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:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                 [1 1 0]     [1 0 0]  
   [h](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                 [0 0 0]     [0 0 0]  ,
   
         [0]
   [0] = [0]
         [1],
   
                 [1 0 1]     [1 0 0]  
   [c](x0, x1) = [0 1 0]x0 + [0 1 0]x1
                 [0 0 0]     [0 0 0]  ,
   
             [1 0 0]  
   [s](x0) = [0 0 1]x0
             [0 0 0]  
  orientation:
                 [1 1 0]    [1 0 1]    [1 0 0]     [1 1 0]    [1 0 1]    [1 0 0]                  
   h(x,c(y,z)) = [0 0 0]x + [0 0 0]y + [0 0 0]z >= [0 0 0]x + [0 0 0]y + [0 0 0]z = h(c(s(y),x),z)
                 [0 0 0]    [0 0 0]    [0 0 0]     [0 0 0]    [0 0 0]    [0 0 0]                  
   
                              [1 0 1]    [1 1 0]    [1 0 0]    [1]    [1 0 1]    [1 1 0]    [1 0 0]                         
   h(c(s(x),c(s(0()),y)),z) = [0 0 0]x + [0 0 0]y + [0 0 0]z + [0] >= [0 0 0]x + [0 0 0]y + [0 0 0]z = h(y,c(s(0()),c(x,z)))
                              [0 0 0]    [0 0 0]    [0 0 0]    [0]    [0 0 0]    [0 0 0]    [0 0 0]                         
  problem:
   h(x,c(y,z)) -> h(c(s(y),x),z)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [h](x0, x1) = 4x0 + 5x1 + 5,
    
    [c](x0, x1) = x0 + x1 + 5,
    
    [s](x0) = x0 + 1
   orientation:
    h(x,c(y,z)) = 4x + 5y + 5z + 30 >= 4x + 4y + 5z + 29 = h(c(s(y),x),z)
   problem:
    
   Qed