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

  r1, r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            app#_A(0,0) = 0
            app#_A(0,1) = 0
            app#_A(1,0) = 0
            app#_A(1,1) = 0
            apply_1_A(0) = 0
            apply_1_A(1) = 0
            apply_2#_A(0,0) = 0
            apply_2#_A(0,1) = 0
            apply_2#_A(1,0) = 0
            apply_2#_A(1,1) = 0
            app_A(0,0) = 0
            app_A(0,1) = 0
            app_A(1,0) = 0
            app_A(1,1) = 0
            apply_0_A = 0
            apply_2_A(0,0) = 0
            apply_2_A(0,1) = 0
            apply_2_A(1,0) = 0
            apply_2_A(1,1) = 0
            app#!_A(0,0) = 1
            app#!_A(0,1) = 1
            app#!_A(1,0) = 0
            app#!_A(1,1) = 0
            apply_1!_A(0) = 1
            apply_1!_A(1) = 0
            apply_2#!_A(0,0) = 1
            apply_2#!_A(0,1) = 1
            apply_2#!_A(1,0) = 0
            apply_2#!_A(1,1) = 0
            app!_A(0,0) = 1
            app!_A(0,1) = 0
            app!_A(1,0) = 0
            app!_A(1,1) = 0
            apply_0!_A = 0
            apply_2!_A(0,0) = 1
            apply_2!_A(0,1) = 0
            apply_2!_A(1,0) = 0
            apply_2!_A(1,1) = 0
  
    partial status:
  
      pi(app#) = [1]
      pi(apply_1) = [1]
      pi(apply_2#) = [1]
      pi(app) = [2]
      pi(apply_0) = []
      pi(apply_2) = [2]

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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:

  (no SCCs)