YES

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

Proof:
 DP Processor:
  DPs:
   f#(a(),a()) -> f#(a(),b())
   f#(a(),b()) -> f#(s(a()),c())
   f#(s(X),c()) -> f#(X,c())
   f#(c(),c()) -> f#(a(),a())
  TRS:
   f(a(),a()) -> f(a(),b())
   f(a(),b()) -> f(s(a()),c())
   f(s(X),c()) -> f(X,c())
   f(c(),c()) -> f(a(),a())
  Usable Rule Processor:
   DPs:
    f#(a(),a()) -> f#(a(),b())
    f#(a(),b()) -> f#(s(a()),c())
    f#(s(X),c()) -> f#(X,c())
    f#(c(),c()) -> f#(a(),a())
   TRS:
    
   Matrix Interpretation Processor: dim=3
    
    usable rules:
     
    interpretation:
     [f#](x0, x1) = [2 0 0]x0 + [0 1 1]x1,
     
           [3]
     [c] = [0]
           [0],
     
               [3 0 0]     [1]
     [s](x0) = [0 1 1]x0 + [1]
               [0 0 2]     [0],
     
           [0]
     [b] = [1]
           [2],
     
           [0]
     [a] = [2]
           [3]
    orientation:
     f#(a(),a()) = 5 >= 3 = f#(a(),b())
     
     f#(a(),b()) = 3 >= 2 = f#(s(a()),c())
     
     f#(s(X),c()) = [6 0 0]X + [2] >= [2 0 0]X = f#(X,c())
     
     f#(c(),c()) = 6 >= 5 = f#(a(),a())
    problem:
     DPs:
      
     TRS:
      
    Qed