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))

-- SCC decomposition.

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

p1: app#(app(F(),app(app(F(),f),x)),x) -> app#(app(F(),app(G(),app(app(F(),f),x))),app(f,x))
p2: app#(app(F(),app(app(F(),f),x)),x) -> app#(F(),app(G(),app(app(F(),f),x)))
p3: app#(app(F(),app(app(F(),f),x)),x) -> app#(G(),app(app(F(),f),x))
p4: app#(app(F(),app(app(F(),f),x)),x) -> app#(f,x)

and R consists of:

r1: app(app(F(),app(app(F(),f),x)),x) -> app(app(F(),app(G(),app(app(F(),f),x))),app(f,x))

The estimated dependency graph contains the following SCCs:

  {p4}


-- Reduction pair.

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

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

and R consists of:

r1: app(app(F(),app(app(F(),f),x)),x) -> app(app(F(),app(G(),app(app(F(),f),x))),app(f,x))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,1,1,0),(1,1,0,1)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x2
      app_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,0,1)) x2
      F_A() = (1,1,1,1)

The next rules are strictly ordered:

  p1
  r1

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