YES

Problem:
 f(x,c(y)) -> f(x,s(f(y,y)))
 f(s(x),y) -> f(x,s(c(y)))

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]  
   [s](x0) = [0 1 1]x0
             [0 0 0]  ,
   
                 [1 0 0]     [1 0 1]     [0]
   [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [1]
                 [0 1 0]     [0 0 0]     [0],
   
             [1 0 0]     [0]
   [c](x0) = [0 0 0]x0 + [0]
             [1 1 1]     [1]
  orientation:
               [1 0 0]    [2 1 1]    [1]    [1 0 0]    [2 0 1]    [0]                 
   f(x,c(y)) = [0 0 0]x + [0 0 0]y + [1] >= [0 0 0]x + [0 0 0]y + [1] = f(x,s(f(y,y)))
               [0 1 0]    [0 0 0]    [0]    [0 1 0]    [0 0 0]    [0]                 
   
               [1 0 0]    [1 0 1]    [0]    [1 0 0]    [1 0 0]    [0]               
   f(s(x),y) = [0 0 0]x + [0 0 0]y + [1] >= [0 0 0]x + [0 0 0]y + [1] = f(x,s(c(y)))
               [0 1 1]    [0 0 0]    [0]    [0 1 0]    [0 0 0]    [0]               
  problem:
   f(s(x),y) -> f(x,s(c(y)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [0]
    [s](x0) = x0 + [1]
                   [0],
    
                  [1 1 0]     [1 0 0]  
    [f](x0, x1) = [1 1 0]x0 + [0 0 0]x1
                  [0 0 0]     [0 0 0]  ,
    
              [1 0 0]  
    [c](x0) = [1 0 0]x0
              [1 0 0]  
   orientation:
                [1 1 0]    [1 0 0]    [1]    [1 1 0]    [1 0 0]                
    f(s(x),y) = [1 1 0]x + [0 0 0]y + [1] >= [1 1 0]x + [0 0 0]y = f(x,s(c(y)))
                [0 0 0]    [0 0 0]    [0]    [0 0 0]    [0 0 0]                
   problem:
    
   Qed