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

Proof:
 Open