YES

Problem:
 app(app(fmap(),fnil()),x) -> nil()
 app(app(fmap(),app(app(fcons(),f),t)),x) -> app(app(cons(),app(f,x)),app(app(fmap(),t),x))

Proof:
 Uncurry Processor:
  fmap2(fnil(),x) -> nil()
  fmap2(fcons2(f,t),x) -> cons2(app(f,x),fmap2(t,x))
  app(fmap1(x3),x4) -> fmap2(x3,x4)
  app(fmap(),x4) -> fmap1(x4)
  app(fcons1(x3),x4) -> fcons2(x3,x4)
  app(fcons(),x4) -> fcons1(x4)
  app(cons1(x3),x4) -> cons2(x3,x4)
  app(cons(),x4) -> cons1(x4)
  DP Processor:
   DPs:
    fmap{2,#}(fcons2(f,t),x) -> fmap{2,#}(t,x)
    fmap{2,#}(fcons2(f,t),x) -> app#(f,x)
    app#(fmap1(x3),x4) -> fmap{2,#}(x3,x4)
   TRS:
    fmap2(fnil(),x) -> nil()
    fmap2(fcons2(f,t),x) -> cons2(app(f,x),fmap2(t,x))
    app(fmap1(x3),x4) -> fmap2(x3,x4)
    app(fmap(),x4) -> fmap1(x4)
    app(fcons1(x3),x4) -> fcons2(x3,x4)
    app(fcons(),x4) -> fcons1(x4)
    app(cons1(x3),x4) -> cons2(x3,x4)
    app(cons(),x4) -> cons1(x4)
   TDG Processor:
    DPs:
     fmap{2,#}(fcons2(f,t),x) -> fmap{2,#}(t,x)
     fmap{2,#}(fcons2(f,t),x) -> app#(f,x)
     app#(fmap1(x3),x4) -> fmap{2,#}(x3,x4)
    TRS:
     fmap2(fnil(),x) -> nil()
     fmap2(fcons2(f,t),x) -> cons2(app(f,x),fmap2(t,x))
     app(fmap1(x3),x4) -> fmap2(x3,x4)
     app(fmap(),x4) -> fmap1(x4)
     app(fcons1(x3),x4) -> fcons2(x3,x4)
     app(fcons(),x4) -> fcons1(x4)
     app(cons1(x3),x4) -> cons2(x3,x4)
     app(cons(),x4) -> cons1(x4)
    graph:
     app#(fmap1(x3),x4) -> fmap{2,#}(x3,x4) ->
     fmap{2,#}(fcons2(f,t),x) -> app#(f,x)
     app#(fmap1(x3),x4) -> fmap{2,#}(x3,x4) ->
     fmap{2,#}(fcons2(f,t),x) -> fmap{2,#}(t,x)
     fmap{2,#}(fcons2(f,t),x) -> app#(f,x) ->
     app#(fmap1(x3),x4) -> fmap{2,#}(x3,x4)
     fmap{2,#}(fcons2(f,t),x) -> fmap{2,#}(t,x) ->
     fmap{2,#}(fcons2(f,t),x) -> app#(f,x)
     fmap{2,#}(fcons2(f,t),x) -> fmap{2,#}(t,x) -> fmap{2,#}(fcons2(f,t),x) -> fmap{2,#}(t,x)
    Subterm Criterion Processor:
     simple projection:
      pi(fmap{2,#}) = 0
      pi(app#) = 0
     problem:
      DPs:
       
      TRS:
       fmap2(fnil(),x) -> nil()
       fmap2(fcons2(f,t),x) -> cons2(app(f,x),fmap2(t,x))
       app(fmap1(x3),x4) -> fmap2(x3,x4)
       app(fmap(),x4) -> fmap1(x4)
       app(fcons1(x3),x4) -> fcons2(x3,x4)
       app(fcons(),x4) -> fcons1(x4)
       app(cons1(x3),x4) -> cons2(x3,x4)
       app(cons(),x4) -> cons1(x4)
     Qed