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, r2, r3, r4

Take the monotone reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          app#_A(x1,x2) = x1 + x2 + 2
          compose_2_A(x1,x2) = x1 + x2 + 4
          compose_3#_A(x1,x2,x3) = x1 + x2 + x3 + 5
          app_A(x1,x2) = x1 + x2 + 2
          compose_0_A() = 2
          compose_1_A(x1) = x1 + 3
          compose_3_A(x1,x2,x3) = x1 + x2 + x3 + 5
  
    base order 2:
  
      precedence:
      
        app# > compose_3# = compose_3 > app > compose_2 > compose_1 > compose_0
      
  
    partial status:
  
      pi(app#) = [1, 2]
      pi(compose_2) = [1, 2]
      pi(compose_3#) = [1, 2, 3]
      pi(app) = [1, 2]
      pi(compose_0) = []
      pi(compose_1) = [1]
      pi(compose_3) = [1, 2, 3]

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)