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(curry(),g),x),y) -> app(app(g,x),y)
  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
  inc() -> app(map(),app(app(curry(),plus()),app(s(),|0|())))
  double() -> app(map(),app(app(curry(),times()),app(s(),app(s(),|0|()))))

Instead we prove termination of the uncurried system:

  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(curry_0(),y) -> curry_1(y)
  app(curry_1(x1),y) -> curry_2(x1,y)
  app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,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)
  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)
  curry_3(g,x,y) -> app(app(g,x),y)
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))
  double_0() -> map_1(curry_2(times_0(),s_1(s_1(|0|_0()))))


-- SCC decomposition.

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

p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: app#(plus_1(x1),y) -> plus_2#(x1,y)
p4: app#(times_1(x1),y) -> times_2#(x1,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: curry_3#(g,x,y) -> app#(app(g,x),y)
p9: curry_3#(g,x,y) -> app#(g,x)
p10: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p11: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: app(times_0(),y) -> times_1(y)
r12: app(times_1(x1),y) -> times_2(x1,y)
r13: plus_2(|0|_0(),y) -> y
r14: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r15: times_2(|0|_0(),y) -> |0|_0()
r16: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r17: curry_3(g,x,y) -> app(app(g,x),y)
r18: map_2(f,nil_0()) -> nil_0()
r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r20: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))
r21: double_0() -> map_1(curry_2(times_0(),s_1(s_1(|0|_0()))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p8, p9, p10, p11}
  {p7}
  {p5}


-- Reduction pair.

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

p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y)
p2: curry_3#(g,x,y) -> app#(g,x)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: curry_3#(g,x,y) -> app#(app(g,x),y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: app(times_0(),y) -> times_1(y)
r12: app(times_1(x1),y) -> times_2(x1,y)
r13: plus_2(|0|_0(),y) -> y
r14: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r15: times_2(|0|_0(),y) -> |0|_0()
r16: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r17: curry_3(g,x,y) -> app(app(g,x),y)
r18: map_2(f,nil_0()) -> nil_0()
r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r20: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))
r21: double_0() -> map_1(curry_2(times_0(),s_1(s_1(|0|_0()))))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = 4
        curry_2_A(x1,x2) = max{x1 - 7, x2 + 13}
        curry_3#_A(x1,x2,x3) = 4
        map_1_A(x1) = x1 + 12
        map_2#_A(x1,x2) = 4
        cons_2_A(x1,x2) = max{5, x1 - 2}
        app_A(x1,x2) = max{21, x1 - 9, x2 + 13}
        cons_0_A = 7
        cons_1_A(x1) = x1 + 8
        curry_0_A = 1
        curry_1_A(x1) = x1 + 2
        curry_3_A(x1,x2,x3) = max{21, x1 - 18, x2 + 4, x3 + 13}
        map_0_A = 1
        map_2_A(x1,x2) = max{20, x1 + 2, x2 + 13}
        plus_0_A = 6
        plus_1_A(x1) = max{8, x1 - 1}
        plus_2_A(x1,x2) = max{x1 - 40, x2 + 11}
        s_0_A = 10
        s_1_A(x1) = max{11, x1 - 39}
        times_0_A = 1
        times_1_A(x1) = max{8, x1 - 26}
        times_2_A(x1,x2) = max{14, x1 - 36, x2 + 12}
        |0|_0_A = 13
        nil_0_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 34, x2 - 4}
          curry_2_A(x1,x2) = max{x1 + 12, x2 + 33}
          curry_3#_A(x1,x2,x3) = max{0, x1 - 34, x2 - 1, x3 - 4}
          map_1_A(x1) = x1 + 16
          map_2#_A(x1,x2) = max{0, x1 - 34, x2 - 17}
          cons_2_A(x1,x2) = max{x1 + 16, x2}
          app_A(x1,x2) = max{x1 - 17, x2 + 33}
          cons_0_A = 1
          cons_1_A(x1) = x1 + 33
          curry_0_A = 51
          curry_1_A(x1) = max{34, x1 + 29}
          curry_3_A(x1,x2,x3) = max{x1 - 5, x2 + 16, x3 + 33}
          map_0_A = 5
          map_2_A(x1,x2) = max{x1 - 1, x2 + 33}
          plus_0_A = 1
          plus_1_A(x1) = x1 + 1
          plus_2_A(x1,x2) = max{32, x1 - 116, x2 + 1}
          s_0_A = 5
          s_1_A(x1) = max{32, x1 - 82}
          times_0_A = 1
          times_1_A(x1) = x1 + 1
          times_2_A(x1,x2) = max{32, x1 - 117, x2 + 1}
          |0|_0_A = 32
          nil_0_A = 33
          app#!_A(x1,x2) = max{x1 - 7, x2 + 26}
          curry_2!_A(x1,x2) = max{x1 - 19, x2 + 27}
          curry_3#!_A(x1,x2,x3) = max{x1 + 4, x2 + 26, x3 + 26}
          map_1!_A(x1) = max{2, x1 - 66}
          map_2#!_A(x1,x2) = max{x1 - 6, x2 + 25}
          cons_2!_A(x1,x2) = max{23, x1 + 4, x2 - 34}
          app!_A(x1,x2) = max{x1 - 2, x2 + 28}
          cons_0!_A = 1
          cons_1!_A(x1) = max{24, x1 - 1}
          curry_0!_A = 1
          curry_1!_A(x1) = 2
          curry_3!_A(x1,x2,x3) = max{x1 - 19, x2 + 31, x3 + 28}
          map_0!_A = 1
          map_2!_A(x1,x2) = max{24, x1 + 13, x2 + 22}
          plus_0!_A = 1
          plus_1!_A(x1) = max{4, x1 - 68}
          plus_2!_A(x1,x2) = max{27, x1 - 67, x2 - 1}
          s_0!_A = 1
          s_1!_A(x1) = max{2, x1 - 34}
          times_0!_A = 1
          times_1!_A(x1) = max{2, x1 - 35}
          times_2!_A(x1,x2) = max{27, x1 - 34, x2 - 1}
          |0|_0!_A = 0
          nil_0!_A = 0
  
    partial status:
  
      pi(app#) = []
      pi(curry_2) = []
      pi(curry_3#) = []
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app) = [2]
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(curry_0) = []
      pi(curry_1) = [1]
      pi(curry_3) = [3]
      pi(map_0) = []
      pi(map_2) = [2]
      pi(plus_0) = []
      pi(plus_1) = []
      pi(plus_2) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_1) = []
      pi(times_2) = []
      pi(|0|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p3, p5



-- SCC decomposition.

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

p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y)
p2: curry_3#(g,x,y) -> app#(g,x)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p4: curry_3#(g,x,y) -> app#(app(g,x),y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: app(times_0(),y) -> times_1(y)
r12: app(times_1(x1),y) -> times_2(x1,y)
r13: plus_2(|0|_0(),y) -> y
r14: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r15: times_2(|0|_0(),y) -> |0|_0()
r16: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r17: curry_3(g,x,y) -> app(app(g,x),y)
r18: map_2(f,nil_0()) -> nil_0()
r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r20: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))
r21: double_0() -> map_1(curry_2(times_0(),s_1(s_1(|0|_0()))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4}
  {p3}


-- Reduction pair.

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

p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y)
p2: curry_3#(g,x,y) -> app#(app(g,x),y)
p3: curry_3#(g,x,y) -> app#(g,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: app(times_0(),y) -> times_1(y)
r12: app(times_1(x1),y) -> times_2(x1,y)
r13: plus_2(|0|_0(),y) -> y
r14: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r15: times_2(|0|_0(),y) -> |0|_0()
r16: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r17: curry_3(g,x,y) -> app(app(g,x),y)
r18: map_2(f,nil_0()) -> nil_0()
r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r20: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))
r21: double_0() -> map_1(curry_2(times_0(),s_1(s_1(|0|_0()))))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{x1 - 8, x2 + 2}
        curry_2_A(x1,x2) = max{x1 + 10, x2 + 25}
        curry_3#_A(x1,x2,x3) = max{x1 + 2, x2 + 17, x3 + 2}
        app_A(x1,x2) = max{x1 - 13, x2 + 25}
        cons_0_A = 11
        cons_1_A(x1) = max{12, x1 - 1}
        cons_2_A(x1,x2) = max{0, x2 - 26}
        curry_0_A = 45
        curry_1_A(x1) = max{31, x1 + 23}
        curry_3_A(x1,x2,x3) = max{x1 - 26, x2 + 12, x3 + 25}
        map_0_A = 1
        map_1_A(x1) = x1 + 15
        map_2_A(x1,x2) = max{10, x1 + 1}
        plus_0_A = 12
        plus_1_A(x1) = x1 + 25
        plus_2_A(x1,x2) = max{3, x1 - 5, x2 + 1}
        s_0_A = 16
        s_1_A(x1) = max{2, x1 - 4}
        times_0_A = 12
        times_1_A(x1) = x1 + 24
        times_2_A(x1,x2) = max{10, x1 - 4, x2 + 4}
        |0|_0_A = 3
        nil_0_A = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{5, x1 - 9, x2 - 6}
          curry_2_A(x1,x2) = max{x1 + 7, x2 + 13}
          curry_3#_A(x1,x2,x3) = max{5, x1 - 6, x2 + 4, x3 - 6}
          app_A(x1,x2) = max{x1 - 6, x2 + 13}
          cons_0_A = 6
          cons_1_A(x1) = max{5, x1 + 4}
          cons_2_A(x1,x2) = max{13, x2 - 1}
          curry_0_A = 4
          curry_1_A(x1) = x1 + 13
          curry_3_A(x1,x2,x3) = max{x1, x2 + 7, x3 + 13}
          map_0_A = 4
          map_1_A(x1) = 4
          map_2_A(x1,x2) = x2 + 13
          plus_0_A = 5
          plus_1_A(x1) = x1 + 4
          plus_2_A(x1,x2) = max{5, x1 - 34, x2 + 1}
          s_0_A = 4
          s_1_A(x1) = max{5, x1 - 33}
          times_0_A = 4
          times_1_A(x1) = x1 + 4
          times_2_A(x1,x2) = max{5, x1 - 35, x2 + 4}
          |0|_0_A = 5
          nil_0_A = 6
          app#!_A(x1,x2) = max{x1 - 1, x2 + 11}
          curry_2!_A(x1,x2) = max{13, x2 - 6}
          curry_3#!_A(x1,x2,x3) = max{x1 - 7, x2 + 12, x3 + 11}
          app!_A(x1,x2) = max{15, x2}
          cons_0!_A = 1
          cons_1!_A(x1) = max{2, x1 - 13}
          cons_2!_A(x1,x2) = max{17, x1 - 13, x2 - 19}
          curry_0!_A = 1
          curry_1!_A(x1) = max{14, x1 - 9}
          curry_3!_A(x1,x2,x3) = max{15, x3}
          map_0!_A = 2
          map_1!_A(x1) = max{1, x1 - 19}
          map_2!_A(x1,x2) = max{18, x1 + 16}
          plus_0!_A = 1
          plus_1!_A(x1) = max{0, x1 - 44}
          plus_2!_A(x1,x2) = max{17, x1 - 45, x2 - 1}
          s_0!_A = 1
          s_1!_A(x1) = max{2, x1 - 84}
          times_0!_A = 1
          times_1!_A(x1) = max{2, x1 - 7}
          times_2!_A(x1,x2) = max{16, x1 - 6, x2 - 42}
          |0|_0!_A = 17
          nil_0!_A = 1
  
    partial status:
  
      pi(app#) = []
      pi(curry_2) = []
      pi(curry_3#) = []
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(curry_0) = []
      pi(curry_1) = [1]
      pi(curry_3) = []
      pi(map_0) = []
      pi(map_1) = [1]
      pi(map_2) = [1]
      pi(plus_0) = []
      pi(plus_1) = [1]
      pi(plus_2) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_1) = []
      pi(times_2) = [2]
      pi(|0|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p3



-- SCC decomposition.

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

p1: app#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y)
p2: curry_3#(g,x,y) -> app#(app(g,x),y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: app(times_0(),y) -> times_1(y)
r12: app(times_1(x1),y) -> times_2(x1,y)
r13: plus_2(|0|_0(),y) -> y
r14: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r15: times_2(|0|_0(),y) -> |0|_0()
r16: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r17: curry_3(g,x,y) -> app(app(g,x),y)
r18: map_2(f,nil_0()) -> nil_0()
r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r20: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))
r21: double_0() -> map_1(curry_2(times_0(),s_1(s_1(|0|_0()))))

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#(curry_2(x1,x2),y) -> curry_3#(x1,x2,y)
p2: curry_3#(g,x,y) -> app#(app(g,x),y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: app(times_0(),y) -> times_1(y)
r12: app(times_1(x1),y) -> times_2(x1,y)
r13: plus_2(|0|_0(),y) -> y
r14: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r15: times_2(|0|_0(),y) -> |0|_0()
r16: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r17: curry_3(g,x,y) -> app(app(g,x),y)
r18: map_2(f,nil_0()) -> nil_0()
r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r20: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))
r21: double_0() -> map_1(curry_2(times_0(),s_1(s_1(|0|_0()))))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19

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}
        curry_2_A(x1,x2) = max{x1 + 18, x2 + 15}
        curry_3#_A(x1,x2,x3) = max{x1 + 7, x2 + 15}
        app_A(x1,x2) = max{x1 + 7, x2 + 15}
        cons_0_A = 17
        cons_1_A(x1) = max{16, x1 + 1}
        cons_2_A(x1,x2) = 3
        curry_0_A = 5
        curry_1_A(x1) = x1 + 11
        curry_3_A(x1,x2,x3) = max{24, x1 + 15, x2 + 22, x3 + 15}
        map_0_A = 18
        map_1_A(x1) = 17
        map_2_A(x1,x2) = max{16, x2 + 14}
        plus_0_A = 1
        plus_1_A(x1) = max{8, x1 - 7}
        plus_2_A(x1,x2) = max{4, x2 + 1}
        s_0_A = 1
        s_1_A(x1) = max{3, x1 - 19}
        times_0_A = 1
        times_1_A(x1) = max{6, x1 + 1}
        times_2_A(x1,x2) = max{14, x1 + 2, x2 + 2}
        |0|_0_A = 3
        nil_0_A = 17
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{1, x1 - 31, x2 - 23}
          curry_2_A(x1,x2) = x2 + 15
          curry_3#_A(x1,x2,x3) = max{1, x2 - 16, x3 - 23}
          app_A(x1,x2) = x2 + 15
          cons_0_A = 12
          cons_1_A(x1) = 0
          cons_2_A(x1,x2) = max{9, x2 - 16}
          curry_0_A = 0
          curry_1_A(x1) = x1 + 15
          curry_3_A(x1,x2,x3) = x3 + 15
          map_0_A = 0
          map_1_A(x1) = max{15, x1 - 19}
          map_2_A(x1,x2) = 9
          plus_0_A = 0
          plus_1_A(x1) = max{12, x1 - 24}
          plus_2_A(x1,x2) = max{5, x2 + 1}
          s_0_A = 0
          s_1_A(x1) = max{2, x1 - 4}
          times_0_A = 0
          times_1_A(x1) = max{11, x1 - 45}
          times_2_A(x1,x2) = max{9, x2 + 1}
          |0|_0_A = 2
          nil_0_A = 9
          app#!_A(x1,x2) = 0
          curry_2!_A(x1,x2) = x1 + 37
          curry_3#!_A(x1,x2,x3) = max{36, x1 + 34}
          app!_A(x1,x2) = max{33, x1 + 23}
          cons_0!_A = 1
          cons_1!_A(x1) = max{34, x1 - 1}
          cons_2!_A(x1,x2) = 2
          curry_0!_A = 1
          curry_1!_A(x1) = max{34, x1 - 1}
          curry_3!_A(x1,x2,x3) = x2 + 38
          map_0!_A = 34
          map_1!_A(x1) = max{33, x1 - 31}
          map_2!_A(x1,x2) = x1 + 3
          plus_0!_A = 37
          plus_1!_A(x1) = 24
          plus_2!_A(x1,x2) = max{36, x1 - 10, x2 - 19}
          s_0!_A = 1
          s_1!_A(x1) = max{0, x1 - 20}
          times_0!_A = 1
          times_1!_A(x1) = max{34, x1 - 22}
          times_2!_A(x1,x2) = max{35, x1 - 21, x2 - 20}
          |0|_0!_A = 0
          nil_0!_A = 1
  
    partial status:
  
      pi(app#) = [1]
      pi(curry_2) = []
      pi(curry_3#) = []
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(cons_2) = []
      pi(curry_0) = []
      pi(curry_1) = [1]
      pi(curry_3) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(map_2) = []
      pi(plus_0) = []
      pi(plus_1) = []
      pi(plus_2) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_1) = [1]
      pi(times_2) = [1, 2]
      pi(|0|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: app(times_0(),y) -> times_1(y)
r12: app(times_1(x1),y) -> times_2(x1,y)
r13: plus_2(|0|_0(),y) -> y
r14: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r15: times_2(|0|_0(),y) -> |0|_0()
r16: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r17: curry_3(g,x,y) -> app(app(g,x),y)
r18: map_2(f,nil_0()) -> nil_0()
r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r20: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))
r21: double_0() -> map_1(curry_2(times_0(),s_1(s_1(|0|_0()))))

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:
      
        map_2#_A(x1,x2) = x2 + 1
        cons_2_A(x1,x2) = max{31, x1 + 6, x2 + 22}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          map_2#_A(x1,x2) = 0
          cons_2_A(x1,x2) = max{30, x1 - 3, x2 - 12}
          map_2#!_A(x1,x2) = x1 + 5
          cons_2!_A(x1,x2) = max{3, x1 - 31, x2 - 17}
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [2]

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

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: app(times_0(),y) -> times_1(y)
r12: app(times_1(x1),y) -> times_2(x1,y)
r13: plus_2(|0|_0(),y) -> y
r14: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r15: times_2(|0|_0(),y) -> |0|_0()
r16: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r17: curry_3(g,x,y) -> app(app(g,x),y)
r18: map_2(f,nil_0()) -> nil_0()
r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r20: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))
r21: double_0() -> map_1(curry_2(times_0(),s_1(s_1(|0|_0()))))

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:
      
        times_2#_A(x1,x2) = max{10, x1}
        s_1_A(x1) = x1 + 10
  
    base order 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{4, x1 - 6}
          times_2#!_A(x1,x2) = 0
          s_1!_A(x1) = max{2, x1 - 28}
  
    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(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(curry_0(),y) -> curry_1(y)
r4: app(curry_1(x1),y) -> curry_2(x1,y)
r5: app(curry_2(x1,x2),y) -> curry_3(x1,x2,y)
r6: app(map_0(),y) -> map_1(y)
r7: app(map_1(x1),y) -> map_2(x1,y)
r8: app(plus_0(),y) -> plus_1(y)
r9: app(plus_1(x1),y) -> plus_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: app(times_0(),y) -> times_1(y)
r12: app(times_1(x1),y) -> times_2(x1,y)
r13: plus_2(|0|_0(),y) -> y
r14: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r15: times_2(|0|_0(),y) -> |0|_0()
r16: times_2(s_1(x),y) -> plus_2(times_2(x,y),y)
r17: curry_3(g,x,y) -> app(app(g,x),y)
r18: map_2(f,nil_0()) -> nil_0()
r19: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r20: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))
r21: double_0() -> map_1(curry_2(times_0(),s_1(s_1(|0|_0()))))

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:
      
        plus_2#_A(x1,x2) = max{9, x1 + 1}
        s_1_A(x1) = x1 + 11
  
    base order 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{4, x1 - 15}
          plus_2#!_A(x1,x2) = x1 + 19
          s_1!_A(x1) = max{2, x1 - 32}
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.