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=2
  
  interpretation:
                 [1 0]     [2 2]     [0]
   [f](x0, x1) = [0 0]x0 + [0 0]x1 + [2],
   
             [1 0]     [0]
   [c](x0) = [2 2]x0 + [1],
   
             [1 0]  
   [s](x0) = [0 0]x0
  orientation:
               [1 0]    [6 4]    [2]    [1 0]    [6 4]    [0]                 
   f(x,c(y)) = [0 0]x + [0 0]y + [2] >= [0 0]x + [0 0]y + [2] = f(x,s(f(y,y)))
   
               [1 0]    [2 2]    [0]    [1 0]    [2 0]    [0]               
   f(s(x),y) = [0 0]x + [0 0]y + [2] >= [0 0]x + [0 0]y + [2] = f(x,s(c(y)))
  problem:
   f(s(x),y) -> f(x,s(c(y)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                  [1 1 1]     [1 0 0]  
    [f](x0, x1) = [0 1 1]x0 + [0 0 1]x1
                  [1 0 0]     [0 0 0]  ,
    
              [1 0 0]  
    [c](x0) = [0 0 1]x0
              [0 0 0]  ,
    
              [1 0 0]     [0]
    [s](x0) = [0 0 0]x0 + [0]
              [0 1 1]     [1]
   orientation:
                [1 1 1]    [1 0 0]    [1]    [1 1 1]    [1 0 0]    [0]               
    f(s(x),y) = [0 1 1]x + [0 0 1]y + [1] >= [0 1 1]x + [0 0 1]y + [1] = f(x,s(c(y)))
                [1 0 0]    [0 0 0]    [0]    [1 0 0]    [0 0 0]    [0]               
   problem:
    
   Qed