YES
We show the termination of the TRS R:

  app(app(app(comp(),f),g),x) -> app(f,app(g,x))
  app(twice(),f) -> app(app(comp(),f),f)

Instead we prove termination of the uncurried system:

  app(comp_0(),y) -> comp_1(y)
  app(comp_1(x1),y) -> comp_2(x1,y)
  app(comp_2(x1,x2),y) -> comp_3(x1,x2,y)
  app(twice_0(),y) -> twice_1(y)
  comp_3(f,g,x) -> app(f,app(g,x))
  twice_1(f) -> comp_2(f,f)


-- SCC decomposition.

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

p1: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y)
p2: app#(twice_0(),y) -> twice_1#(y)
p3: comp_3#(f,g,x) -> app#(f,app(g,x))
p4: comp_3#(f,g,x) -> app#(g,x)

and R consists of:

r1: app(comp_0(),y) -> comp_1(y)
r2: app(comp_1(x1),y) -> comp_2(x1,y)
r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y)
r4: app(twice_0(),y) -> twice_1(y)
r5: comp_3(f,g,x) -> app(f,app(g,x))
r6: twice_1(f) -> comp_2(f,f)

The estimated dependency graph contains the following SCCs:

  {p1, p3, p4}


-- Reduction pair.

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

p1: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y)
p2: comp_3#(f,g,x) -> app#(g,x)
p3: comp_3#(f,g,x) -> app#(f,app(g,x))

and R consists of:

r1: app(comp_0(),y) -> comp_1(y)
r2: app(comp_1(x1),y) -> comp_2(x1,y)
r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y)
r4: app(twice_0(),y) -> twice_1(y)
r5: comp_3(f,g,x) -> app(f,app(g,x))
r6: twice_1(f) -> comp_2(f,f)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{1, x1 - 1}
        comp_2_A(x1,x2) = max{7, x1 + 4, x2 + 4}
        comp_3#_A(x1,x2,x3) = max{5, x1 + 2, x2 + 2}
        app_A(x1,x2) = 6
  
    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) = 0
            comp_2_A(x1,x2) = x1 + 4
            comp_3#_A(x1,x2,x3) = 0
            app_A(x1,x2) = x2 + 2
            app#!_A(x1,x2) = 4
            comp_2!_A(x1,x2) = x1 + x2 + 1
            comp_3#!_A(x1,x2,x3) = x2 + x3 + 3
            app!_A(x1,x2) = 2
  
    partial status:
  
      pi(app#) = []
      pi(comp_2) = [1, 2]
      pi(comp_3#) = []
      pi(app) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.