If, in a Boolean lattice $(X,\vee,\wedge,0,1,')$ (i.e. a complemented distributive lattice), we define $x+y=(x'\wedge y)\vee(x\wedge y')$, is there an elegant way to prove that $(x+y)+z=x+(y+z)$ rather than just opening everything?
For example, in a ring with identity $(X,+,0,-,\cdot,1)$, if we define $x\vee y=x+y-xy$, we easily get the associativity of $\vee$ by opening everything, but we can also do this: notice that $1-(x\vee y)=1-x-y+xy=(1-x)(1-y)$, therefore:
$1-(x\vee(y\vee z))=(1-x)(1-(y\vee z))=(1-x)((1-y)(1-z))=((1-x)(1-y))(1-z)=(1-(x\vee y))(1-z)=1-((x\vee y)\vee z)$
Treating true as $1$ and false as $0$, the xor of $x,\,y$ has the same parity as their sum. (If you want to work in algebra, it's modulo $2$ arithmetic.) Since addition associates, xor associates up to parity. But no distinct truth values are of the same parity, so xor associates.