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)))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
    [h#](x0, x1) = [0 1]x0 + [1 0]x1,
    
          [0]
    [0] = [2],
    
              [0 2]  
    [s](x0) = [0 0]x0,
    
                  [0 0]     [0 0]     [1]
    [h](x0, x1) = [0 1]x0 + [1 0]x1 + [0],
    
                  [0 2]          [1]
    [c](x0, x1) = [1 0]x0 + x1 + [0]
   orientation:
    h#(x,c(y,z)) = [0 1]x + [0 2]y + [1 0]z + [1] >= [0 1]x + [0 2]y + [1 0]z = h#(c(s(y),x),z)
    
    h#(c(s(x),c(s(0()),y)),z) = [0 2]x + [0 1]y + [1 0]z + [4] >= [0 2]x + [0 1]y + [1 0]z + [2] = h#(y,c(s(0()),c(x,z)))
    
                  [0 0]    [0 0]    [0 0]    [1]    [0 0]    [0 0]    [0 0]    [1]                 
    h(x,c(y,z)) = [0 1]x + [0 2]y + [1 0]z + [1] >= [0 1]x + [0 2]y + [1 0]z + [0] = h(c(s(y),x),z)
    
                               [0 0]    [0 0]    [0 0]    [1]    [0 0]    [0 0]    [0 0]    [1]                        
    h(c(s(x),c(s(0()),y)),z) = [0 2]x + [0 1]y + [1 0]z + [4] >= [0 2]x + [0 1]y + [1 0]z + [2] = h(y,c(s(0()),c(x,z)))
   problem:
    DPs:
     
    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)))
   Qed