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

  (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{3, x1 - 2, x2 - 14}
        F_1_A(x1) = max{6, x1 - 1}
        F_2#_A(x1,x2) = max{4, x1 - 5, x2 - 14}
        F_2_A(x1,x2) = max{x1 + 4, x2 - 8}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 7, x2 - 5}
          F_1_A(x1) = x1 + 4
          F_2#_A(x1,x2) = max{0, x1 - 3, x2 - 5}
          F_2_A(x1,x2) = max{2, x1 - 4, x2 - 2}
          app#!_A(x1,x2) = max{1, x1 - 12, x2 - 11}
          F_1!_A(x1) = max{2, x1 - 7}
          F_2#!_A(x1,x2) = max{1, x1 - 8, x2 - 11}
          F_2!_A(x1,x2) = max{3, x1 - 4, x2 - 10}
  
    partial status:
  
      pi(app#) = []
      pi(F_1) = []
      pi(F_2#) = []
      pi(F_2) = [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)