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 reduction pair:

  generalized weighted path order (>=MG, >G)
  
    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 + 6
          compose_3#_A(x1,x2,x3) = x1 + x2 + x3 + 7
          app_A(x1,x2) = x1 + x2 + 2
          compose_0_A() = 6
          compose_1_A(x1) = x1 + 5
          compose_3_A(x1,x2,x3) = x1 + x2 + x3 + 5
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            app#_A(x1,x2) = x1 + x2
            compose_2_A(x1,x2) = x1 + x2 + 4
            compose_3#_A(x1,x2,x3) = x1 + x2 + x3 + 4
            app_A(x1,x2) = x1 + x2 + 4
            compose_0_A() = 1
            compose_1_A(x1) = x1 + 1
            compose_3_A(x1,x2,x3) = x1 + x2 + x3 + 8
            app#!_A(x1,x2) = x2 + 4
            compose_2!_A(x1,x2) = 2
            compose_3#!_A(x1,x2,x3) = 3
            app!_A(x1,x2) = x1 + x2
            compose_0!_A() = 1
            compose_1!_A(x1) = 0
            compose_3!_A(x1,x2,x3) = 3
  
    partial status:
  
      pi(app#) = []
      pi(compose_2) = [2]
      pi(compose_3#) = []
      pi(app) = []
      pi(compose_0) = []
      pi(compose_1) = [1]
      pi(compose_3) = [1]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.