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:
         [0]
   [0] = [1]
         [0],
   
             [1 0 0]  
   [s](x0) = [0 0 0]x0
             [0 1 0]  ,
   
                 [1 0 1]     [1 0 0]  
   [h](x0, x1) = [1 0 1]x0 + [1 0 0]x1
                 [0 0 0]     [0 0 0]  ,
   
                 [1 1 0]     [1 0 0]  
   [c](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                 [0 0 1]     [0 0 1]  
  orientation:
                 [1 0 1]    [1 1 0]    [1 0 0]     [1 0 1]    [1 1 0]    [1 0 0]                  
   h(x,c(y,z)) = [1 0 1]x + [1 1 0]y + [1 0 0]z >= [1 0 1]x + [1 1 0]y + [1 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 1 0]    [1 0 1]    [1 0 0]    [1]    [1 1 0]    [1 0 1]    [1 0 0]                         
   h(c(s(x),c(s(0()),y)),z) = [1 1 0]x + [1 0 1]y + [1 0 0]z + [1] >= [1 1 0]x + [1 0 1]y + [1 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=3
   
   interpretation:
              [1 0 0]  
    [s](x0) = [0 0 0]x0
              [0 0 0]  ,
    
                  [1 0 0]     [1 1 0]  
    [h](x0, x1) = [0 1 0]x0 + [0 0 1]x1
                  [0 0 0]     [0 0 1]  ,
    
                  [1 0 0]          [0]
    [c](x0, x1) = [0 0 0]x0 + x1 + [1]
                  [0 0 0]          [1]
   orientation:
                  [1 0 0]    [1 0 0]    [1 1 0]    [1]    [1 0 0]    [1 0 0]    [1 1 0]    [0]                 
    h(x,c(y,z)) = [0 1 0]x + [0 0 0]y + [0 0 1]z + [1] >= [0 1 0]x + [0 0 0]y + [0 0 1]z + [1] = h(c(s(y),x),z)
                  [0 0 0]    [0 0 0]    [0 0 1]    [1]    [0 0 0]    [0 0 0]    [0 0 1]    [0]                 
   problem:
    
   Qed