YES

We show the termination of the TRS R:

  app(app(app(compose(),f),g),x) -> app(f,app(g,x))

-- SCC decomposition.

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

p1: app#(app(app(compose(),f),g),x) -> app#(f,app(g,x))
p2: app#(app(app(compose(),f),g),x) -> app#(g,x)

and R consists of:

r1: app(app(app(compose(),f),g),x) -> app(f,app(g,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#(app(app(compose(),f),g),x) -> app#(f,app(g,x))
p2: app#(app(app(compose(),f),g),x) -> app#(g,x)

and R consists of:

r1: app(app(app(compose(),f),g),x) -> app(f,app(g,x))

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        app#_A(x1,x2) = ((0,1),(0,1)) x1
        app_A(x1,x2) = x1 + x2 + (1,1)
        compose_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        app#_A(x1,x2) = (0,0)
        app_A(x1,x2) = (1,1)
        compose_A() = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        app#_A(x1,x2) = (0,0)
        app_A(x1,x2) = (1,1)
        compose_A() = (1,1)
    

The next rules are strictly ordered:

  p1, p2

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