YES

We show the termination of the TRS R:

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

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: app#(app(apply(),f),x) -> app#(f,x)

and R consists of:

r1: app(app(apply(),f),x) -> app(f,x)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: app#(app(apply(),f),x) -> app#(f,x)

and R consists of:

r1: app(app(apply(),f),x) -> app(f,x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 + 1, x2 + 1}
          app_A(x1,x2) = x2
          apply_A = 2
    
      precedence:
      
        app# = app = apply
      
      partial status:
    
        pi(app#) = [1]
        pi(app) = [2]
        pi(apply) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 1}
          app_A(x1,x2) = x2
          apply_A = 3
    
      precedence:
      
        app# = app = apply
      
      partial status:
    
        pi(app#) = []
        pi(app) = [2]
        pi(apply) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.