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

  r1, r2, r4, r12, r14, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_3#_A(x1,x2,x3) = x2 + 9
        false_0_A = 4
        ap#_A(x1,x2) = max{8, x1 + 2}
        last_1_A(x1) = max{10, x1 - 27}
        map_1_A(x1) = max{12, x1 + 11}
        map_2#_A(x1,x2) = x1 + 9
        isEmpty_1_A(x1) = max{11, x1 - 13}
        dropLast_1_A(x1) = max{5, x1 - 28}
        if_2_A(x1,x2) = max{12, x1 + 5, x2 + 10}
        ap_A(x1,x2) = max{11, x1 + 3, x2 - 13}
        cons_0_A = 1
        cons_1_A(x1) = max{10, x1 - 21}
        cons_2_A(x1,x2) = max{12, x1 - 20, x2 - 14}
        if_0_A = 1
        if_1_A(x1) = max{5, x1 - 14}
        if_3_A(x1,x2,x3) = max{16, x1 - 28, x2 - 16, x3 - 10}
        true_0_A = 0
        nil_0_A = 1
  
    base order 2:
  
      precedence:
      
        if_3 > if_2 = if_0 > ap = if_1 > cons_2 > last_1 = isEmpty_1 = dropLast_1 = cons_1 > if_3# = map_2# = cons_0 = nil_0 > ap# = map_1 = true_0 > false_0
      
  
    partial status:
  
      pi(if_3#) = [2]
      pi(false_0) = []
      pi(ap#) = []
      pi(last_1) = []
      pi(map_1) = [1]
      pi(map_2#) = [1]
      pi(isEmpty_1) = []
      pi(dropLast_1) = []
      pi(if_2) = [1]
      pi(ap) = [1]
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_3) = []
      pi(true_0) = []
      pi(nil_0) = []

The following rules are removed:

  p5



-- 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: 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))

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#(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))

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

  r1, r4, r5, r12, r14, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_3#_A(x1,x2,x3) = x2 + 15
        false_0_A = 11
        ap#_A(x1,x2) = max{13, x1 + 1}
        last_1_A(x1) = max{2, x1 - 14}
        map_1_A(x1) = x1 + 16
        map_2#_A(x1,x2) = x1 + 15
        isEmpty_1_A(x1) = max{10, x1 + 7}
        dropLast_1_A(x1) = max{2, x1 - 8}
        ap_A(x1,x2) = max{12, x1 - 17, x2 + 8}
        cons_0_A = 1
        cons_1_A(x1) = max{2, x1 - 5}
        if_0_A = 9
        if_1_A(x1) = max{10, x1 - 1}
        if_2_A(x1,x2) = max{11, x1 - 20, x2 - 3}
        if_3_A(x1,x2,x3) = max{13, x1 - 19, x2 - 2, x3 - 2}
        true_0_A = 0
        nil_0_A = 1
  
    base order 2:
  
      precedence:
      
        if_3 > ap > last_1 = if_0 = if_2 > ap# = map_1 = cons_1 = if_1 > if_3# = map_2# = cons_0 > dropLast_1 > false_0 = nil_0 > isEmpty_1 > true_0
      
  
    partial status:
  
      pi(if_3#) = []
      pi(false_0) = []
      pi(ap#) = [1]
      pi(last_1) = []
      pi(map_1) = []
      pi(map_2#) = []
      pi(isEmpty_1) = []
      pi(dropLast_1) = []
      pi(ap) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_2) = []
      pi(if_3) = []
      pi(true_0) = []
      pi(nil_0) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: ap#(map_1(x1),y) -> map_2#(x1,y)
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))

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, p3}


-- Reduction pair.

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

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

  r4, r5, r12, r14, r15, r18, r19, r20

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = x2 + 10
        if_3#_A(x1,x2,x3) = max{10, x1 + 7, x3 + 1}
        isEmpty_1_A(x1) = max{3, x1 - 6}
        false_0_A = 6
        dropLast_1_A(x1) = max{2, x1 - 10}
        ap_A(x1,x2) = max{0, x1 - 9, x2 - 5}
        if_0_A = 8
        if_1_A(x1) = max{0, x1 - 5}
        if_2_A(x1,x2) = max{0, x1 - 16, x2 - 6}
        if_3_A(x1,x2,x3) = max{x1 - 15, x2 + 14, x3 - 1}
        true_0_A = 0
        nil_0_A = 1
        cons_2_A(x1,x2) = x2 + 13
  
    base order 2:
  
      precedence:
      
        if_3 > dropLast_1 = ap = cons_2 > isEmpty_1 = if_2 = nil_0 > map_2# = if_1 = true_0 > if_3# = false_0 = if_0
      
  
    partial status:
  
      pi(map_2#) = []
      pi(if_3#) = [3]
      pi(isEmpty_1) = []
      pi(false_0) = []
      pi(dropLast_1) = []
      pi(ap) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_2) = []
      pi(if_3) = []
      pi(true_0) = []
      pi(nil_0) = []
      pi(cons_2) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: 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:

  (no SCCs)

-- 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

  r1, r4, r5, r9, r12, r15, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        dropLast_1#_A(x1) = x1 + 9
        cons_2_A(x1,x2) = max{34, x1 - 7, x2 + 10}
        ap_A(x1,x2) = max{6, x1 - 13, x2 - 7}
        cons_0_A = 4
        cons_1_A(x1) = max{5, x1 - 8}
        if_0_A = 3
        if_1_A(x1) = max{4, x1 - 8}
        if_2_A(x1,x2) = max{5, x1 - 26, x2 - 13}
        map_0_A = 4
        map_1_A(x1) = max{5, x1 - 10}
        if_3_A(x1,x2,x3) = max{35, x1 - 25, x2 - 12, x3 - 12}
        true_0_A = 0
        nil_0_A = 2
        isEmpty_1_A(x1) = max{7, x1 - 35}
        false_0_A = 5
        dropLast_1_A(x1) = max{6, x1 - 13}
  
    base order 2:
  
      precedence:
      
        if_3 > cons_2 = ap > dropLast_1# = cons_1 = if_1 = if_2 = isEmpty_1 = dropLast_1 > cons_0 = if_0 = nil_0 = false_0 > true_0 > map_1 > map_0
      
  
    partial status:
  
      pi(dropLast_1#) = [1]
      pi(cons_2) = []
      pi(ap) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_2) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(if_3) = []
      pi(true_0) = []
      pi(nil_0) = []
      pi(isEmpty_1) = []
      pi(false_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

  r1, r4, r5, r9, r12, r15, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        last_1#_A(x1) = max{7, x1 + 4}
        cons_2_A(x1,x2) = max{25, x1 + 8, x2 + 5}
        ap_A(x1,x2) = max{22, x1 - 19, x2 - 10}
        cons_0_A = 17
        cons_1_A(x1) = max{18, x1 - 11}
        if_0_A = 16
        if_1_A(x1) = max{17, x1 - 11}
        if_2_A(x1,x2) = max{18, x1 - 48, x2 - 11}
        map_0_A = 38
        map_1_A(x1) = max{18, x1 - 11}
        if_3_A(x1,x2,x3) = max{31, x1 - 49, x2 - 10, x3 - 10}
        true_0_A = 0
        nil_0_A = 7
        isEmpty_1_A(x1) = max{21, x1 - 26}
        false_0_A = 9
        dropLast_1_A(x1) = max{21, x1 - 11}
  
    base order 2:
  
      precedence:
      
        last_1# = if_3 > cons_2 = if_0 = map_0 > ap > cons_1 = if_2 = isEmpty_1 = dropLast_1 > cons_0 = if_1 = nil_0 = false_0 > true_0 > map_1
      
  
    partial status:
  
      pi(last_1#) = [1]
      pi(cons_2) = [1]
      pi(ap) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_2) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(if_3) = []
      pi(true_0) = []
      pi(nil_0) = []
      pi(isEmpty_1) = []
      pi(false_0) = []
      pi(dropLast_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.