YES
We show the termination of the TRS R:

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

Instead we prove termination of the uncurried system:

  app(compose_0(),y) -> compose_1(y)
  app(compose_1(x1),y) -> compose_2(x1,y)
  app(compose_2(x1,x2),y) -> compose_3(x1,x2,y)
  compose_3(f,g,x) -> app(f,app(g,x))


-- SCC decomposition.

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

p1: app#(compose_2(x1,x2),y) -> compose_3#(x1,x2,y)
p2: compose_3#(f,g,x) -> app#(f,app(g,x))
p3: compose_3#(f,g,x) -> app#(g,x)

and R consists of:

r1: app(compose_0(),y) -> compose_1(y)
r2: app(compose_1(x1),y) -> compose_2(x1,y)
r3: app(compose_2(x1,x2),y) -> compose_3(x1,x2,y)
r4: compose_3(f,g,x) -> app(f,app(g,x))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: app#(compose_2(x1,x2),y) -> compose_3#(x1,x2,y)
p2: compose_3#(f,g,x) -> app#(g,x)
p3: compose_3#(f,g,x) -> app#(f,app(g,x))

and R consists of:

r1: app(compose_0(),y) -> compose_1(y)
r2: app(compose_1(x1),y) -> compose_2(x1,y)
r3: app(compose_2(x1,x2),y) -> compose_3(x1,x2,y)
r4: compose_3(f,g,x) -> app(f,app(g,x))

The set of usable rules consists of

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{2, x1 - 5}
        compose_2_A(x1,x2) = max{x1 + 8, x2 + 1}
        compose_3#_A(x1,x2,x3) = max{2, x1 + 1, x2 - 5}
        app_A(x1,x2) = max{3, x1 - 11, x2 - 4}
        compose_0_A = 1
        compose_1_A(x1) = max{2, x1 - 5}
  
    base order 2:
  
      precedence:
      
        compose_2 > app > compose_3# = compose_1 > app# = compose_0
      
  
    partial status:
  
      pi(app#) = []
      pi(compose_2) = [2]
      pi(compose_3#) = [1]
      pi(app) = []
      pi(compose_0) = []
      pi(compose_1) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: compose_3#(f,g,x) -> app#(g,x)
p2: compose_3#(f,g,x) -> app#(f,app(g,x))

and R consists of:

r1: app(compose_0(),y) -> compose_1(y)
r2: app(compose_1(x1),y) -> compose_2(x1,y)
r3: app(compose_2(x1,x2),y) -> compose_3(x1,x2,y)
r4: compose_3(f,g,x) -> app(f,app(g,x))

The estimated dependency graph contains the following SCCs:

  (no SCCs)