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

  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) = max{x1 - 23, x2 + 18, x3 + 16}
        false_0_A = 54
        ap#_A(x1,x2) = max{x1 - 4, x2 + 18}
        last_1_A(x1) = max{12, x1 - 4}
        map_1_A(x1) = max{56, x1 + 41}
        map_2#_A(x1,x2) = max{x1 + 18, x2 + 16}
        isEmpty_1_A(x1) = max{41, x1 + 39}
        dropLast_1_A(x1) = max{15, x1}
        if_2_A(x1,x2) = max{x1 + 19, x2 + 22}
        nil_0_A = 14
        true_0_A = 13
        cons_2_A(x1,x2) = max{15, x1 + 5, x2 + 5}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            if_3#_A(x1,x2,x3) = x2 + x3 + 3
            false_0_A() = 1
            ap#_A(x1,x2) = x1 + x2
            last_1_A(x1) = x1 + 3
            map_1_A(x1) = x1 + 3
            map_2#_A(x1,x2) = x1 + x2 + 3
            isEmpty_1_A(x1) = x1 + 3
            dropLast_1_A(x1) = x1
            if_2_A(x1,x2) = x1 + x2 + 5
            nil_0_A() = 9
            true_0_A() = 1
            cons_2_A(x1,x2) = x1 + x2 + 5
            if_3#!_A(x1,x2,x3) = 0
            false_0!_A() = 2
            ap#!_A(x1,x2) = 1
            last_1!_A(x1) = 1
            map_1!_A(x1) = 4
            map_2#!_A(x1,x2) = 0
            isEmpty_1!_A(x1) = 3
            dropLast_1!_A(x1) = 4
            if_2!_A(x1,x2) = x1 + x2 + 1
            nil_0!_A() = 4
            true_0!_A() = 0
            cons_2!_A(x1,x2) = 1
  
    partial status:
  
      pi(if_3#) = []
      pi(false_0) = []
      pi(ap#) = [2]
      pi(last_1) = []
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(isEmpty_1) = []
      pi(dropLast_1) = []
      pi(if_2) = [2]
      pi(nil_0) = []
      pi(true_0) = []
      pi(cons_2) = []

The following rules are removed:

  p1, p2, p5



-- 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)
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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- 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

  r12, r14, r15, r18, r19, r20

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 + 8
        if_3#_A(x1,x2,x3) = max{x1 - 8, x3 + 7}
        isEmpty_1_A(x1) = max{7, x1 - 3}
        false_0_A = 62
        dropLast_1_A(x1) = max{45, x1 - 2}
        if_3_A(x1,x2,x3) = max{67, x1 + 2, x2 - 65, x3 - 3}
        true_0_A = 41
        nil_0_A = 45
        cons_2_A(x1,x2) = max{66, x2 + 22}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            map_2#_A(x1,x2) = 0
            if_3#_A(x1,x2,x3) = 0
            isEmpty_1_A(x1) = x1 + 2
            false_0_A() = 1
            dropLast_1_A(x1) = x1 + 1
            if_3_A(x1,x2,x3) = 4
            true_0_A() = 1
            nil_0_A() = 0
            cons_2_A(x1,x2) = 5
            map_2#!_A(x1,x2) = x1 + x2 + 8
            if_3#!_A(x1,x2,x3) = 7
            isEmpty_1!_A(x1) = 1
            false_0!_A() = 7
            dropLast_1!_A(x1) = 6
            if_3!_A(x1,x2,x3) = x3 + 2
            true_0!_A() = 0
            nil_0!_A() = 5
            cons_2!_A(x1,x2) = 4
  
    partial status:
  
      pi(map_2#) = []
      pi(if_3#) = []
      pi(isEmpty_1) = []
      pi(false_0) = []
      pi(dropLast_1) = []
      pi(if_3) = [1]
      pi(true_0) = []
      pi(nil_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

  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{53, x1 + 20}
        cons_2_A(x1,x2) = max{34, x1 - 81, x2 + 18}
        dropLast_1_A(x1) = max{17, x1 + 3}
        nil_0_A = 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            dropLast_1#_A(x1) = x1
            cons_2_A(x1,x2) = x1 + x2 + 5
            dropLast_1_A(x1) = x1 + 3
            nil_0_A() = 8
            dropLast_1#!_A(x1) = 8
            cons_2!_A(x1,x2) = x2 + 7
            dropLast_1!_A(x1) = x1 + 1
            nil_0!_A() = 1
  
    partial status:
  
      pi(dropLast_1#) = [1]
      pi(cons_2) = []
      pi(dropLast_1) = [1]
      pi(nil_0) = []

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

  r18

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        last_1#_A(x1) = 9
        cons_2_A(x1,x2) = max{39, x1 + 14, x2 + 5}
        dropLast_1_A(x1) = x1 + 6
        nil_0_A = 10
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            last_1#_A(x1) = x1
            cons_2_A(x1,x2) = x1 + x2 + 5
            dropLast_1_A(x1) = x1 + 3
            nil_0_A() = 8
            last_1#!_A(x1) = x1 + 8
            cons_2!_A(x1,x2) = x2 + 7
            dropLast_1!_A(x1) = x1 + 1
            nil_0!_A() = 1
  
    partial status:
  
      pi(last_1#) = []
      pi(cons_2) = []
      pi(dropLast_1) = [1]
      pi(nil_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.