YES

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

Proof:
 DP Processor:
  DPs:
   +#(s(x),y) -> +#(x,y)
   -#(s(x),s(y)) -> -#(x,y)
  TRS:
   +(0(),y) -> y
   +(s(x),y) -> s(+(x,y))
   -(0(),y) -> 0()
   -(x,0()) -> x
   -(s(x),s(y)) -> -(x,y)
  Usable Rule Processor:
   DPs:
    +#(s(x),y) -> +#(x,y)
    -#(s(x),s(y)) -> -#(x,y)
   TRS:
    
   TDG Processor:
    DPs:
     +#(s(x),y) -> +#(x,y)
     -#(s(x),s(y)) -> -#(x,y)
    TRS:
     
    graph:
     -#(s(x),s(y)) -> -#(x,y) -> -#(s(x),s(y)) -> -#(x,y)
     +#(s(x),y) -> +#(x,y) -> +#(s(x),y) -> +#(x,y)
    CDG Processor:
     DPs:
      +#(s(x),y) -> +#(x,y)
      -#(s(x),s(y)) -> -#(x,y)
     TRS:
      
     graph:
      
     Qed