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)
  EDG 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)
   graph:
    f#(s(X),X) -> f#(X,a(X)) -> f#(s(X),X) -> f#(X,a(X))
    f#(X,c(X)) -> f#(s(X),X) -> f#(s(X),X) -> f#(X,a(X))
    f#(X,c(X)) -> f#(s(X),X) -> f#(X,c(X)) -> f#(s(X),X)
   SCC Processor:
    #sccs: 2
    #rules: 2
    #arcs: 3/4
    DPs:
     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) = -12x0 + x1,
      
      [c](x0) = 2x0,
      
      [a](x0) = x0,
      
      [f](x0, x1) = x0 + 4x1,
      
      [s](x0) = 5x0
     orientation:
      f#(X,c(X)) = 2X >= X = f#(s(X),X)
      
      f(s(X),X) = 5X >= 4X = f(X,a(X))
      
      f(X,c(X)) = 6X >= 5X = f(s(X),X)
      
      f(X,X) = 4X >= 2X = 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
    
    DPs:
     f#(s(X),X) -> f#(X,a(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) = x0 + 4x1 + 0,
      
      [c](x0) = 1x0 + 2,
      
      [a](x0) = 0,
      
      [f](x0, x1) = x0 + 4x1 + 4,
      
      [s](x0) = x0 + 5
     orientation:
      f#(s(X),X) = 4X + 5 >= X + 4 = f#(X,a(X))
      
      f(s(X),X) = 4X + 5 >= X + 4 = f(X,a(X))
      
      f(X,c(X)) = 5X + 6 >= 4X + 5 = f(s(X),X)
      
      f(X,X) = 4X + 4 >= 1X + 2 = 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