YES
We show the termination of the TRS R:

  app(app(F(),app(app(F(),f),x)),x) -> app(app(F(),app(G(),app(app(F(),f),x))),app(f,x))

Instead we prove termination of the uncurried system:

  app(F_0(),y) -> F_1(y)
  app(F_1(x1),y) -> F_2(x1,y)
  app(G_0(),y) -> G_1(y)
  F_2(F_2(f,x),x) -> F_2(G_1(F_2(f,x)),app(f,x))


-- SCC decomposition.

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

p1: app#(F_1(x1),y) -> F_2#(x1,y)
p2: F_2#(F_2(f,x),x) -> F_2#(G_1(F_2(f,x)),app(f,x))
p3: F_2#(F_2(f,x),x) -> app#(f,x)

and R consists of:

r1: app(F_0(),y) -> F_1(y)
r2: app(F_1(x1),y) -> F_2(x1,y)
r3: app(G_0(),y) -> G_1(y)
r4: F_2(F_2(f,x),x) -> F_2(G_1(F_2(f,x)),app(f,x))

The estimated dependency graph contains the following SCCs:

  {p1, p3}


-- Reduction pair.

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

p1: app#(F_1(x1),y) -> F_2#(x1,y)
p2: F_2#(F_2(f,x),x) -> app#(f,x)

and R consists of:

r1: app(F_0(),y) -> F_1(y)
r2: app(F_1(x1),y) -> F_2(x1,y)
r3: app(G_0(),y) -> G_1(y)
r4: F_2(F_2(f,x),x) -> F_2(G_1(F_2(f,x)),app(f,x))

The set of usable rules consists of

  r1, r2, r3, r4

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
            F_1_A(0) = 1
            F_1_A(1) = 1
            F_2#_A(0,0) = 0
            F_2#_A(0,1) = 0
            F_2#_A(1,0) = 0
            F_2#_A(1,1) = 0
            F_2_A(0,0) = 1
            F_2_A(0,1) = 1
            F_2_A(1,0) = 1
            F_2_A(1,1) = 1
            app_A(0,0) = 1
            app_A(0,1) = 1
            app_A(1,0) = 1
            app_A(1,1) = 1
            F_0_A = 1
            G_0_A = 1
            G_1_A(0) = 1
            G_1_A(1) = 1
            app#!_A(0,0) = 1
            app#!_A(0,1) = 0
            app#!_A(1,0) = 1
            app#!_A(1,1) = 1
            F_1!_A(0) = 0
            F_1!_A(1) = 0
            F_2#!_A(0,0) = 1
            F_2#!_A(0,1) = 1
            F_2#!_A(1,0) = 1
            F_2#!_A(1,1) = 1
            F_2!_A(0,0) = 0
            F_2!_A(0,1) = 0
            F_2!_A(1,0) = 1
            F_2!_A(1,1) = 1
            app!_A(0,0) = 0
            app!_A(0,1) = 1
            app!_A(1,0) = 1
            app!_A(1,1) = 1
            F_0!_A = 1
            G_0!_A = 1
            G_1!_A(0) = 1
            G_1!_A(1) = 0
  
    partial status:
  
      pi(app#) = [1, 2]
      pi(F_1) = [1]
      pi(F_2#) = [1, 2]
      pi(F_2) = [1, 2]
      pi(app) = [1, 2]
      pi(F_0) = []
      pi(G_0) = []
      pi(G_1) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: app#(F_1(x1),y) -> F_2#(x1,y)

and R consists of:

r1: app(F_0(),y) -> F_1(y)
r2: app(F_1(x1),y) -> F_2(x1,y)
r3: app(G_0(),y) -> G_1(y)
r4: F_2(F_2(f,x),x) -> F_2(G_1(F_2(f,x)),app(f,x))

The estimated dependency graph contains the following SCCs:

  (no SCCs)