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.