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)))
  Usable Rule Processor:
   DPs:
    f#(c(X,s(Y))) -> f#(c(s(X),Y))
    g#(c(s(X),Y)) -> f#(c(X,s(Y)))
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [g#](x0) = 2x0 + -16,
     
     [f#](x0) = x0,
     
     [c](x0, x1) = x1,
     
     [s](x0) = 1x0
    orientation:
     f#(c(X,s(Y))) = 1Y >= Y = f#(c(s(X),Y))
     
     g#(c(s(X),Y)) = 2Y + -16 >= 1Y = f#(c(X,s(Y)))
    problem:
     DPs:
      
     TRS:
      
    Qed