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

Proof:
 Open