YES
We show the termination of the TRS R:

  ap(ap(map(),f),xs) -> ap(ap(ap(if(),ap(isEmpty(),xs)),f),xs)
  ap(ap(ap(if(),true()),f),xs) -> nil()
  ap(ap(ap(if(),false()),f),xs) -> ap(ap(cons(),ap(f,ap(last(),xs))),ap(ap(map(),f),ap(dropLast(),xs)))
  ap(isEmpty(),nil()) -> true()
  ap(isEmpty(),ap(ap(cons(),x),xs)) -> false()
  ap(last(),ap(ap(cons(),x),nil())) -> x
  ap(last(),ap(ap(cons(),x),ap(ap(cons(),y),ys))) -> ap(last(),ap(ap(cons(),y),ys))
  ap(dropLast(),nil()) -> nil()
  ap(dropLast(),ap(ap(cons(),x),nil())) -> nil()
  ap(dropLast(),ap(ap(cons(),x),ap(ap(cons(),y),ys))) -> ap(ap(cons(),x),ap(dropLast(),ap(ap(cons(),y),ys)))

Instead we prove termination of the uncurried system:

  ap(cons_0(),y) -> cons_1(y)
  ap(cons_1(x1),y) -> cons_2(x1,y)
  ap(dropLast_0(),y) -> dropLast_1(y)
  ap(if_0(),y) -> if_1(y)
  ap(if_1(x1),y) -> if_2(x1,y)
  ap(if_2(x1,x2),y) -> if_3(x1,x2,y)
  ap(isEmpty_0(),y) -> isEmpty_1(y)
  ap(last_0(),y) -> last_1(y)
  ap(map_0(),y) -> map_1(y)
  ap(map_1(x1),y) -> map_2(x1,y)
  map_2(f,xs) -> if_3(isEmpty_1(xs),f,xs)
  if_3(true_0(),f,xs) -> nil_0()
  if_3(false_0(),f,xs) -> cons_2(ap(f,last_1(xs)),map_2(f,dropLast_1(xs)))
  isEmpty_1(nil_0()) -> true_0()
  isEmpty_1(cons_2(x,xs)) -> false_0()
  last_1(cons_2(x,nil_0())) -> x
  last_1(cons_2(x,cons_2(y,ys))) -> last_1(cons_2(y,ys))
  dropLast_1(nil_0()) -> nil_0()
  dropLast_1(cons_2(x,nil_0())) -> nil_0()
  dropLast_1(cons_2(x,cons_2(y,ys))) -> cons_2(x,dropLast_1(cons_2(y,ys)))


-- SCC decomposition.

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

p1: ap#(dropLast_0(),y) -> dropLast_1#(y)
p2: ap#(if_2(x1,x2),y) -> if_3#(x1,x2,y)
p3: ap#(isEmpty_0(),y) -> isEmpty_1#(y)
p4: ap#(last_0(),y) -> last_1#(y)
p5: ap#(map_1(x1),y) -> map_2#(x1,y)
p6: map_2#(f,xs) -> if_3#(isEmpty_1(xs),f,xs)
p7: map_2#(f,xs) -> isEmpty_1#(xs)
p8: if_3#(false_0(),f,xs) -> ap#(f,last_1(xs))
p9: if_3#(false_0(),f,xs) -> last_1#(xs)
p10: if_3#(false_0(),f,xs) -> map_2#(f,dropLast_1(xs))
p11: if_3#(false_0(),f,xs) -> dropLast_1#(xs)
p12: last_1#(cons_2(x,cons_2(y,ys))) -> last_1#(cons_2(y,ys))
p13: dropLast_1#(cons_2(x,cons_2(y,ys))) -> dropLast_1#(cons_2(y,ys))

and R consists of:

r1: ap(cons_0(),y) -> cons_1(y)
r2: ap(cons_1(x1),y) -> cons_2(x1,y)
r3: ap(dropLast_0(),y) -> dropLast_1(y)
r4: ap(if_0(),y) -> if_1(y)
r5: ap(if_1(x1),y) -> if_2(x1,y)
r6: ap(if_2(x1,x2),y) -> if_3(x1,x2,y)
r7: ap(isEmpty_0(),y) -> isEmpty_1(y)
r8: ap(last_0(),y) -> last_1(y)
r9: ap(map_0(),y) -> map_1(y)
r10: ap(map_1(x1),y) -> map_2(x1,y)
r11: map_2(f,xs) -> if_3(isEmpty_1(xs),f,xs)
r12: if_3(true_0(),f,xs) -> nil_0()
r13: if_3(false_0(),f,xs) -> cons_2(ap(f,last_1(xs)),map_2(f,dropLast_1(xs)))
r14: isEmpty_1(nil_0()) -> true_0()
r15: isEmpty_1(cons_2(x,xs)) -> false_0()
r16: last_1(cons_2(x,nil_0())) -> x
r17: last_1(cons_2(x,cons_2(y,ys))) -> last_1(cons_2(y,ys))
r18: dropLast_1(nil_0()) -> nil_0()
r19: dropLast_1(cons_2(x,nil_0())) -> nil_0()
r20: dropLast_1(cons_2(x,cons_2(y,ys))) -> cons_2(x,dropLast_1(cons_2(y,ys)))

The estimated dependency graph contains the following SCCs:

  {p2, p5, p6, p8, p10}
  {p13}
  {p12}


-- Reduction pair.

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

p1: if_3#(false_0(),f,xs) -> ap#(f,last_1(xs))
p2: ap#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(f,xs) -> if_3#(isEmpty_1(xs),f,xs)
p4: if_3#(false_0(),f,xs) -> map_2#(f,dropLast_1(xs))
p5: ap#(if_2(x1,x2),y) -> if_3#(x1,x2,y)

and R consists of:

r1: ap(cons_0(),y) -> cons_1(y)
r2: ap(cons_1(x1),y) -> cons_2(x1,y)
r3: ap(dropLast_0(),y) -> dropLast_1(y)
r4: ap(if_0(),y) -> if_1(y)
r5: ap(if_1(x1),y) -> if_2(x1,y)
r6: ap(if_2(x1,x2),y) -> if_3(x1,x2,y)
r7: ap(isEmpty_0(),y) -> isEmpty_1(y)
r8: ap(last_0(),y) -> last_1(y)
r9: ap(map_0(),y) -> map_1(y)
r10: ap(map_1(x1),y) -> map_2(x1,y)
r11: map_2(f,xs) -> if_3(isEmpty_1(xs),f,xs)
r12: if_3(true_0(),f,xs) -> nil_0()
r13: if_3(false_0(),f,xs) -> cons_2(ap(f,last_1(xs)),map_2(f,dropLast_1(xs)))
r14: isEmpty_1(nil_0()) -> true_0()
r15: isEmpty_1(cons_2(x,xs)) -> false_0()
r16: last_1(cons_2(x,nil_0())) -> x
r17: last_1(cons_2(x,cons_2(y,ys))) -> last_1(cons_2(y,ys))
r18: dropLast_1(nil_0()) -> nil_0()
r19: dropLast_1(cons_2(x,nil_0())) -> nil_0()
r20: dropLast_1(cons_2(x,cons_2(y,ys))) -> cons_2(x,dropLast_1(cons_2(y,ys)))

The set of usable rules consists of

  r7, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_3#_A(x1,x2,x3) = 1
        false_0_A = 0
        ap#_A(x1,x2) = 1
        last_1_A(x1) = max{5, x1 - 6}
        map_1_A(x1) = max{3, x1 - 26}
        map_2#_A(x1,x2) = 1
        isEmpty_1_A(x1) = max{2, x1 - 15}
        dropLast_1_A(x1) = max{0, x1 - 6}
        if_2_A(x1,x2) = max{2, x1 - 26, x2 - 26}
        ap_A(x1,x2) = max{4, x1 - 15, x2 - 9}
        isEmpty_0_A = 1
        nil_0_A = 0
        true_0_A = 0
        cons_2_A(x1,x2) = max{0, x2 - 9}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_3#_A(x1,x2,x3) = max{1, x1 - 8, x2 - 137}
          false_0_A = 1
          ap#_A(x1,x2) = max{1, x1 - 137}
          last_1_A(x1) = max{93, x1 - 1}
          map_1_A(x1) = max{136, x1 + 130}
          map_2#_A(x1,x2) = max{1, x1 - 137}
          isEmpty_1_A(x1) = 1
          dropLast_1_A(x1) = max{21, x1 - 23}
          if_2_A(x1,x2) = max{130, x1 + 129, x2 + 129}
          ap_A(x1,x2) = max{57, x1 - 199, x2 - 36}
          isEmpty_0_A = 165
          nil_0_A = 22
          true_0_A = 1
          cons_2_A(x1,x2) = max{22, x1 - 35, x2 - 35}
          if_3#!_A(x1,x2,x3) = max{4, x1 - 2, x2 - 130}
          false_0!_A = 3
          ap#!_A(x1,x2) = max{4, x1 - 131}
          last_1!_A(x1) = max{8, x1 - 152}
          map_1!_A(x1) = max{5, x1 - 129}
          map_2#!_A(x1,x2) = max{4, x1 - 130}
          isEmpty_1!_A(x1) = max{5, x1 - 133}
          dropLast_1!_A(x1) = max{2, x1 - 111}
          if_2!_A(x1,x2) = max{5, x1 - 1, x2 - 33}
          ap!_A(x1,x2) = max{7, x1 - 164, x2 - 128}
          isEmpty_0!_A = 1
          nil_0!_A = 1
          true_0!_A = 0
          cons_2!_A(x1,x2) = max{4, x1 - 90, x2 - 129}
  
    partial status:
  
      pi(if_3#) = []
      pi(false_0) = []
      pi(ap#) = []
      pi(last_1) = []
      pi(map_1) = []
      pi(map_2#) = []
      pi(isEmpty_1) = []
      pi(dropLast_1) = []
      pi(if_2) = []
      pi(ap) = []
      pi(isEmpty_0) = []
      pi(nil_0) = []
      pi(true_0) = []
      pi(cons_2) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: if_3#(false_0(),f,xs) -> ap#(f,last_1(xs))
p2: map_2#(f,xs) -> if_3#(isEmpty_1(xs),f,xs)
p3: if_3#(false_0(),f,xs) -> map_2#(f,dropLast_1(xs))
p4: ap#(if_2(x1,x2),y) -> if_3#(x1,x2,y)

and R consists of:

r1: ap(cons_0(),y) -> cons_1(y)
r2: ap(cons_1(x1),y) -> cons_2(x1,y)
r3: ap(dropLast_0(),y) -> dropLast_1(y)
r4: ap(if_0(),y) -> if_1(y)
r5: ap(if_1(x1),y) -> if_2(x1,y)
r6: ap(if_2(x1,x2),y) -> if_3(x1,x2,y)
r7: ap(isEmpty_0(),y) -> isEmpty_1(y)
r8: ap(last_0(),y) -> last_1(y)
r9: ap(map_0(),y) -> map_1(y)
r10: ap(map_1(x1),y) -> map_2(x1,y)
r11: map_2(f,xs) -> if_3(isEmpty_1(xs),f,xs)
r12: if_3(true_0(),f,xs) -> nil_0()
r13: if_3(false_0(),f,xs) -> cons_2(ap(f,last_1(xs)),map_2(f,dropLast_1(xs)))
r14: isEmpty_1(nil_0()) -> true_0()
r15: isEmpty_1(cons_2(x,xs)) -> false_0()
r16: last_1(cons_2(x,nil_0())) -> x
r17: last_1(cons_2(x,cons_2(y,ys))) -> last_1(cons_2(y,ys))
r18: dropLast_1(nil_0()) -> nil_0()
r19: dropLast_1(cons_2(x,nil_0())) -> nil_0()
r20: dropLast_1(cons_2(x,cons_2(y,ys))) -> cons_2(x,dropLast_1(cons_2(y,ys)))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: if_3#(false_0(),f,xs) -> ap#(f,last_1(xs))
p2: ap#(if_2(x1,x2),y) -> if_3#(x1,x2,y)
p3: if_3#(false_0(),f,xs) -> map_2#(f,dropLast_1(xs))
p4: map_2#(f,xs) -> if_3#(isEmpty_1(xs),f,xs)

and R consists of:

r1: ap(cons_0(),y) -> cons_1(y)
r2: ap(cons_1(x1),y) -> cons_2(x1,y)
r3: ap(dropLast_0(),y) -> dropLast_1(y)
r4: ap(if_0(),y) -> if_1(y)
r5: ap(if_1(x1),y) -> if_2(x1,y)
r6: ap(if_2(x1,x2),y) -> if_3(x1,x2,y)
r7: ap(isEmpty_0(),y) -> isEmpty_1(y)
r8: ap(last_0(),y) -> last_1(y)
r9: ap(map_0(),y) -> map_1(y)
r10: ap(map_1(x1),y) -> map_2(x1,y)
r11: map_2(f,xs) -> if_3(isEmpty_1(xs),f,xs)
r12: if_3(true_0(),f,xs) -> nil_0()
r13: if_3(false_0(),f,xs) -> cons_2(ap(f,last_1(xs)),map_2(f,dropLast_1(xs)))
r14: isEmpty_1(nil_0()) -> true_0()
r15: isEmpty_1(cons_2(x,xs)) -> false_0()
r16: last_1(cons_2(x,nil_0())) -> x
r17: last_1(cons_2(x,cons_2(y,ys))) -> last_1(cons_2(y,ys))
r18: dropLast_1(nil_0()) -> nil_0()
r19: dropLast_1(cons_2(x,nil_0())) -> nil_0()
r20: dropLast_1(cons_2(x,cons_2(y,ys))) -> cons_2(x,dropLast_1(cons_2(y,ys)))

The set of usable rules consists of

  r7, r14, r15, r16, r17, r18, r19, r20

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_3#_A(x1,x2,x3) = 0
        false_0_A = 8
        ap#_A(x1,x2) = 0
        last_1_A(x1) = max{51, x1 + 35}
        if_2_A(x1,x2) = max{18, x1 - 106, x2 - 15}
        map_2#_A(x1,x2) = 0
        dropLast_1_A(x1) = max{27, x1 + 11}
        isEmpty_1_A(x1) = max{15, x1 - 53}
        ap_A(x1,x2) = max{16, x1 - 53, x2 - 52}
        isEmpty_0_A = 1
        nil_0_A = 29
        true_0_A = 0
        cons_2_A(x1,x2) = max{17, x1 - 21, x2 + 9}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_3#_A(x1,x2,x3) = max{0, x1 - 9, x2 - 29, x3 - 13}
          false_0_A = 0
          ap#_A(x1,x2) = max{0, x1 - 29, x2 - 9}
          last_1_A(x1) = max{1, x1 - 4}
          if_2_A(x1,x2) = max{x1 + 20, x2 + 2}
          map_2#_A(x1,x2) = max{0, x1 - 29, x2 - 13}
          dropLast_1_A(x1) = max{12, x1}
          isEmpty_1_A(x1) = max{0, x1 - 34}
          ap_A(x1,x2) = max{0, x1 - 45, x2 - 19}
          isEmpty_0_A = 42
          nil_0_A = 27
          true_0_A = 0
          cons_2_A(x1,x2) = max{26, x1 + 17, x2 + 5}
          if_3#!_A(x1,x2,x3) = max{16, x1 - 3, x2 - 2, x3 + 2}
          false_0!_A = 15
          ap#!_A(x1,x2) = max{x1 - 3, x2 + 3}
          last_1!_A(x1) = max{16, x1 - 42}
          if_2!_A(x1,x2) = max{19, x1 - 23}
          map_2#!_A(x1,x2) = max{16, x1 - 2, x2 + 2}
          dropLast_1!_A(x1) = max{21, x1 - 33}
          isEmpty_1!_A(x1) = max{16, x1 - 56}
          ap!_A(x1,x2) = max{18, x1 - 43, x2 - 44}
          isEmpty_0!_A = 1
          nil_0!_A = 1
          true_0!_A = 0
          cons_2!_A(x1,x2) = max{15, x1 - 21, x2 - 34}
  
    partial status:
  
      pi(if_3#) = []
      pi(false_0) = []
      pi(ap#) = []
      pi(last_1) = []
      pi(if_2) = []
      pi(map_2#) = []
      pi(dropLast_1) = []
      pi(isEmpty_1) = []
      pi(ap) = []
      pi(isEmpty_0) = []
      pi(nil_0) = []
      pi(true_0) = []
      pi(cons_2) = [2]

The following rules are removed:

  p1, p2



-- SCC decomposition.

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

p1: if_3#(false_0(),f,xs) -> map_2#(f,dropLast_1(xs))
p2: map_2#(f,xs) -> if_3#(isEmpty_1(xs),f,xs)

and R consists of:

r1: ap(cons_0(),y) -> cons_1(y)
r2: ap(cons_1(x1),y) -> cons_2(x1,y)
r3: ap(dropLast_0(),y) -> dropLast_1(y)
r4: ap(if_0(),y) -> if_1(y)
r5: ap(if_1(x1),y) -> if_2(x1,y)
r6: ap(if_2(x1,x2),y) -> if_3(x1,x2,y)
r7: ap(isEmpty_0(),y) -> isEmpty_1(y)
r8: ap(last_0(),y) -> last_1(y)
r9: ap(map_0(),y) -> map_1(y)
r10: ap(map_1(x1),y) -> map_2(x1,y)
r11: map_2(f,xs) -> if_3(isEmpty_1(xs),f,xs)
r12: if_3(true_0(),f,xs) -> nil_0()
r13: if_3(false_0(),f,xs) -> cons_2(ap(f,last_1(xs)),map_2(f,dropLast_1(xs)))
r14: isEmpty_1(nil_0()) -> true_0()
r15: isEmpty_1(cons_2(x,xs)) -> false_0()
r16: last_1(cons_2(x,nil_0())) -> x
r17: last_1(cons_2(x,cons_2(y,ys))) -> last_1(cons_2(y,ys))
r18: dropLast_1(nil_0()) -> nil_0()
r19: dropLast_1(cons_2(x,nil_0())) -> nil_0()
r20: dropLast_1(cons_2(x,cons_2(y,ys))) -> cons_2(x,dropLast_1(cons_2(y,ys)))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: if_3#(false_0(),f,xs) -> map_2#(f,dropLast_1(xs))
p2: map_2#(f,xs) -> if_3#(isEmpty_1(xs),f,xs)

and R consists of:

r1: ap(cons_0(),y) -> cons_1(y)
r2: ap(cons_1(x1),y) -> cons_2(x1,y)
r3: ap(dropLast_0(),y) -> dropLast_1(y)
r4: ap(if_0(),y) -> if_1(y)
r5: ap(if_1(x1),y) -> if_2(x1,y)
r6: ap(if_2(x1,x2),y) -> if_3(x1,x2,y)
r7: ap(isEmpty_0(),y) -> isEmpty_1(y)
r8: ap(last_0(),y) -> last_1(y)
r9: ap(map_0(),y) -> map_1(y)
r10: ap(map_1(x1),y) -> map_2(x1,y)
r11: map_2(f,xs) -> if_3(isEmpty_1(xs),f,xs)
r12: if_3(true_0(),f,xs) -> nil_0()
r13: if_3(false_0(),f,xs) -> cons_2(ap(f,last_1(xs)),map_2(f,dropLast_1(xs)))
r14: isEmpty_1(nil_0()) -> true_0()
r15: isEmpty_1(cons_2(x,xs)) -> false_0()
r16: last_1(cons_2(x,nil_0())) -> x
r17: last_1(cons_2(x,cons_2(y,ys))) -> last_1(cons_2(y,ys))
r18: dropLast_1(nil_0()) -> nil_0()
r19: dropLast_1(cons_2(x,nil_0())) -> nil_0()
r20: dropLast_1(cons_2(x,cons_2(y,ys))) -> cons_2(x,dropLast_1(cons_2(y,ys)))

The set of usable rules consists of

  r3, r14, r15, r18, r19, r20

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_3#_A(x1,x2,x3) = max{x1 - 2, x3 + 37}
        false_0_A = 51
        map_2#_A(x1,x2) = max{49, x2 + 41}
        dropLast_1_A(x1) = max{7, x1 - 4}
        isEmpty_1_A(x1) = x1 + 40
        ap_A(x1,x2) = x2 + 35
        dropLast_0_A = 37
        nil_0_A = 6
        true_0_A = 2
        cons_2_A(x1,x2) = x2 + 11
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_3#_A(x1,x2,x3) = max{x1 - 33, x2 + 5}
          false_0_A = 34
          map_2#_A(x1,x2) = max{x1 + 5, x2 - 9}
          dropLast_1_A(x1) = 14
          isEmpty_1_A(x1) = x1 + 24
          ap_A(x1,x2) = max{14, x1 - 42, x2 - 22}
          dropLast_0_A = 19
          nil_0_A = 13
          true_0_A = 34
          cons_2_A(x1,x2) = 13
          if_3#!_A(x1,x2,x3) = max{0, x1 - 26}
          false_0!_A = 5
          map_2#!_A(x1,x2) = max{7, x2 - 10}
          dropLast_1!_A(x1) = x1 + 24
          isEmpty_1!_A(x1) = x1 + 37
          ap!_A(x1,x2) = max{3, x1 - 18, x2 - 26}
          dropLast_0!_A = 1
          nil_0!_A = 0
          true_0!_A = 5
          cons_2!_A(x1,x2) = 0
  
    partial status:
  
      pi(if_3#) = []
      pi(false_0) = []
      pi(map_2#) = []
      pi(dropLast_1) = []
      pi(isEmpty_1) = [1]
      pi(ap) = [2]
      pi(dropLast_0) = []
      pi(nil_0) = []
      pi(true_0) = []
      pi(cons_2) = []

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: dropLast_1#(cons_2(x,cons_2(y,ys))) -> dropLast_1#(cons_2(y,ys))

and R consists of:

r1: ap(cons_0(),y) -> cons_1(y)
r2: ap(cons_1(x1),y) -> cons_2(x1,y)
r3: ap(dropLast_0(),y) -> dropLast_1(y)
r4: ap(if_0(),y) -> if_1(y)
r5: ap(if_1(x1),y) -> if_2(x1,y)
r6: ap(if_2(x1,x2),y) -> if_3(x1,x2,y)
r7: ap(isEmpty_0(),y) -> isEmpty_1(y)
r8: ap(last_0(),y) -> last_1(y)
r9: ap(map_0(),y) -> map_1(y)
r10: ap(map_1(x1),y) -> map_2(x1,y)
r11: map_2(f,xs) -> if_3(isEmpty_1(xs),f,xs)
r12: if_3(true_0(),f,xs) -> nil_0()
r13: if_3(false_0(),f,xs) -> cons_2(ap(f,last_1(xs)),map_2(f,dropLast_1(xs)))
r14: isEmpty_1(nil_0()) -> true_0()
r15: isEmpty_1(cons_2(x,xs)) -> false_0()
r16: last_1(cons_2(x,nil_0())) -> x
r17: last_1(cons_2(x,cons_2(y,ys))) -> last_1(cons_2(y,ys))
r18: dropLast_1(nil_0()) -> nil_0()
r19: dropLast_1(cons_2(x,nil_0())) -> nil_0()
r20: dropLast_1(cons_2(x,cons_2(y,ys))) -> cons_2(x,dropLast_1(cons_2(y,ys)))

The set of usable rules consists of

  r12, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        dropLast_1#_A(x1) = max{13, x1 - 1}
        cons_2_A(x1,x2) = max{13, x2 + 4}
        if_3_A(x1,x2,x3) = max{11, x1 - 57, x2 - 17, x3 - 1}
        true_0_A = 0
        nil_0_A = 1
        dropLast_1_A(x1) = max{10, x1 - 18}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          dropLast_1#_A(x1) = 0
          cons_2_A(x1,x2) = max{x1 - 9, x2 + 10}
          if_3_A(x1,x2,x3) = max{11, x1 - 42, x2 - 93, x3 - 33}
          true_0_A = 2
          nil_0_A = 10
          dropLast_1_A(x1) = 48
          dropLast_1#!_A(x1) = max{17, x1 - 4}
          cons_2!_A(x1,x2) = max{x1 + 68, x2 + 6}
          if_3!_A(x1,x2,x3) = max{x1 - 3, x2 - 14, x3 + 46}
          true_0!_A = 0
          nil_0!_A = 70
          dropLast_1!_A(x1) = 69
  
    partial status:
  
      pi(dropLast_1#) = []
      pi(cons_2) = []
      pi(if_3) = []
      pi(true_0) = []
      pi(nil_0) = []
      pi(dropLast_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: last_1#(cons_2(x,cons_2(y,ys))) -> last_1#(cons_2(y,ys))

and R consists of:

r1: ap(cons_0(),y) -> cons_1(y)
r2: ap(cons_1(x1),y) -> cons_2(x1,y)
r3: ap(dropLast_0(),y) -> dropLast_1(y)
r4: ap(if_0(),y) -> if_1(y)
r5: ap(if_1(x1),y) -> if_2(x1,y)
r6: ap(if_2(x1,x2),y) -> if_3(x1,x2,y)
r7: ap(isEmpty_0(),y) -> isEmpty_1(y)
r8: ap(last_0(),y) -> last_1(y)
r9: ap(map_0(),y) -> map_1(y)
r10: ap(map_1(x1),y) -> map_2(x1,y)
r11: map_2(f,xs) -> if_3(isEmpty_1(xs),f,xs)
r12: if_3(true_0(),f,xs) -> nil_0()
r13: if_3(false_0(),f,xs) -> cons_2(ap(f,last_1(xs)),map_2(f,dropLast_1(xs)))
r14: isEmpty_1(nil_0()) -> true_0()
r15: isEmpty_1(cons_2(x,xs)) -> false_0()
r16: last_1(cons_2(x,nil_0())) -> x
r17: last_1(cons_2(x,cons_2(y,ys))) -> last_1(cons_2(y,ys))
r18: dropLast_1(nil_0()) -> nil_0()
r19: dropLast_1(cons_2(x,nil_0())) -> nil_0()
r20: dropLast_1(cons_2(x,cons_2(y,ys))) -> cons_2(x,dropLast_1(cons_2(y,ys)))

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:
      
        last_1#_A(x1) = max{12, x1}
        cons_2_A(x1,x2) = max{x1 + 12, x2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          last_1#_A(x1) = 0
          cons_2_A(x1,x2) = max{0, x1 - 9, x2 - 9}
          last_1#!_A(x1) = x1 + 20
          cons_2!_A(x1,x2) = max{x1 - 14, x2 + 28}
  
    partial status:
  
      pi(last_1#) = [1]
      pi(cons_2) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.