YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(X),X) -> f#(X,a(X))
   f#(X,c(X)) -> f#(s(X),X)
  TRS:
   f(s(X),X) -> f(X,a(X))
   f(X,c(X)) -> f(s(X),X)
   f(X,X) -> c(X)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0, x1) = -4x0 + x1 + -16,
    
    [c](x0) = 1x0 + 3,
    
    [a](x0) = 1,
    
    [f](x0, x1) = x0 + 4x1 + 4,
    
    [s](x0) = x0 + 6
   orientation:
    f#(s(X),X) = X + 2 >= -4X + 1 = f#(X,a(X))
    
    f#(X,c(X)) = 1X + 3 >= X + 2 = f#(s(X),X)
    
    f(s(X),X) = 4X + 6 >= X + 5 = f(X,a(X))
    
    f(X,c(X)) = 5X + 7 >= 4X + 6 = f(s(X),X)
    
    f(X,X) = 4X + 4 >= 1X + 3 = c(X)
   problem:
    DPs:
     
    TRS:
     f(s(X),X) -> f(X,a(X))
     f(X,c(X)) -> f(s(X),X)
     f(X,X) -> c(X)
   Qed