence, the solution x exists.
The second part seems too simple. Assuming that x can take two values, x1 and x2... Proof of uniqueness:
\begin{align*}
m + x1 &= n\\
m + x2 &= n\\
m + x1 &= m + x2\\
(-m) + (m + x1) &= (-m) + (m + x2)\\
(-m + m) + x1 &= (-m + m) + x2\\
0 + x1 &= 0 + x2\\
x1 &= x2
\end{align*}
Hence, the solution x is unique. What do you think? Thank you!
 A: Suppose $m+x_1=n$ and $m+x_2=n$. Then $m+x_1=m+x_2$ and so $x_1=x_2$ by left-cancellation. I think you did a great deal more work than you really needed to here, but it never hurts to try to be thorough. 
A: You did a great job with your proof! Why? 
Your strategy is in the page 58 of the edition 2014 of the book "A Transition to Advanced Mathematics", written by Smith, Eggen and Andre.
"PROOF OF (∃!x)P(x): 
(i) Prove that (∃x) P(x) is true. Use any method. 
(ii) Prove that (∀y)(∀z)[P(y) ∧ P(z) ⇒ y = z]. 
Assume that y and z are objects in the universe such that P(y) and P(z) are true. 
From (i) and (ii) conclude that (∃!x)P(x) is true."
My proof would be:
Assume m, n, x, y, and z are integers; and m + x = n. 
There exists a solution because:
-m + m + x = -m + n (-m is the inverse of m under addition.)
x = -m + n (-m + m = 0 because 0 is the identity element under addition.)
x = n - m (commutative property).
As a final step we will be prove the uniqueness of the solution.
Assume we have two differents solutions y and z. 
y = n - m, and z = n - m
then
y = z.
For that reason, the solution is unique. Q.E.D.
A: In your proof of existence you basically show that the equation 
$$m+x=n$$
is equivalent to the equation
$$x=(-m)+n$$
But it is clear that this last equation has one unique solution.
Hence you have actually already proved the uniqueness.
