Using field axioms for a simple proof 
Question: If $F$ is a field, and $a, b, c \in F$, then prove that if $a+b = a+c$, then $b=c$ by using the axioms for a field.
Relevant information:
  Field Axioms (for $a, b, c \in F$):
Addition:
  $a+b = b+a$ (Commutativity)
  $a+(b+c) = (a+b)+c$ (Associativity)
  $a+0 = a$ (Identity element exists)
  $a+(-a) = 0$ (Inverse exists)
Multiplication:
  $ab = ba$ (Commutativity)
  $a(bc) = (ab)c$ (Associativity)
  $a1 = a$ (Identity element exists)
  $aa^{-1} = 1$ (Inverse exists)
Distributive Property:
  $a(b+c) = ab + ac$
Attempt at solution:
  I'm not sure where I can begin. Is it ok to start with adding the inverse of a to both sides, as in the following?
  $(a+b)+(-a) = (a+c)+(-a)$ (Justification?)
  $(b+a)+(-a) = (c+a)+(-a)$ (Commutativity)
  $b+(a+(-a)) = c+(a+(-a))$ (Associativity)
  $b+0 = c+0$ (Definition of additive inverse)
  $b = c$ (Definition of additive identity)

I'm wondering about my very first step. Specifically, the axioms don't mention anything about doing something to both sides of an equation simultaneously. Is there some other axiom I can use to justify this step?
This is Exercise 1, part b in Section 1 on page 2 of Halmos, Finite Dimensional Vector Spaces (reading book for fun--this is not homework (probably too easy to be a homework problem anyway!)). In part a, I proved that $0+a = a$, in case that is somehow helpful in this problem. Thanks!

The first step is justified by the existence of $-a$. No further axiom is required in order to deduce that $(a+b)+(-a)=(a+c)+(-a)$. To see that, just write $a+b=y=a+c$ (after all, it is given that $a+b=a+c$). Now, one of the information defining a field is the function of addition: $(u,v)\mapsto u+v$. Applying it to $u=y$ and $v=(-a)$, yields the element $y+(-a)$. But, since $y=a+b$, $y+(-a)=(a+b)+(-a)$. Similarly, since $y=a+c$, $y+(-a)=(a+c)+(-a)$. By transitivity of equality, it follows that $(a+b)+(-a)=(a+c)+(-a)$.
Note that it would be a lot easier to add $(-a)$ on the left rather than on the right. It will save you using commutativity. 
