NO

Problem:
 app(app(const(),x),y) -> x
 app(app(app(subst(),f),g),x) -> app(app(f,x),app(g,x))
 app(app(fix(),f),x) -> app(app(f,app(fix(),f)),x)

Proof:
 Extended Uncurrying Processor:
  application symbol: app
  symbol table:
   fix ==> fix0/0 fix1/1 fix2/2
   subst ==> subst0/0 subst1/1 subst2/2 subst3/3
   const ==> const0/0 const1/1 const2/2
   
  uncurry-rules:
   app(const1(x4),x5) -> const2(x4,x5)
   app(const0(),x4) -> const1(x4)
   app(subst2(x7,x8),x9) -> subst3(x7,x8,x9)
   app(subst1(x7),x8) -> subst2(x7,x8)
   app(subst0(),x7) -> subst1(x7)
   app(fix1(x11),x12) -> fix2(x11,x12)
   app(fix0(),x11) -> fix1(x11)
  eta-rules:
   
  problem:
   const2(x,y) -> x
   subst3(f,g,x) -> app(app(f,x),app(g,x))
   fix2(f,x) -> app(app(f,fix1(f)),x)
   app(const1(x4),x5) -> const2(x4,x5)
   app(const0(),x4) -> const1(x4)
   app(subst2(x7,x8),x9) -> subst3(x7,x8,x9)
   app(subst1(x7),x8) -> subst2(x7,x8)
   app(subst0(),x7) -> subst1(x7)
   app(fix1(x11),x12) -> fix2(x11,x12)
   app(fix0(),x11) -> fix1(x11)
  Unfolding Processor:
   loop length: 5
   terms:
    fix2(subst1(x85658),x86396)
    app(app(subst1(x85658),fix1(subst1(x85658))),x86396)
    app(subst2(x85658,fix1(subst1(x85658))),x86396)
    subst3(x85658,fix1(subst1(x85658)),x86396)
    app(app(x85658,x86396),app(fix1(subst1(x85658)),x86396))
   context: app(app(x85658,x86396),[])
   substitution:
    x85658 -> x85658
    x86396 -> x86396
   Qed