YES
We show the termination of the TRS R:

  app(app(plus(),|0|()),y) -> y
  app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
  app(app(times(),|0|()),y) -> |0|()
  app(app(times(),app(s(),x)),y) -> app(app(plus(),app(app(times(),x),y)),y)
  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(plus_0(),y) -> plus_1(y)
  app(plus_1(x1),y) -> plus_2(x1,y)
  app(s_0(),y) -> s_1(y)
  app(times_0(),y) -> times_1(y)
  app(times_1(x1),y) -> times_2(x1,y)
  app(twice_0(),y) -> twice_1(y)
  plus_2(|0|_0(),y) -> y
  plus_2(s_1(x),y) -> s_1(plus_2(x,y))
  times_2(|0|_0(),y) -> |0|_0()
  times_2(s_1(x),y) -> plus_2(times_2(x,y),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#(plus_1(x1),y) -> plus_2#(x1,y)
p3: app#(times_1(x1),y) -> times_2#(x1,y)
p4: app#(twice_0(),y) -> twice_1#(y)
p5: plus_2#(s_1(x),y) -> plus_2#(x,y)
p6: times_2#(s_1(x),y) -> plus_2#(times_2(x,y),y)
p7: times_2#(s_1(x),y) -> times_2#(x,y)
p8: comp_3#(f,g,x) -> app#(f,app(g,x))
p9: 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(plus_0(),y) -> plus_1(y)
r5: app(plus_1(x1),y) -> plus_2(x1,y)
r6: app(s_0(),y) -> s_1(y)
r7: app(times_0(),y) -> times_1(y)
r8: app(times_1(x1),y) -> times_2(x1,y)
r9: app(twice_0(),y) -> twice_1(y)
r10: plus_2(|0|_0(),y) -> y
r11: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r12: times_2(|0|_0(),y) -> |0|_0()
r13: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r14: comp_3(f,g,x) -> app(f,app(g,x))
r15: twice_1(f) -> comp_2(f,f)

The estimated dependency graph contains the following SCCs:

  {p1, p8, p9}
  {p7}
  {p5}


-- 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(plus_0(),y) -> plus_1(y)
r5: app(plus_1(x1),y) -> plus_2(x1,y)
r6: app(s_0(),y) -> s_1(y)
r7: app(times_0(),y) -> times_1(y)
r8: app(times_1(x1),y) -> times_2(x1,y)
r9: app(twice_0(),y) -> twice_1(y)
r10: plus_2(|0|_0(),y) -> y
r11: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r12: times_2(|0|_0(),y) -> |0|_0()
r13: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r14: comp_3(f,g,x) -> app(f,app(g,x))
r15: twice_1(f) -> comp_2(f,f)

The set of usable rules consists of

  r1, r2, r12, r15

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!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 1
            comp_2_A(x1,x2) = 4
            comp_3#_A(x1,x2,x3) = 1
            app_A(x1,x2) = max{4, x2}
            comp_0_A = 4
            comp_1_A(x1) = max{4, x1}
            times_2_A(x1,x2) = max{4, x2}
            |0|_0_A = 1
            twice_1_A(x1) = 4
            app#!_A(x1,x2) = 0
            comp_2!_A(x1,x2) = 4
            comp_3#!_A(x1,x2,x3) = 1
            app!_A(x1,x2) = 4
            comp_0!_A = 1
            comp_1!_A(x1) = 2
            times_2!_A(x1,x2) = max{6, x1 - 4, x2 - 7}
            |0|_0!_A = 5
            twice_1!_A(x1) = x1 + 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 1
            comp_2_A(x1,x2) = 1
            comp_3#_A(x1,x2,x3) = 1
            app_A(x1,x2) = 23
            comp_0_A = 0
            comp_1_A(x1) = 1
            times_2_A(x1,x2) = max{9, x2 + 2}
            |0|_0_A = 3
            twice_1_A(x1) = max{1, x1 - 6}
            app#!_A(x1,x2) = 0
            comp_2!_A(x1,x2) = 1
            comp_3#!_A(x1,x2,x3) = max{0, x1 - 1, x2 - 1}
            app!_A(x1,x2) = 3
            comp_0!_A = 1
            comp_1!_A(x1) = max{4, x1 - 1}
            times_2!_A(x1,x2) = max{13, x1 + 6}
            |0|_0!_A = 13
            twice_1!_A(x1) = 4
        
  
    partial status:
  
      pi(app#) = [1]
      pi(comp_2) = [1, 2]
      pi(comp_3#) = [1, 2]
      pi(app) = [1, 2]
      pi(comp_0) = []
      pi(comp_1) = [1]
      pi(times_2) = [1, 2]
      pi(|0|_0) = []
      pi(twice_1) = [1]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- Reduction pair.

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

p1: times_2#(s_1(x),y) -> times_2#(x,y)

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(plus_0(),y) -> plus_1(y)
r5: app(plus_1(x1),y) -> plus_2(x1,y)
r6: app(s_0(),y) -> s_1(y)
r7: app(times_0(),y) -> times_1(y)
r8: app(times_1(x1),y) -> times_2(x1,y)
r9: app(twice_0(),y) -> twice_1(y)
r10: plus_2(|0|_0(),y) -> y
r11: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r12: times_2(|0|_0(),y) -> |0|_0()
r13: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r14: comp_3(f,g,x) -> app(f,app(g,x))
r15: 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:
  
      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!):
        
          max/plus interpretation on natural numbers:
          
            times_2#_A(x1,x2) = 0
            s_1_A(x1) = max{7, x1 - 9}
            times_2#!_A(x1,x2) = max{0, x1 - 18}
            s_1!_A(x1) = max{1, x1 - 17}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times_2#_A(x1,x2) = 0
            s_1_A(x1) = max{7, x1 - 21}
            times_2#!_A(x1,x2) = max{4, x1 - 8, x2 - 1}
            s_1!_A(x1) = max{3, x1 - 47}
        
  
    partial status:
  
      pi(times_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: plus_2#(s_1(x),y) -> plus_2#(x,y)

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(plus_0(),y) -> plus_1(y)
r5: app(plus_1(x1),y) -> plus_2(x1,y)
r6: app(s_0(),y) -> s_1(y)
r7: app(times_0(),y) -> times_1(y)
r8: app(times_1(x1),y) -> times_2(x1,y)
r9: app(twice_0(),y) -> twice_1(y)
r10: plus_2(|0|_0(),y) -> y
r11: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r12: times_2(|0|_0(),y) -> |0|_0()
r13: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r14: comp_3(f,g,x) -> app(f,app(g,x))
r15: 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:
  
      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!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{7, x1 - 9}
            plus_2#!_A(x1,x2) = max{0, x1 - 10}
            s_1!_A(x1) = max{3, x1 - 9}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{7, x1 - 21}
            plus_2#!_A(x1,x2) = max{3, x1 - 8, x2 - 1}
            s_1!_A(x1) = max{2, x1 - 47}
        
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.