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)
  Usable Rule Processor:
   DPs:
    f#(s(X),X) -> f#(X,a(X))
    f#(X,c(X)) -> f#(s(X),X)
   TRS:
    
   CDG Processor:
    DPs:
     f#(s(X),X) -> f#(X,a(X))
     f#(X,c(X)) -> f#(s(X),X)
    TRS:
     
    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)
    Restore Modifier:
     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)
     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)
      Matrix Interpretation Processor:
       dimension: 1
       interpretation:
        [f#](x0, x1) = x1,
        
        [c](x0) = x0 + 1,
        
        [a](x0) = 0,
        
        [f](x0, x1) = x0 + x1 + 1,
        
        [s](x0) = 1
       orientation:
        f#(X,c(X)) = X + 1 >= X = f#(s(X),X)
        
        f(s(X),X) = X + 2 >= X + 1 = f(X,a(X))
        
        f(X,c(X)) = 2X + 2 >= X + 2 = f(s(X),X)
        
        f(X,X) = 2X + 1 >= X + 1 = 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)
      Matrix Interpretation Processor:
       dimension: 1
       interpretation:
        [f#](x0, x1) = x0 + x1,
        
        [c](x0) = 1,
        
        [a](x0) = 0,
        
        [f](x0, x1) = x0 + x1 + 1,
        
        [s](x0) = 1
       orientation:
        f#(s(X),X) = X + 1 >= X = f#(X,a(X))
        
        f(s(X),X) = X + 2 >= X + 1 = f(X,a(X))
        
        f(X,c(X)) = X + 2 >= X + 2 = f(s(X),X)
        
        f(X,X) = 2X + 1 >= 1 = 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