YES

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

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