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. lexicographic path order with precedence:
    
      precedence:
      
        app# > app > compose
      
      argument filter:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(compose) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        app > app# > compose
      
      argument filter:
    
        pi(app#) = 2
        pi(app) = 2
        pi(compose) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        app > app# > compose
      
      argument filter:
    
        pi(app#) = []
        pi(app) = [2]
        pi(compose) = []
    

The next rules are strictly ordered:

  p1, p2

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