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.