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, r4, r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{9, x1 - 10}
        comp_2_A(x1,x2) = max{x1 + 21, x2 + 9}
        comp_3#_A(x1,x2,x3) = max{x1 + 10, x2 - 2}
        app_A(x1,x2) = max{6, x1 - 23, x2 - 23}
        comp_0_A = 1
        comp_1_A(x1) = max{2, x1 - 24}
        plus_0_A = 4
        plus_1_A(x1) = max{5, x1 - 24}
        s_0_A = 1
        s_1_A(x1) = max{2, x1 - 49}
        times_0_A = 3
        times_1_A(x1) = max{4, x1 - 24}
  
    base order 2:
  
      precedence:
      
        times_0 > comp_2 = times_1 > app > comp_3# = comp_1 > app# = comp_0 = plus_1 = s_1 > plus_0 = s_0
      
  
    partial status:
  
      pi(app#) = []
      pi(comp_2) = [2]
      pi(comp_3#) = []
      pi(app) = []
      pi(comp_0) = []
      pi(comp_1) = []
      pi(plus_0) = []
      pi(plus_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_1) = []

The following rules are removed:

  p2



-- 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: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- 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#(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, r4, r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{2, x1 - 8}
        comp_2_A(x1,x2) = max{x1 + 17, x2 - 2}
        comp_3#_A(x1,x2,x3) = max{8, x1 + 1, x2 - 18}
        app_A(x1,x2) = max{6, x1 - 18, x2 - 18}
        comp_0_A = 4
        comp_1_A(x1) = max{5, x1 - 19}
        plus_0_A = 4
        plus_1_A(x1) = max{5, x1 - 19}
        s_0_A = 1
        s_1_A(x1) = max{2, x1 - 39}
        times_0_A = 3
        times_1_A(x1) = max{4, x1 - 19}
  
    base order 2:
  
      precedence:
      
        app# = comp_2 > comp_3# = app > comp_1 > comp_0 = times_1 > plus_1 = s_1 = times_0 > plus_0 = s_0
      
  
    partial status:
  
      pi(app#) = []
      pi(comp_2) = [1]
      pi(comp_3#) = []
      pi(app) = []
      pi(comp_0) = []
      pi(comp_1) = []
      pi(plus_0) = []
      pi(plus_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_1) = []

The following rules are removed:

  p2



-- 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)

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:

  (no SCCs)

-- 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

  r1, r2, r4, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times_2#_A(x1,x2) = max{x1 + 18, x2 + 19}
        s_1_A(x1) = x1 + 17
        app_A(x1,x2) = max{6, x1 - 7, x2 - 7}
        comp_0_A = 6
        comp_1_A(x1) = max{1, x1 - 7}
        comp_2_A(x1,x2) = max{5, x1 - 15, x2 - 8}
        plus_0_A = 1
        plus_1_A(x1) = max{2, x1 - 8}
        times_0_A = 5
        times_1_A(x1) = max{6, x1 - 7}
  
    base order 2:
  
      precedence:
      
        app > times_2# = comp_2 > s_1 = comp_1 = times_1 > comp_0 = plus_1 = times_0 > plus_0
      
  
    partial status:
  
      pi(times_2#) = [1]
      pi(s_1) = []
      pi(app) = []
      pi(comp_0) = []
      pi(comp_1) = []
      pi(comp_2) = []
      pi(plus_0) = []
      pi(plus_1) = []
      pi(times_0) = []
      pi(times_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

  r1, r2, r4, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus_2#_A(x1,x2) = x1 + 7
        s_1_A(x1) = x1 + 7
        app_A(x1,x2) = max{5, x1 - 6, x2 - 6}
        comp_0_A = 5
        comp_1_A(x1) = max{3, x1 - 7}
        comp_2_A(x1,x2) = max{4, x1 - 14, x2 - 7}
        plus_0_A = 1
        plus_1_A(x1) = 2
        times_0_A = 4
        times_1_A(x1) = max{5, x1 - 6}
  
    base order 2:
  
      precedence:
      
        app > comp_2 > s_1 = comp_1 = times_1 > plus_2# = comp_0 = plus_1 = times_0 > plus_0
      
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = []
      pi(app) = []
      pi(comp_0) = []
      pi(comp_1) = []
      pi(comp_2) = []
      pi(plus_0) = []
      pi(plus_1) = []
      pi(times_0) = []
      pi(times_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.