What properties of numbers allow us to remove parentheses from expressions? I've seen it asserted in several places (e.g., Spivak's Calculus, p.3) that the fact that "parentheses can be freely rearranged" in expressions involving only addition ($+$) is based solely on (P1) associtivity of addition, $$a+\left(b+c\right) =\left(a+b\right)+c,$$ 
and can see that this is the case in every example I've tired. However it is also asserted that one can eliminate parenthesis altogether, so that, for example $$a+b+c$$ is identical to the above expressions.
I can't see how to prove this with using only P1. The proof would seem to require (P2) additive identity, $$a+0=a,$$ (P3) additive inverse, $$a+\left(-a\right)=0,$$ and (P4) commutativity, $$a+b=b+a.$$
For example, the proof
$$\left(a+b\right)+c$$
$$=\left(a+b\right)+c+0$$
$$=0+\left(a+b\right)+c$$
$$=a+\left(-a\right)+\left(a+b\right)+c$$
$$=a+\left(\left(-a\right)+a\right)+b+c$$
$$=a+\left(a+\left(-a\right)\right)+b+c$$
$$=a+0+b+c$$
$$=a+b+c$$
requires P2, P4, P3, P1, P4, P3, and P2.
Am I missing something that allows one to conclude that $$\left(a+b\right)+c=a+b+c$$ based solely on P1? Perhaps there something subtle about what parentheses represent that I'm missing.
 A: Hint $ $ It's easy: keep pushing ')'s rightward using the rewrite rule $\rm\ (x+y)+z\ \to\:\ x+(y+z).\,$ An easy induction shows that this process terminates with the right-associated normal form where all ')'s are at the right end,
e.g. $\rm\ a+(b+(c+(d+\:\cdots\:))).\, $  By associativity, the rewrite rule preserves equality, so every possible bracketing of the summands is equal to the normalized bracketing. Thus we can omit the brackets, yielding a well-defined $\rm\:n$-ary addition operation
$\rm\ a_1+ a_2+\: \cdots\:+a_n\:. $ 
However, for nonassociative operations, different bracketings need not yield equal values, so the brackets are required in order to uniquely specify the intended value. 
It might aid intuition to think of the expressions presented as (parse) trees, e.g. below. 

A: The point is that no matter how you parenthesize $a+b+c$, the result is the same, so the expression without parentheses is unambiguous. Thus, it doesn’t matter whether you define $a+b+c$ to be $(a+b)+c$ or $a+(b+c)$. (Note that some such definition is ultimately required, since $+$ is originally defined only as a binary operation.)
Compare this with $a-b-c$: in general $(a-b)-c\ne a-(b-c)$, so the expression $a-b-c$ is uninterpretable without some convention, e.g., work from left to right. When the operation is associative, no such convention is required.
A: $+$ is initially defined only as a binary operation.  Then we can define $a+b+c$ as either $(a+b)+c$ or $a+(b+c)$; associativity says it doesn't matter which one we use, since they are equal.  But until that definition is made, the expression $a+b+c$ has no meaning.
