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:
  
      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:
          
            if_3#_A(x1,x2,x3) = max{0, x1 - 60, x2 - 51, x3 - 50}
            false_0_A = 0
            ap#_A(x1,x2) = max{0, x1 - 51, x2 - 50}
            last_1_A(x1) = x1
            map_1_A(x1) = max{50, x1 + 46}
            map_2#_A(x1,x2) = max{0, x1 - 51, x2 - 50}
            isEmpty_1_A(x1) = max{11, x1 + 1}
            dropLast_1_A(x1) = max{35, x1}
            if_2_A(x1,x2) = max{x1 - 9, x2 + 46}
            nil_0_A = 65
            true_0_A = 0
            cons_2_A(x1,x2) = max{64, x1 + 51, x2 + 31}
            if_3#!_A(x1,x2,x3) = max{5, x1 - 55, x2 - 46, x3 - 54}
            false_0!_A = 1
            ap#!_A(x1,x2) = max{5, x1 - 46, x2 - 54}
            last_1!_A(x1) = max{9, x1 - 99}
            map_1!_A(x1) = max{1, x1 - 101}
            map_2#!_A(x1,x2) = max{5, x1 - 46, x2 - 54}
            isEmpty_1!_A(x1) = max{6, x1 - 101}
            dropLast_1!_A(x1) = max{6, x1 - 53}
            if_2!_A(x1,x2) = max{2, x1 - 58, x2 - 46}
            nil_0!_A = 7
            true_0!_A = 0
            cons_2!_A(x1,x2) = max{8, x2 - 36}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_3#_A(x1,x2,x3) = max{0, x1 - 28, x2 - 28, x3 - 59}
            false_0_A = 22
            ap#_A(x1,x2) = max{0, x1 - 29, x2 - 59}
            last_1_A(x1) = max{58, x1 - 2}
            map_1_A(x1) = x1 + 22
            map_2#_A(x1,x2) = max{0, x1 - 28, x2 - 59}
            isEmpty_1_A(x1) = max{1, x1 - 49}
            dropLast_1_A(x1) = max{54, x1}
            if_2_A(x1,x2) = max{12, x1 + 1, x2 + 1}
            nil_0_A = 144
            true_0_A = 2
            cons_2_A(x1,x2) = max{x1 + 34, x2 + 71}
            if_3#!_A(x1,x2,x3) = max{7, x1 - 12, x2 - 12, x3 - 60}
            false_0!_A = 15
            ap#!_A(x1,x2) = max{9, x1 - 13, x2 - 59}
            last_1!_A(x1) = max{17, x1 - 125}
            map_1!_A(x1) = max{1, x1 - 132}
            map_2#!_A(x1,x2) = max{8, x1 - 12, x2 - 60}
            isEmpty_1!_A(x1) = max{16, x1 - 145}
            dropLast_1!_A(x1) = max{6, x1 - 129}
            if_2!_A(x1,x2) = 3
            nil_0!_A = 5
            true_0!_A = 15
            cons_2!_A(x1,x2) = max{16, x1 - 94, x2 - 37}
        
  
    partial status:
  
      pi(if_3#) = [2]
      pi(false_0) = []
      pi(ap#) = [1]
      pi(last_1) = [1]
      pi(map_1) = [1]
      pi(map_2#) = [1]
      pi(isEmpty_1) = []
      pi(dropLast_1) = [1]
      pi(if_2) = [2]
      pi(nil_0) = []
      pi(true_0) = []
      pi(cons_2) = [1, 2]

The following rules are removed:

  p1, 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)
p2: if_3#(false_0(),f,xs) -> map_2#(f,dropLast_1(xs))
p3: 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}


-- 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

  r14, r15, r18, r19, r20

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) = max{2, x1 - 1, x2 + 1}
            if_3#_A(x1,x2,x3) = max{2, x1 - 4, x2 - 1, x3 + 1}
            isEmpty_1_A(x1) = 5
            false_0_A = 3
            dropLast_1_A(x1) = 0
            nil_0_A = 0
            true_0_A = 1
            cons_2_A(x1,x2) = 0
            map_2#!_A(x1,x2) = x2 + 9
            if_3#!_A(x1,x2,x3) = max{x1 + 3, x3 + 9}
            isEmpty_1!_A(x1) = max{14, x1 + 10}
            false_0!_A = 11
            dropLast_1!_A(x1) = 0
            nil_0!_A = 0
            true_0!_A = 0
            cons_2!_A(x1,x2) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{0, x1 - 14, x2 - 10}
            if_3#_A(x1,x2,x3) = max{0, x1 - 23, x2 - 14, x3 - 10}
            isEmpty_1_A(x1) = max{10, x1 + 5}
            false_0_A = 21
            dropLast_1_A(x1) = max{9, x1 - 8}
            nil_0_A = 4
            true_0_A = 10
            cons_2_A(x1,x2) = x2 + 17
            map_2#!_A(x1,x2) = x2 + 7
            if_3#!_A(x1,x2,x3) = max{7, x1 - 4, x3}
            isEmpty_1!_A(x1) = max{26, x1 - 71}
            false_0!_A = 25
            dropLast_1!_A(x1) = 10
            nil_0!_A = 1
            true_0!_A = 0
            cons_2!_A(x1,x2) = 8
        
  
    partial status:
  
      pi(map_2#) = []
      pi(if_3#) = []
      pi(isEmpty_1) = [1]
      pi(false_0) = []
      pi(dropLast_1) = []
      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: 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

  (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:
          
            dropLast_1#_A(x1) = 0
            cons_2_A(x1,x2) = max{12, x1 - 1, x2 - 6}
            dropLast_1#!_A(x1) = max{1, x1 - 14}
            cons_2!_A(x1,x2) = max{7, x1 - 16, x2 - 13}
        
        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{37, x1 - 30, x2 - 25}
            dropLast_1#!_A(x1) = max{1, x1 - 106}
            cons_2!_A(x1,x2) = max{6, x1 - 15, x2 - 38}
        
  
    partial status:
  
      pi(dropLast_1#) = [1]
      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: 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:
  
      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:
          
            last_1#_A(x1) = 0
            cons_2_A(x1,x2) = max{13, x1 - 7, x2 - 7}
            last_1#!_A(x1) = max{1, x1 - 35}
            cons_2!_A(x1,x2) = max{4, x1 - 42, x2 - 17}
        
        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{6, x1 - 6, x2 - 6}
            last_1#!_A(x1) = max{1, x1 - 20}
            cons_2!_A(x1,x2) = max{5, x1 - 27, x2 - 7}
        
  
    partial status:
  
      pi(last_1#) = [1]
      pi(cons_2) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.