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:

  max/plus interpretations on natural numbers:
  
    app#_A(x1,x2) = x1 + 3
    app_A(x1,x2) = max{4, x1 + 1, x2}
    compose_A = 1

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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}


-- Reduction pair.

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

p1: 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

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    app#_A(x1,x2) = x1
    app_A(x1,x2) = max{x1 + 1, x2 + 1}
    compose_A = 0

The next rules are strictly ordered:

  p1

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