YES

Problem:
 f(s(X),X) -> f(X,a(X))
 f(X,c(X)) -> f(s(X),X)
 f(X,X) -> c(X)

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