YES

Problem:
 app(app(apply(),f),x) -> app(f,x)

Proof:
 DP Processor:
  DPs:
   app#(app(apply(),f),x) -> app#(f,x)
  TRS:
   app(app(apply(),f),x) -> app(f,x)
  Subterm Criterion Processor:
   simple projection:
    pi(app#) = 0
   problem:
    DPs:
     
    TRS:
     app(app(apply(),f),x) -> app(f,x)
   Qed