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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. 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 + 3
              compose_3#_A(x1,x2,x3) = x1 + x2 + x3 + 1
              app_A(x1,x2) = x1 + x2 + 1
              compose_0_A() = 3
              compose_1_A(x1) = x1 + 4
              compose_3_A(x1,x2,x3) = x1 + x2 + x3 + 2
              app#!_A(x1,x2) = 1
              compose_2!_A(x1,x2) = x2 + 3
              compose_3#!_A(x1,x2,x3) = x2 + 2
              app!_A(x1,x2) = x1 + x2
              compose_0!_A() = 0
              compose_1!_A(x1) = 1
              compose_3!_A(x1,x2,x3) = x1 + x2 + x3 + 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 6, x2 - 8}
            compose_2_A(x1,x2) = max{5, x1, x2}
            compose_3#_A(x1,x2,x3) = max{0, x1 - 6, x2 - 6, x3 - 8}
            app_A(x1,x2) = max{7, x1 + 1, x2}
            compose_0_A = 1
            compose_1_A(x1) = max{3, x1}
            compose_3_A(x1,x2,x3) = max{7, x1 + 1, x2 + 1, x3}
            app#!_A(x1,x2) = max{6, x1 - 6}
            compose_2!_A(x1,x2) = max{2, x1 - 2, x2 - 2}
            compose_3#!_A(x1,x2,x3) = max{7, x1 + 1, x2 - 3, x3 + 5}
            app!_A(x1,x2) = max{4, x2}
            compose_0!_A = 2
            compose_1!_A(x1) = max{1, x1 - 3}
            compose_3!_A(x1,x2,x3) = max{3, x1 - 1, x2 - 1, x3 - 1}
        
  
    partial status:
  
      pi(app#) = [1]
      pi(compose_2) = [1, 2]
      pi(compose_3#) = [1, 2]
      pi(app) = [1, 2]
      pi(compose_0) = []
      pi(compose_1) = [1]
      pi(compose_3) = [1, 2, 3]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.