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(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
  app(app(app(curry(),g),x),y) -> app(app(g,x),y)
  inc() -> app(map(),app(app(curry(),plus()),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)
  plus_2(|0|_0(),y) -> y
  plus_2(s_1(x),y) -> s_1(plus_2(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))
  curry_3(g,x,y) -> app(app(g,x),y)
  inc_0() -> map_1(curry_2(plus_0(),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: plus_2#(s_1(x),y) -> plus_2#(x,y)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p7: curry_3#(g,x,y) -> app#(app(g,x),y)
p8: 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: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p5, p6, p7, p8}
  {p4}


-- 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: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = 12
        curry_2_A(x1,x2) = max{12, x2 - 5}
        curry_3#_A(x1,x2,x3) = 12
        map_1_A(x1) = max{0, x1 - 5}
        map_2#_A(x1,x2) = 12
        cons_2_A(x1,x2) = x2 + 14
        app_A(x1,x2) = x2 + 15
        cons_0_A = 1
        cons_1_A(x1) = max{11, x1 - 5}
        curry_0_A = 12
        curry_1_A(x1) = max{11, x1 - 5}
        curry_3_A(x1,x2,x3) = x3 + 15
        map_0_A = 1
        map_2_A(x1,x2) = x2
        plus_0_A = 6
        plus_1_A(x1) = max{7, x1 - 5}
        plus_2_A(x1,x2) = x2 + 10
        s_0_A = 10
        s_1_A(x1) = max{9, x1 - 11}
        |0|_0_A = 10
        nil_0_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 35, x2 - 32}
          curry_2_A(x1,x2) = max{34, x1 + 1, x2 + 10}
          curry_3#_A(x1,x2,x3) = max{0, x1 - 35, x2 - 25, x3 - 32}
          map_1_A(x1) = max{6, x1 + 3}
          map_2#_A(x1,x2) = max{0, x1 - 35, x2 - 32}
          cons_2_A(x1,x2) = max{10, x1 + 3, x2}
          app_A(x1,x2) = max{x1 - 7, x2 + 10}
          cons_0_A = 23
          cons_1_A(x1) = max{16, x1 + 10}
          curry_0_A = 48
          curry_1_A(x1) = max{41, x1 + 8}
          curry_3_A(x1,x2,x3) = max{11, x1 - 6, x2 + 3, x3 + 10}
          map_0_A = 12
          map_2_A(x1,x2) = max{x1 - 4, x2 + 10}
          plus_0_A = 29
          plus_1_A(x1) = max{22, x1 + 6}
          plus_2_A(x1,x2) = max{15, x1 - 15, x2 + 1}
          s_0_A = 6
          s_1_A(x1) = max{10, x1 - 14}
          |0|_0_A = 0
          nil_0_A = 6
          app#!_A(x1,x2) = max{0, x1 - 32}
          curry_2!_A(x1,x2) = max{2, x1 - 30, x2 - 21}
          curry_3#!_A(x1,x2,x3) = max{1, x1 - 31, x2 - 22}
          map_1!_A(x1) = max{2, x1 - 35}
          map_2#!_A(x1,x2) = max{0, x1 - 29}
          cons_2!_A(x1,x2) = 8
          app!_A(x1,x2) = 4
          cons_0!_A = 1
          cons_1!_A(x1) = max{5, x1 - 1}
          curry_0!_A = 1
          curry_1!_A(x1) = 5
          curry_3!_A(x1,x2,x3) = 4
          map_0!_A = 1
          map_2!_A(x1,x2) = max{7, x1 + 5, x2 - 1}
          plus_0!_A = 1
          plus_1!_A(x1) = max{3, x1 - 17}
          plus_2!_A(x1,x2) = max{2, x1 - 17, x2 - 15}
          s_0!_A = 2
          s_1!_A(x1) = max{1, x1 - 16}
          |0|_0!_A = 2
          nil_0!_A = 6
  
    partial status:
  
      pi(app#) = []
      pi(curry_2) = []
      pi(curry_3#) = []
      pi(map_1) = []
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(curry_0) = []
      pi(curry_1) = []
      pi(curry_3) = []
      pi(map_0) = []
      pi(map_2) = [2]
      pi(plus_0) = []
      pi(plus_1) = []
      pi(plus_2) = [2]
      pi(s_0) = []
      pi(s_1) = []
      pi(|0|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p2



-- 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: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p4: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: 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: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, 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#(app(g,x),y)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: 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: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = 24
        curry_2_A(x1,x2) = max{x1 - 6, x2 + 13}
        curry_3#_A(x1,x2,x3) = 24
        app_A(x1,x2) = max{x1 - 8, x2 + 13}
        map_1_A(x1) = max{23, x1 + 13}
        map_2#_A(x1,x2) = 24
        cons_2_A(x1,x2) = 3
        cons_0_A = 9
        cons_1_A(x1) = max{10, x1 + 7}
        curry_0_A = 9
        curry_1_A(x1) = max{7, x1 + 2}
        curry_3_A(x1,x2,x3) = max{x1 - 14, x2 + 5, x3 + 13}
        map_0_A = 31
        map_2_A(x1,x2) = max{14, x1 + 4}
        plus_0_A = 53
        plus_1_A(x1) = 44
        plus_2_A(x1,x2) = max{35, x2 + 1}
        s_0_A = 6
        s_1_A(x1) = max{7, x1 - 27}
        |0|_0_A = 0
        nil_0_A = 13
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 32, x2 - 21}
          curry_2_A(x1,x2) = max{x1 - 1, x2 + 31}
          curry_3#_A(x1,x2,x3) = max{0, x1 - 33, x2 - 1, x3 - 21}
          app_A(x1,x2) = max{x1 - 27, x2 + 31}
          map_1_A(x1) = max{20, x1 + 4}
          map_2#_A(x1,x2) = max{0, x1 - 28, x2 - 21}
          cons_2_A(x1,x2) = max{16, x1 + 4, x2}
          cons_0_A = 15
          cons_1_A(x1) = x1 + 31
          curry_0_A = 13
          curry_1_A(x1) = x1 + 26
          curry_3_A(x1,x2,x3) = max{x1 - 54, x2 + 4, x3 + 31}
          map_0_A = 13
          map_2_A(x1,x2) = max{x1 - 23, x2 + 31}
          plus_0_A = 16
          plus_1_A(x1) = max{15, x1 + 13}
          plus_2_A(x1,x2) = max{18, x1 - 14, x2 + 1}
          s_0_A = 22
          s_1_A(x1) = max{13, x1 - 16}
          |0|_0_A = 0
          nil_0_A = 16
          app#!_A(x1,x2) = max{4, x1 + 2}
          curry_2!_A(x1,x2) = max{10, x2 - 14}
          curry_3#!_A(x1,x2,x3) = max{x1 - 15, x2 + 33}
          app!_A(x1,x2) = max{8, x1 - 14, x2 - 3}
          map_1!_A(x1) = max{0, x1 - 45}
          map_2#!_A(x1,x2) = max{7, x1 + 5}
          cons_2!_A(x1,x2) = max{3, x1 - 4, x2 - 32}
          cons_0!_A = 1
          cons_1!_A(x1) = max{2, x1 - 5}
          curry_0!_A = 1
          curry_1!_A(x1) = max{9, x1 + 1}
          curry_3!_A(x1,x2,x3) = max{x1 - 15, x2 + 17, x3 - 3}
          map_0!_A = 1
          map_2!_A(x1,x2) = max{7, x1}
          plus_0!_A = 1
          plus_1!_A(x1) = 1
          plus_2!_A(x1,x2) = max{7, x1 - 2, x2 - 4}
          s_0!_A = 6
          s_1!_A(x1) = max{7, x1 - 37}
          |0|_0!_A = 0
          nil_0!_A = 1
  
    partial status:
  
      pi(app#) = []
      pi(curry_2) = [2]
      pi(curry_3#) = []
      pi(app) = [2]
      pi(map_1) = []
      pi(map_2#) = []
      pi(cons_2) = []
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(curry_0) = []
      pi(curry_1) = []
      pi(curry_3) = [3]
      pi(map_0) = []
      pi(map_2) = [1]
      pi(plus_0) = []
      pi(plus_1) = []
      pi(plus_2) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(|0|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p3, p4



-- 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)
p3: 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: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),s_1(|0|_0())))

The estimated dependency graph contains the following SCCs:

  {p1, p2}
  {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)

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: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = x1
        curry_2_A(x1,x2) = max{x1, x2}
        curry_3#_A(x1,x2,x3) = max{x1, x2}
        app_A(x1,x2) = max{x1, x2}
        cons_0_A = 0
        cons_1_A(x1) = x1
        cons_2_A(x1,x2) = 0
        curry_0_A = 1
        curry_1_A(x1) = max{1, x1}
        curry_3_A(x1,x2,x3) = max{x1, x2, x3}
        map_0_A = 1
        map_1_A(x1) = x1
        map_2_A(x1,x2) = max{0, x1 - 2}
        plus_0_A = 1
        plus_1_A(x1) = 1
        plus_2_A(x1,x2) = max{1, x2}
        s_0_A = 1
        s_1_A(x1) = max{0, x1 - 2}
        |0|_0_A = 1
        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{1, x1 - 3, x2 - 3}
          curry_2_A(x1,x2) = max{x1 - 12, x2 + 11}
          curry_3#_A(x1,x2,x3) = max{x1 - 15, x2 + 8, x3 - 3}
          app_A(x1,x2) = max{x1 - 12, x2 + 11}
          cons_0_A = 4
          cons_1_A(x1) = x1 + 4
          cons_2_A(x1,x2) = 10
          curry_0_A = 2
          curry_1_A(x1) = max{11, x1 + 9}
          curry_3_A(x1,x2,x3) = max{x1 - 24, x2 - 1, x3 + 11}
          map_0_A = 0
          map_1_A(x1) = max{11, x1 + 4}
          map_2_A(x1,x2) = x2 + 11
          plus_0_A = 2
          plus_1_A(x1) = x1 + 2
          plus_2_A(x1,x2) = max{11, x2 + 10}
          s_0_A = 4
          s_1_A(x1) = max{11, x1 - 33}
          |0|_0_A = 11
          nil_0_A = 10
          app#!_A(x1,x2) = 0
          curry_2!_A(x1,x2) = max{7, x1 + 5}
          curry_3#!_A(x1,x2,x3) = x1 + 4
          app!_A(x1,x2) = max{3, 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) = 2
          curry_3!_A(x1,x2,x3) = max{x1 - 15, x2 + 8}
          map_0!_A = 1
          map_1!_A(x1) = 2
          map_2!_A(x1,x2) = 0
          plus_0!_A = 2
          plus_1!_A(x1) = 1
          plus_2!_A(x1,x2) = max{2, x1 - 12}
          s_0!_A = 0
          s_1!_A(x1) = 0
          |0|_0!_A = 0
          nil_0!_A = 0
  
    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) = []
      pi(map_2) = []
      pi(plus_0) = []
      pi(plus_1) = []
      pi(plus_2) = [2]
      pi(s_0) = []
      pi(s_1) = []
      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: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),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) = max{4, x2}
        cons_2_A(x1,x2) = max{4, x2}
  
    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{6, x1 - 2, x2 - 4}
          map_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = max{3, x1 - 6, x2 - 3}
  
    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: 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: plus_2(|0|_0(),y) -> y
r12: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r13: map_2(f,nil_0()) -> nil_0()
r14: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r15: curry_3(g,x,y) -> app(app(g,x),y)
r16: inc_0() -> map_1(curry_2(plus_0(),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{x1 + 4, x2 + 5}
        s_1_A(x1) = x1 + 3
  
    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{3, x1 - 3}
          plus_2#!_A(x1,x2) = max{3, x2 + 1}
          s_1!_A(x1) = max{2, x1 - 10}
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.