YES

Problem:
 f(c(X,s(Y))) -> f(c(s(X),Y))
 g(c(s(X),Y)) -> f(c(X,s(Y)))

Proof:
 DP Processor:
  DPs:
   f#(c(X,s(Y))) -> f#(c(s(X),Y))
   g#(c(s(X),Y)) -> f#(c(X,s(Y)))
  TRS:
   f(c(X,s(Y))) -> f(c(s(X),Y))
   g(c(s(X),Y)) -> f(c(X,s(Y)))
  TDG Processor:
   DPs:
    f#(c(X,s(Y))) -> f#(c(s(X),Y))
    g#(c(s(X),Y)) -> f#(c(X,s(Y)))
   TRS:
    f(c(X,s(Y))) -> f(c(s(X),Y))
    g(c(s(X),Y)) -> f(c(X,s(Y)))
   graph:
    g#(c(s(X),Y)) -> f#(c(X,s(Y))) -> f#(c(X,s(Y))) -> f#(c(s(X),Y))
    f#(c(X,s(Y))) -> f#(c(s(X),Y)) -> f#(c(X,s(Y))) -> f#(c(s(X),Y))
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(c(X,s(Y))) -> f#(c(s(X),Y))
    TRS:
     f(c(X,s(Y))) -> f(c(s(X),Y))
     g(c(s(X),Y)) -> f(c(X,s(Y)))
    CDG Processor:
     DPs:
      f#(c(X,s(Y))) -> f#(c(s(X),Y))
     TRS:
      f(c(X,s(Y))) -> f(c(s(X),Y))
      g(c(s(X),Y)) -> f(c(X,s(Y)))
     graph:
      
     Qed