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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((1,1),(0,1)) x1 + (5,6)
          app_A(x1,x2) = x1 + x2 + (2,2)
          compose_A() = (2,1)
  
    precedence:
    
      app > app# = compose
    
    partial status:
  
      pi(app#) = [1]
      pi(app) = []
      pi(compose) = []

The next rules are strictly ordered:

  p2

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#(f,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#(f,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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((0,1),(0,0)) x1 + (1,3)
          app_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(1,1)) x2 + (2,0)
          compose_A() = (0,0)
  
    precedence:
    
      compose > app# = app
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(compose) = []

The next rules are strictly ordered:

  p1

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