YES

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

Proof:
 DP Processor:
  DPs:
   add#(s(x),y) -> add#(x,y)
  TRS:
   add(0(),x) -> x
   add(s(x),y) -> s(add(x,y))
  Usable Rule Processor:
   DPs:
    add#(s(x),y) -> add#(x,y)
   TRS:
    
   Restore Modifier:
    DPs:
     add#(s(x),y) -> add#(x,y)
    TRS:
     add(0(),x) -> x
     add(s(x),y) -> s(add(x,y))
    SCC Processor:
     #sccs: 1
     #rules: 1
     #arcs: 1/1
     DPs:
      add#(s(x),y) -> add#(x,y)
     TRS:
      add(0(),x) -> x
      add(s(x),y) -> s(add(x,y))
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [add#](x0, x1) = x0 + 1,
       
       [s](x0) = x0 + 1,
       
       [add](x0, x1) = x0 + x1,
       
       [0] = 0
      orientation:
       add#(s(x),y) = x + 2 >= x + 1 = add#(x,y)
       
       add(0(),x) = x >= x = x
       
       add(s(x),y) = x + y + 1 >= x + y + 1 = s(add(x,y))
      problem:
       DPs:
        
       TRS:
        add(0(),x) -> x
        add(s(x),y) -> s(add(x,y))
      Qed