YES

Problem:
 f(c(s(x),y)) -> f(c(x,s(y)))
 g(c(x,s(y))) -> g(c(s(x),y))
 g(s(f(x))) -> g(f(x))

Proof:
 DP Processor:
  DPs:
   f#(c(s(x),y)) -> f#(c(x,s(y)))
   g#(c(x,s(y))) -> g#(c(s(x),y))
   g#(s(f(x))) -> g#(f(x))
  TRS:
   f(c(s(x),y)) -> f(c(x,s(y)))
   g(c(x,s(y))) -> g(c(s(x),y))
   g(s(f(x))) -> g(f(x))
  CDG Processor:
   DPs:
    f#(c(s(x),y)) -> f#(c(x,s(y)))
    g#(c(x,s(y))) -> g#(c(s(x),y))
    g#(s(f(x))) -> g#(f(x))
   TRS:
    f(c(s(x),y)) -> f(c(x,s(y)))
    g(c(x,s(y))) -> g(c(s(x),y))
    g(s(f(x))) -> g(f(x))
   graph:
    
   Qed