Problem:
 +(x,0()) -> x
 +(x,s(y)) -> s(+(x,y))
 +(0(),y) -> y
 +(s(x),y) -> s(+(x,y))
 +(+(x,y),z) -> +(x,+(y,z))
 +(x,y) -> +(y,x)
 s(s(x)) -> s(x)
 s(x) -> s(s(x))

Proof:
 Open