YES
We show the termination of the TRS R:

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

Instead we prove termination of the uncurried system:

  app(apply_0(),y) -> apply_1(y)
  app(apply_1(x1),y) -> apply_2(x1,y)
  apply_2(f,x) -> app(f,x)


-- SCC decomposition.

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

p1: app#(apply_1(x1),y) -> apply_2#(x1,y)
p2: apply_2#(f,x) -> app#(f,x)

and R consists of:

r1: app(apply_0(),y) -> apply_1(y)
r2: app(apply_1(x1),y) -> apply_2(x1,y)
r3: apply_2(f,x) -> app(f,x)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: app#(apply_1(x1),y) -> apply_2#(x1,y)
p2: apply_2#(f,x) -> app#(f,x)

and R consists of:

r1: app(apply_0(),y) -> apply_1(y)
r2: app(apply_1(x1),y) -> apply_2(x1,y)
r3: apply_2(f,x) -> app(f,x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{2, x1}
        apply_1_A(x1) = x1 + 5
        apply_2#_A(x1,x2) = max{3, x1 + 1}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            app#_A(x1,x2) = 0
            apply_1_A(x1) = 2
            apply_2#_A(x1,x2) = 0
            app#!_A(x1,x2) = x1 + x2 + 3
            apply_1!_A(x1) = 1
            apply_2#!_A(x1,x2) = 2
  
    partial status:
  
      pi(app#) = []
      pi(apply_1) = []
      pi(apply_2#) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.