YES

Problem:
 -(x,s(y)) -> -(d(x),y)
 d(s(x)) -> x
 -(s(x),s(y)) -> -(x,y)
 -(d(x),y) -> -(x,s(y))

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  d(s(x)) -> x
  -(s(x),s(y)) -> -(x,y)
  -(x,s(y)) -> -(d(x),y)
  -(d(x),y) -> -(x,s(y))
  
   T' = (P union S) with
  
   TRS P:-(x,s(y)) -> -(d(x),y)
         -(d(x),y) -> -(x,s(y))
  
   TRS S:d(s(x)) -> x
         -(s(x),s(y)) -> -(x,y)
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  
   CP(S,P union P^-1) = 
  -(x118,y) = -(s(x118),s(y)), -(x119,y) = -(d(s(x119)),y)
  
   CP(P union P^-1,S) = 
  -(d(s(x)),y) = -(x,y)
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [-](x0, x1) = x0 + x1,
    
    [s](x0) = x0,
    
    [d](x0) = x0 + 1
   orientation:
    d(s(x)) = x + 1 >= x = x
    
    -(s(x),s(y)) = x + y >= x + y = -(x,y)
   problem:
    -(s(x),s(y)) -> -(x,y)
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [-](x0, x1) = 3x0 + 6x1,
     
     [s](x0) = x0 + 3
    orientation:
     -(s(x),s(y)) = 3x + 6y + 27 >= 3x + 6y = -(x,y)
    problem:
     
    Qed