YES

Problem:
 h(x,c(y,z)) -> h(c(s(y),x),z)
 h(c(s(x),c(s(0()),y)),z) -> h(y,c(s(0()),c(x,z)))

Proof:
 DP Processor:
  DPs:
   h#(x,c(y,z)) -> h#(c(s(y),x),z)
   h#(c(s(x),c(s(0()),y)),z) -> h#(y,c(s(0()),c(x,z)))
  TRS:
   h(x,c(y,z)) -> h(c(s(y),x),z)
   h(c(s(x),c(s(0()),y)),z) -> h(y,c(s(0()),c(x,z)))
  Usable Rule Processor:
   DPs:
    h#(x,c(y,z)) -> h#(c(s(y),x),z)
    h#(c(s(x),c(s(0()),y)),z) -> h#(y,c(s(0()),c(x,z)))
   TRS:
    
   Matrix Interpretation Processor: dim=2
    
    usable rules:
     
    interpretation:
     [h#](x0, x1) = [2 0]x0 + [0 1]x1,
     
           [0]
     [0] = [3],
     
               [0 1]  
     [s](x0) = [0 0]x0,
     
                   [1 0]          [0]
     [c](x0, x1) = [0 2]x0 + x1 + [2]
    orientation:
     h#(x,c(y,z)) = [2 0]x + [0 2]y + [0 1]z + [2] >= [2 0]x + [0 2]y + [0 1]z = h#(c(s(y),x),z)
     
     h#(c(s(x),c(s(0()),y)),z) = [0 2]x + [2 0]y + [0 1]z + [6] >= [0 2]x + [2 0]y + [0 1]z + [4] = h#(y,c(s(0()),c(x,z)))
    problem:
     DPs:
      
     TRS:
      
    Qed