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:
  
      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) = max{0, x1 - 87, x2 - 79}
            curry_2_A(x1,x2) = max{x1 + 7, x2 + 86}
            curry_3#_A(x1,x2,x3) = max{0, x1 - 87, x2 - 1, x3 - 79}
            map_1_A(x1) = x1 + 70
            map_2#_A(x1,x2) = max{0, x1 - 17, x2 - 79}
            cons_2_A(x1,x2) = max{11, x1 + 7, x2}
            app_A(x1,x2) = max{x1 - 79, x2 + 86}
            cons_0_A = 75
            cons_1_A(x1) = x1 + 86
            curry_0_A = 73
            curry_1_A(x1) = x1 + 86
            curry_3_A(x1,x2,x3) = max{x1 - 158, x2 + 7, x3 + 86}
            map_0_A = 78
            map_2_A(x1,x2) = max{x1 - 72, x2 + 86}
            plus_0_A = 131
            plus_1_A(x1) = x1 + 71
            plus_2_A(x1,x2) = max{11, x2 + 10}
            s_0_A = 66
            s_1_A(x1) = 11
            times_0_A = 144
            times_1_A(x1) = max{66, x1 + 54}
            times_2_A(x1,x2) = x2 + 24
            |0|_0_A = 24
            nil_0_A = 11
            app#!_A(x1,x2) = max{x1 - 62, x2 + 5}
            curry_2!_A(x1,x2) = max{12, x1 - 132, x2 - 42}
            curry_3#!_A(x1,x2,x3) = max{x1 - 56, x2 + 24, x3 + 5}
            map_1!_A(x1) = max{10, x1 - 128}
            map_2#!_A(x1,x2) = max{3, x1 - 61, x2 - 1}
            cons_2!_A(x1,x2) = max{1, x1 - 92}
            app!_A(x1,x2) = max{6, x1 - 67, x2 + 4}
            cons_0!_A = 8
            cons_1!_A(x1) = 7
            curry_0!_A = 6
            curry_1!_A(x1) = max{5, x1 - 131}
            curry_3!_A(x1,x2,x3) = max{x1 - 146, x2 + 19, x3 + 4}
            map_0!_A = 11
            map_2!_A(x1,x2) = max{2, x1 - 66, x2 - 12}
            plus_0!_A = 8
            plus_1!_A(x1) = max{7, x1 - 1}
            plus_2!_A(x1,x2) = max{5, x2 + 3}
            s_0!_A = 1
            s_1!_A(x1) = max{0, x1 - 12}
            times_0!_A = 2
            times_1!_A(x1) = max{1, x1 - 24}
            times_2!_A(x1,x2) = max{5, x1 - 23, x2 + 3}
            |0|_0!_A = 2
            nil_0!_A = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 31, x2 - 29}
            curry_2_A(x1,x2) = max{x1, x2 + 30}
            curry_3#_A(x1,x2,x3) = max{0, x1 - 31, x2 - 1, x3 - 29}
            map_1_A(x1) = max{20, x1 + 9}
            map_2#_A(x1,x2) = max{0, x1 - 22, x2 - 29}
            cons_2_A(x1,x2) = max{x1 + 9, x2}
            app_A(x1,x2) = max{x1 - 21, x2 + 30}
            cons_0_A = 0
            cons_1_A(x1) = x1 + 30
            curry_0_A = 20
            curry_1_A(x1) = x1 + 22
            curry_3_A(x1,x2,x3) = max{x1 - 21, x2 + 9, x3 + 30}
            map_0_A = 0
            map_2_A(x1,x2) = max{x1 - 12, x2 + 30}
            plus_0_A = 10
            plus_1_A(x1) = x1 + 20
            plus_2_A(x1,x2) = max{10, x1 - 50, x2 + 1}
            s_0_A = 20
            s_1_A(x1) = max{10, x1 - 50}
            times_0_A = 20
            times_1_A(x1) = max{20, x1 - 33}
            times_2_A(x1,x2) = max{x1 - 54, x2 + 10}
            |0|_0_A = 0
            nil_0_A = 5
            app#!_A(x1,x2) = max{2, x2 + 1}
            curry_2!_A(x1,x2) = max{0, x1 - 11, x2 - 1}
            curry_3#!_A(x1,x2,x3) = max{2, x3 + 1}
            map_1!_A(x1) = max{25, x1 + 19}
            map_2#!_A(x1,x2) = max{26, x2 - 6}
            cons_2!_A(x1,x2) = max{1, x1 - 1, x2 - 31}
            app!_A(x1,x2) = 0
            cons_0!_A = 1
            cons_1!_A(x1) = 0
            curry_0!_A = 2
            curry_1!_A(x1) = max{1, x1 - 12}
            curry_3!_A(x1,x2,x3) = 0
            map_0!_A = 1
            map_2!_A(x1,x2) = max{26, x2 - 10}
            plus_0!_A = 1
            plus_1!_A(x1) = max{2, x1 - 1}
            plus_2!_A(x1,x2) = max{5, x2 + 3}
            s_0!_A = 1
            s_1!_A(x1) = 4
            times_0!_A = 1
            times_1!_A(x1) = max{2, x1 - 12}
            times_2!_A(x1,x2) = max{x1 - 11, x2 + 6}
            |0|_0!_A = 7
            nil_0!_A = 1
        
  
    partial status:
  
      pi(app#) = []
      pi(curry_2) = []
      pi(curry_3#) = []
      pi(map_1) = []
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app) = [2]
      pi(cons_0) = []
      pi(cons_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) = [1]
      pi(plus_2) = [2]
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_1) = [1]
      pi(times_2) = [2]
      pi(|0|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p2, 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: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p3: 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, p3}
  {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:
  
      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) = max{12, x1 - 9, x2 - 2}
            curry_2_A(x1,x2) = x2 + 20
            curry_3#_A(x1,x2,x3) = max{12, x2 + 11, x3 - 2}
            app_A(x1,x2) = x2 + 20
            cons_0_A = 8
            cons_1_A(x1) = x1 + 8
            cons_2_A(x1,x2) = 20
            curry_0_A = 8
            curry_1_A(x1) = x1 + 20
            curry_3_A(x1,x2,x3) = x3 + 20
            map_0_A = 8
            map_1_A(x1) = max{9, x1 + 4}
            map_2_A(x1,x2) = x2 + 20
            plus_0_A = 4
            plus_1_A(x1) = x1 + 8
            plus_2_A(x1,x2) = max{16, x2 + 11}
            s_0_A = 2
            s_1_A(x1) = max{16, x1 - 5}
            times_0_A = 4
            times_1_A(x1) = x1 + 8
            times_2_A(x1,x2) = max{16, x2 + 15}
            |0|_0_A = 16
            nil_0_A = 19
            app#!_A(x1,x2) = 0
            curry_2!_A(x1,x2) = max{16, x1 + 10}
            curry_3#!_A(x1,x2,x3) = x1 + 9
            app!_A(x1,x2) = max{8, x1 - 3}
            cons_0!_A = 0
            cons_1!_A(x1) = 0
            cons_2!_A(x1,x2) = 0
            curry_0!_A = 1
            curry_1!_A(x1) = 0
            curry_3!_A(x1,x2,x3) = x2 + 17
            map_0!_A = 1
            map_1!_A(x1) = 6
            map_2!_A(x1,x2) = 0
            plus_0!_A = 1
            plus_1!_A(x1) = 2
            plus_2!_A(x1,x2) = 6
            s_0!_A = 5
            s_1!_A(x1) = 4
            times_0!_A = 2
            times_1!_A(x1) = 1
            times_2!_A(x1,x2) = max{7, x1 - 17}
            |0|_0!_A = 1
            nil_0!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 16, x2 - 16}
            curry_2_A(x1,x2) = max{x1 + 6, x2 + 15}
            curry_3#_A(x1,x2,x3) = max{0, x1 - 16, x2 - 1, x3 - 16}
            app_A(x1,x2) = max{x1 - 8, x2 + 15}
            cons_0_A = 2
            cons_1_A(x1) = x1 + 4
            cons_2_A(x1,x2) = 0
            curry_0_A = 4
            curry_1_A(x1) = x1 + 14
            curry_3_A(x1,x2,x3) = max{x1 - 16, x2 + 7, x3 + 15}
            map_0_A = 0
            map_1_A(x1) = x1 + 7
            map_2_A(x1,x2) = max{x1 - 1, x2 + 15}
            plus_0_A = 9
            plus_1_A(x1) = max{15, x1 + 7}
            plus_2_A(x1,x2) = max{2, x1 - 1, x2 + 1}
            s_0_A = 4
            s_1_A(x1) = x1
            times_0_A = 9
            times_1_A(x1) = x1 + 4
            times_2_A(x1,x2) = max{2, x2 + 1}
            |0|_0_A = 0
            nil_0_A = 0
            app#!_A(x1,x2) = 0
            curry_2!_A(x1,x2) = 10
            curry_3#!_A(x1,x2,x3) = max{3, x1 - 2, x3 - 1}
            app!_A(x1,x2) = max{2, x1 - 3}
            cons_0!_A = 1
            cons_1!_A(x1) = max{0, x1 - 1}
            cons_2!_A(x1,x2) = 2
            curry_0!_A = 1
            curry_1!_A(x1) = 0
            curry_3!_A(x1,x2,x3) = max{x1 + 1, x2 + 12}
            map_0!_A = 1
            map_1!_A(x1) = 2
            map_2!_A(x1,x2) = max{2, x1 - 5}
            plus_0!_A = 10
            plus_1!_A(x1) = max{0, x1 - 1}
            plus_2!_A(x1,x2) = 3
            s_0!_A = 1
            s_1!_A(x1) = max{0, x1 - 3}
            times_0!_A = 1
            times_1!_A(x1) = 6
            times_2!_A(x1,x2) = max{7, x1 - 4, x2 - 1}
            |0|_0!_A = 8
            nil_0!_A = 1
        
  
    partial status:
  
      pi(app#) = [1]
      pi(curry_2) = [2]
      pi(curry_3#) = [2]
      pi(app) = [2]
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(cons_2) = []
      pi(curry_0) = []
      pi(curry_1) = []
      pi(curry_3) = [3]
      pi(map_0) = []
      pi(map_1) = [1]
      pi(map_2) = []
      pi(plus_0) = []
      pi(plus_1) = []
      pi(plus_2) = [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:

  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:
  
      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:
          
            map_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{3, x1 - 3, x2 - 3}
            map_2#!_A(x1,x2) = max{0, x2 - 15}
            cons_2!_A(x1,x2) = max{3, x1 - 16, x2 - 14}
        
        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{3, x1 - 3, x2 - 3}
            map_2#!_A(x1,x2) = max{2, x1 - 1, x2 - 25}
            cons_2!_A(x1,x2) = max{1, x1 - 26, x2 - 24}
        
  
    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:
  
      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{4, x1 - 6}
            times_2#!_A(x1,x2) = 0
            s_1!_A(x1) = max{4, x1 - 53}
        
        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) = max{5, x1 - 5, x2 - 1}
            s_1!_A(x1) = max{4, x1 - 24}
        
  
    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:
  
      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{6, x1 - 8}
            plus_2#!_A(x1,x2) = 0
            s_1!_A(x1) = max{4, x1 - 30}
        
        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 - 6}
            plus_2#!_A(x1,x2) = max{1, x1 - 17, x2 - 1}
            s_1!_A(x1) = max{4, x1 - 17}
        
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.