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 reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((0,1),(0,0)) x1 + x2 + (2,4)
          app_A(x1,x2) = ((0,1),(1,1)) x1 + x2 + (0,1)
          F_A() = (1,1)
  
    precedence:
    
      app > app# = F
    
    partial status:
  
      pi(app#) = []
      pi(app) = [2]
      pi(F) = []

The next rules are strictly ordered:

  p1

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