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

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if_3#_A(x1,x2,x3) = x2
              false_0_A() = 1
              ap#_A(x1,x2) = x1
              last_1_A(x1) = x1 + 3
              map_1_A(x1) = x1 + 3
              map_2#_A(x1,x2) = x1
              isEmpty_1_A(x1) = 3
              dropLast_1_A(x1) = 3
              if_2_A(x1,x2) = x1 + x2 + 5
              nil_0_A() = 8
              true_0_A() = 1
              cons_2_A(x1,x2) = x1 + 7
              if_3#!_A(x1,x2,x3) = x2 + 1
              false_0!_A() = 6
              ap#!_A(x1,x2) = x1
              last_1!_A(x1) = 5
              map_1!_A(x1) = x1 + 1
              map_2#!_A(x1,x2) = x1 + 1
              isEmpty_1!_A(x1) = 7
              dropLast_1!_A(x1) = 5
              if_2!_A(x1,x2) = 4
              nil_0!_A() = 0
              true_0!_A() = 1
              cons_2!_A(x1,x2) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_3#_A(x1,x2,x3) = max{1, x1 - 63, x2 - 58}
            false_0_A = 28
            ap#_A(x1,x2) = max{1, x1 - 58}
            last_1_A(x1) = max{22, x1 - 4}
            map_1_A(x1) = x1 + 57
            map_2#_A(x1,x2) = max{1, x1 - 58}
            isEmpty_1_A(x1) = 62
            dropLast_1_A(x1) = max{1, x1 - 25}
            if_2_A(x1,x2) = max{x1 + 15, x2 + 48}
            nil_0_A = 19
            true_0_A = 1
            cons_2_A(x1,x2) = max{x1 - 41, x2 + 11}
            if_3#!_A(x1,x2,x3) = max{31, x1 - 92, x2 + 29}
            false_0!_A = 10
            ap#!_A(x1,x2) = max{31, x1 - 16}
            last_1!_A(x1) = max{7, x1 - 14}
            map_1!_A(x1) = max{31, x1 + 30}
            map_2#!_A(x1,x2) = max{31, x1 + 29}
            isEmpty_1!_A(x1) = 28
            dropLast_1!_A(x1) = max{25, x1 - 31}
            if_2!_A(x1,x2) = 9
            nil_0!_A = 26
            true_0!_A = 27
            cons_2!_A(x1,x2) = x2 + 13
        
  
    partial status:
  
      pi(if_3#) = []
      pi(false_0) = []
      pi(ap#) = []
      pi(last_1) = []
      pi(map_1) = []
      pi(map_2#) = []
      pi(isEmpty_1) = [1]
      pi(dropLast_1) = []
      pi(if_2) = [1]
      pi(nil_0) = []
      pi(true_0) = []
      pi(cons_2) = [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

  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!):
        
          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) = 2
              false_0_A() = 2
              dropLast_1_A(x1) = x1 + 1
              nil_0_A() = 5
              true_0_A() = 2
              cons_2_A(x1,x2) = x1 + 5
              map_2#!_A(x1,x2) = 2
              if_3#!_A(x1,x2,x3) = x1
              isEmpty_1!_A(x1) = x1 + 3
              false_0!_A() = 9
              dropLast_1!_A(x1) = x1 + 6
              nil_0!_A() = 1
              true_0!_A() = 0
              cons_2!_A(x1,x2) = 10
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 1, x2 - 16}
            if_3#_A(x1,x2,x3) = max{1, x2 - 1, x3 - 16}
            isEmpty_1_A(x1) = x1 + 5
            false_0_A = 26
            dropLast_1_A(x1) = max{17, x1 - 4}
            nil_0_A = 17
            true_0_A = 0
            cons_2_A(x1,x2) = max{21, x2 + 11}
            map_2#!_A(x1,x2) = x2 + 4
            if_3#!_A(x1,x2,x3) = max{x1 - 4, x3 + 3}
            isEmpty_1!_A(x1) = max{27, x1 - 33}
            false_0!_A = 27
            dropLast_1!_A(x1) = max{26, x1 - 12}
            nil_0!_A = 21
            true_0!_A = 22
            cons_2!_A(x1,x2) = 21
        
  
    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:

  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

  (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!):
        
          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
              cons_2!_A(x1,x2) = 6
        
        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{16, x1 - 23, x2 - 23}
            dropLast_1#!_A(x1) = x1 + 45
            cons_2!_A(x1,x2) = max{60, x1 - 43}
        
  
    partial status:
  
      pi(dropLast_1#) = [1]
      pi(cons_2) = [1, 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!):
        
          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
              last_1#!_A(x1) = x1
              cons_2!_A(x1,x2) = 1
        
        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{3, x1 - 2, x2 - 2}
            last_1#!_A(x1) = max{5, x1 - 4}
            cons_2!_A(x1,x2) = max{4, x1 - 9}
        
  
    partial status:
  
      pi(last_1#) = [1]
      pi(cons_2) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.