YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = 4x0 + 2,
   
   [f](x0, x1) = 4x0 + 2x1,
   
   [s](x0) = 2x0
  orientation:
   f(s(x),y) = 8x + 2y >= 8x = f(x,s(x))
   
   f(x,s(y)) = 4x + 4y >= 2x + 4y = f(y,x)
   
   f(c(x),y) = 16x + 2y + 8 >= 8x = f(x,s(x))
  problem:
   f(s(x),y) -> f(x,s(x))
   f(x,s(y)) -> f(y,x)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [f](x0, x1) = 2x0 + x1,
    
    [s](x0) = 2x0 + 4
   orientation:
    f(s(x),y) = 4x + y + 8 >= 4x + 4 = f(x,s(x))
    
    f(x,s(y)) = 2x + 2y + 4 >= x + 2y = f(y,x)
   problem:
    
   Qed