Cancellation Law for Addition Proof This is a question I have regard the proof of the cancellation law for addition in Apostol's Calculus. We are told that the sum of two real numbers x and y is x+y and that it is uniquely determined. We are also given a few axioms. The ones that matter are the commutative law, associative law and the existence of negatives.

Cancellation law for addition:
If $a + b = a + c$, then $b = c$.

We assume that $a + b = a + c$. By the Existence of Negatives Axiom, we know that there is a number $y$ such that $y + a = 0$. 

My question regards this step: 

Since sums are uniquely determined we have $y + (a + b) = y + (a + c)$. Why is it allowed to do this step? What is the reasoning behind it?
Cancellation law for addition:-
If $x,y,z\in\mathbb{R}$ and $x+y=x+z$, then $y=z$

Proof: $$x+y= x+z \qquad \qquad (1)$$

Since, $x\in\mathbb{R}$ (existence of additive inverse). Now adding $-x$ to both sides of $(1)$, we get $(-x)+(x+y) = (-x)+(x+z)$

$\Rightarrow (-x+x)+y =(-x+x)+z$       (by associativity of addition)

$\Rightarrow 0+y = 0+z$               ( by property of additive inverse)

$\Rightarrow  y = z$                        ( by property of additive identity)
Cancellation law for addition:-
If $x,y,z\in\mathbb{R}$ and $x+y=x+z$, then $y=z$

Proof: $$x+y= x+z \qquad \qquad (1)$$

Since, $x\in\mathbb{R}$ (existence of additive inverse). Now adding $-x$ to both sides of $(1)$, we get $(-x)+(x+y) = (-x)+(x+z)$

$\Rightarrow (-x+x)+y =(-x+x)+z$       (by associativity of addition)

$\Rightarrow 0+y = 0+z$               ( by property of additive inverse)

$\Rightarrow  y = z$                        ( by property of additive identity)