YES

We show the termination of the TRS R:

  a__pairNs() -> cons(|0|(),incr(oddNs()))
  a__oddNs() -> a__incr(a__pairNs())
  a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
  a__take(|0|(),XS) -> nil()
  a__take(s(N),cons(X,XS)) -> cons(mark(X),take(N,XS))
  a__zip(nil(),XS) -> nil()
  a__zip(X,nil()) -> nil()
  a__zip(cons(X,XS),cons(Y,YS)) -> cons(pair(mark(X),mark(Y)),zip(XS,YS))
  a__tail(cons(X,XS)) -> mark(XS)
  a__repItems(nil()) -> nil()
  a__repItems(cons(X,XS)) -> cons(mark(X),cons(X,repItems(XS)))
  mark(pairNs()) -> a__pairNs()
  mark(incr(X)) -> a__incr(mark(X))
  mark(oddNs()) -> a__oddNs()
  mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
  mark(zip(X1,X2)) -> a__zip(mark(X1),mark(X2))
  mark(tail(X)) -> a__tail(mark(X))
  mark(repItems(X)) -> a__repItems(mark(X))
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(|0|()) -> |0|()
  mark(s(X)) -> s(mark(X))
  mark(nil()) -> nil()
  mark(pair(X1,X2)) -> pair(mark(X1),mark(X2))
  a__pairNs() -> pairNs()
  a__incr(X) -> incr(X)
  a__oddNs() -> oddNs()
  a__take(X1,X2) -> take(X1,X2)
  a__zip(X1,X2) -> zip(X1,X2)
  a__tail(X) -> tail(X)
  a__repItems(X) -> repItems(X)

-- SCC decomposition.

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

p1: a__oddNs#() -> a__incr#(a__pairNs())
p2: a__oddNs#() -> a__pairNs#()
p3: a__incr#(cons(X,XS)) -> mark#(X)
p4: a__take#(s(N),cons(X,XS)) -> mark#(X)
p5: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(X)
p6: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(Y)
p7: a__tail#(cons(X,XS)) -> mark#(XS)
p8: a__repItems#(cons(X,XS)) -> mark#(X)
p9: mark#(pairNs()) -> a__pairNs#()
p10: mark#(incr(X)) -> a__incr#(mark(X))
p11: mark#(incr(X)) -> mark#(X)
p12: mark#(oddNs()) -> a__oddNs#()
p13: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p14: mark#(take(X1,X2)) -> mark#(X1)
p15: mark#(take(X1,X2)) -> mark#(X2)
p16: mark#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p17: mark#(zip(X1,X2)) -> mark#(X1)
p18: mark#(zip(X1,X2)) -> mark#(X2)
p19: mark#(tail(X)) -> a__tail#(mark(X))
p20: mark#(tail(X)) -> mark#(X)
p21: mark#(repItems(X)) -> a__repItems#(mark(X))
p22: mark#(repItems(X)) -> mark#(X)
p23: mark#(cons(X1,X2)) -> mark#(X1)
p24: mark#(s(X)) -> mark#(X)
p25: mark#(pair(X1,X2)) -> mark#(X1)
p26: mark#(pair(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__pairNs() -> cons(|0|(),incr(oddNs()))
r2: a__oddNs() -> a__incr(a__pairNs())
r3: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r4: a__take(|0|(),XS) -> nil()
r5: a__take(s(N),cons(X,XS)) -> cons(mark(X),take(N,XS))
r6: a__zip(nil(),XS) -> nil()
r7: a__zip(X,nil()) -> nil()
r8: a__zip(cons(X,XS),cons(Y,YS)) -> cons(pair(mark(X),mark(Y)),zip(XS,YS))
r9: a__tail(cons(X,XS)) -> mark(XS)
r10: a__repItems(nil()) -> nil()
r11: a__repItems(cons(X,XS)) -> cons(mark(X),cons(X,repItems(XS)))
r12: mark(pairNs()) -> a__pairNs()
r13: mark(incr(X)) -> a__incr(mark(X))
r14: mark(oddNs()) -> a__oddNs()
r15: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r16: mark(zip(X1,X2)) -> a__zip(mark(X1),mark(X2))
r17: mark(tail(X)) -> a__tail(mark(X))
r18: mark(repItems(X)) -> a__repItems(mark(X))
r19: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r20: mark(|0|()) -> |0|()
r21: mark(s(X)) -> s(mark(X))
r22: mark(nil()) -> nil()
r23: mark(pair(X1,X2)) -> pair(mark(X1),mark(X2))
r24: a__pairNs() -> pairNs()
r25: a__incr(X) -> incr(X)
r26: a__oddNs() -> oddNs()
r27: a__take(X1,X2) -> take(X1,X2)
r28: a__zip(X1,X2) -> zip(X1,X2)
r29: a__tail(X) -> tail(X)
r30: a__repItems(X) -> repItems(X)

The estimated dependency graph contains the following SCCs:

  {p1, p3, p4, p5, p6, p7, p8, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21, p22, p23, p24, p25, p26}


-- Reduction pair.

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

p1: a__oddNs#() -> a__incr#(a__pairNs())
p2: a__incr#(cons(X,XS)) -> mark#(X)
p3: mark#(pair(X1,X2)) -> mark#(X2)
p4: mark#(pair(X1,X2)) -> mark#(X1)
p5: mark#(s(X)) -> mark#(X)
p6: mark#(cons(X1,X2)) -> mark#(X1)
p7: mark#(repItems(X)) -> mark#(X)
p8: mark#(repItems(X)) -> a__repItems#(mark(X))
p9: a__repItems#(cons(X,XS)) -> mark#(X)
p10: mark#(tail(X)) -> mark#(X)
p11: mark#(tail(X)) -> a__tail#(mark(X))
p12: a__tail#(cons(X,XS)) -> mark#(XS)
p13: mark#(zip(X1,X2)) -> mark#(X2)
p14: mark#(zip(X1,X2)) -> mark#(X1)
p15: mark#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p16: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(Y)
p17: mark#(take(X1,X2)) -> mark#(X2)
p18: mark#(take(X1,X2)) -> mark#(X1)
p19: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p20: a__take#(s(N),cons(X,XS)) -> mark#(X)
p21: mark#(oddNs()) -> a__oddNs#()
p22: mark#(incr(X)) -> mark#(X)
p23: mark#(incr(X)) -> a__incr#(mark(X))
p24: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(X)

and R consists of:

r1: a__pairNs() -> cons(|0|(),incr(oddNs()))
r2: a__oddNs() -> a__incr(a__pairNs())
r3: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r4: a__take(|0|(),XS) -> nil()
r5: a__take(s(N),cons(X,XS)) -> cons(mark(X),take(N,XS))
r6: a__zip(nil(),XS) -> nil()
r7: a__zip(X,nil()) -> nil()
r8: a__zip(cons(X,XS),cons(Y,YS)) -> cons(pair(mark(X),mark(Y)),zip(XS,YS))
r9: a__tail(cons(X,XS)) -> mark(XS)
r10: a__repItems(nil()) -> nil()
r11: a__repItems(cons(X,XS)) -> cons(mark(X),cons(X,repItems(XS)))
r12: mark(pairNs()) -> a__pairNs()
r13: mark(incr(X)) -> a__incr(mark(X))
r14: mark(oddNs()) -> a__oddNs()
r15: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r16: mark(zip(X1,X2)) -> a__zip(mark(X1),mark(X2))
r17: mark(tail(X)) -> a__tail(mark(X))
r18: mark(repItems(X)) -> a__repItems(mark(X))
r19: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r20: mark(|0|()) -> |0|()
r21: mark(s(X)) -> s(mark(X))
r22: mark(nil()) -> nil()
r23: mark(pair(X1,X2)) -> pair(mark(X1),mark(X2))
r24: a__pairNs() -> pairNs()
r25: a__incr(X) -> incr(X)
r26: a__oddNs() -> oddNs()
r27: a__take(X1,X2) -> take(X1,X2)
r28: a__zip(X1,X2) -> zip(X1,X2)
r29: a__tail(X) -> tail(X)
r30: a__repItems(X) -> repItems(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__oddNs#_A = 52
          a__incr#_A(x1) = max{40, x1 + 11}
          a__pairNs_A = 40
          cons_A(x1,x2) = max{25, x1 + 14, x2}
          mark#_A(x1) = max{24, x1 + 15}
          pair_A(x1,x2) = max{30, x1 + 27, x2 + 25}
          s_A(x1) = max{25, x1}
          repItems_A(x1) = max{8, x1 + 4}
          a__repItems#_A(x1) = max{24, x1 + 17}
          mark_A(x1) = max{3, x1}
          tail_A(x1) = max{10, x1 + 7}
          a__tail#_A(x1) = max{16, x1 + 15}
          zip_A(x1,x2) = max{34, x1 + 29, x2 + 31}
          a__zip#_A(x1,x2) = max{23, x1 + 20, x2 + 1}
          take_A(x1,x2) = max{56, x1 + 31, x2 + 26}
          a__take#_A(x1,x2) = max{x1 + 32, x2 + 27}
          oddNs_A = 40
          incr_A(x1) = max{39, x1}
          a__oddNs_A = 40
          a__incr_A(x1) = max{39, x1}
          a__take_A(x1,x2) = max{56, x1 + 31, x2 + 26}
          |0|_A = 26
          nil_A = 0
          a__zip_A(x1,x2) = max{34, x1 + 29, x2 + 31}
          a__tail_A(x1) = max{10, x1 + 7}
          a__repItems_A(x1) = max{8, x1 + 4}
          pairNs_A = 40
    
      precedence:
      
        a__zip# > a__incr# > pair = mark = a__oddNs = a__zip = a__tail > a__incr > a__take = nil > a__pairNs = a__tail# = zip = a__repItems > pairNs > oddNs > take > s > a__oddNs# = cons = mark# > repItems = incr = |0| > tail > a__repItems# = a__take#
      
      partial status:
    
        pi(a__oddNs#) = []
        pi(a__incr#) = [1]
        pi(a__pairNs) = []
        pi(cons) = []
        pi(mark#) = [1]
        pi(pair) = [2]
        pi(s) = [1]
        pi(repItems) = [1]
        pi(a__repItems#) = [1]
        pi(mark) = [1]
        pi(tail) = [1]
        pi(a__tail#) = [1]
        pi(zip) = []
        pi(a__zip#) = [1, 2]
        pi(take) = [1, 2]
        pi(a__take#) = [1, 2]
        pi(oddNs) = []
        pi(incr) = [1]
        pi(a__oddNs) = []
        pi(a__incr) = [1]
        pi(a__take) = [1, 2]
        pi(|0|) = []
        pi(nil) = []
        pi(a__zip) = [2]
        pi(a__tail) = []
        pi(a__repItems) = [1]
        pi(pairNs) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__oddNs#_A = 247
          a__incr#_A(x1) = max{206, x1 + 142}
          a__pairNs_A = 104
          cons_A(x1,x2) = 105
          mark#_A(x1) = x1 + 239
          pair_A(x1,x2) = max{42, x2}
          s_A(x1) = x1 + 103
          repItems_A(x1) = max{114, x1 + 93}
          a__repItems#_A(x1) = max{206, x1 + 142}
          mark_A(x1) = max{102, x1 + 60}
          tail_A(x1) = x1 + 104
          a__tail#_A(x1) = max{206, x1 + 122}
          zip_A(x1,x2) = 40
          a__zip#_A(x1,x2) = max{x1 + 141, x2 + 142}
          take_A(x1,x2) = max{x1 + 123, x2 + 6}
          a__take#_A(x1,x2) = max{x1 + 142, x2 + 142}
          oddNs_A = 105
          incr_A(x1) = max{205, x1 + 161}
          a__oddNs_A = 103
          a__incr_A(x1) = 206
          a__take_A(x1,x2) = 65
          |0|_A = 105
          nil_A = 116
          a__zip_A(x1,x2) = 101
          a__tail_A(x1) = 103
          a__repItems_A(x1) = 115
          pairNs_A = 105
    
      precedence:
      
        a__oddNs# = a__incr# = a__pairNs = s = repItems = a__repItems# = mark = tail = a__tail# = take = a__zip > mark# = pair > zip > a__zip# = a__take# = oddNs = incr = a__oddNs = a__incr = a__take = nil = a__tail = a__repItems = pairNs > cons > |0|
      
      partial status:
    
        pi(a__oddNs#) = []
        pi(a__incr#) = []
        pi(a__pairNs) = []
        pi(cons) = []
        pi(mark#) = [1]
        pi(pair) = [2]
        pi(s) = [1]
        pi(repItems) = [1]
        pi(a__repItems#) = [1]
        pi(mark) = [1]
        pi(tail) = [1]
        pi(a__tail#) = []
        pi(zip) = []
        pi(a__zip#) = [2]
        pi(take) = [1, 2]
        pi(a__take#) = [2]
        pi(oddNs) = []
        pi(incr) = [1]
        pi(a__oddNs) = []
        pi(a__incr) = []
        pi(a__take) = []
        pi(|0|) = []
        pi(nil) = []
        pi(a__zip) = []
        pi(a__tail) = []
        pi(a__repItems) = []
        pi(pairNs) = []
    

The next rules are strictly ordered:

  p1, p2, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21, p22, p23, p24

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(pair(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__pairNs() -> cons(|0|(),incr(oddNs()))
r2: a__oddNs() -> a__incr(a__pairNs())
r3: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r4: a__take(|0|(),XS) -> nil()
r5: a__take(s(N),cons(X,XS)) -> cons(mark(X),take(N,XS))
r6: a__zip(nil(),XS) -> nil()
r7: a__zip(X,nil()) -> nil()
r8: a__zip(cons(X,XS),cons(Y,YS)) -> cons(pair(mark(X),mark(Y)),zip(XS,YS))
r9: a__tail(cons(X,XS)) -> mark(XS)
r10: a__repItems(nil()) -> nil()
r11: a__repItems(cons(X,XS)) -> cons(mark(X),cons(X,repItems(XS)))
r12: mark(pairNs()) -> a__pairNs()
r13: mark(incr(X)) -> a__incr(mark(X))
r14: mark(oddNs()) -> a__oddNs()
r15: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r16: mark(zip(X1,X2)) -> a__zip(mark(X1),mark(X2))
r17: mark(tail(X)) -> a__tail(mark(X))
r18: mark(repItems(X)) -> a__repItems(mark(X))
r19: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r20: mark(|0|()) -> |0|()
r21: mark(s(X)) -> s(mark(X))
r22: mark(nil()) -> nil()
r23: mark(pair(X1,X2)) -> pair(mark(X1),mark(X2))
r24: a__pairNs() -> pairNs()
r25: a__incr(X) -> incr(X)
r26: a__oddNs() -> oddNs()
r27: a__take(X1,X2) -> take(X1,X2)
r28: a__zip(X1,X2) -> zip(X1,X2)
r29: a__tail(X) -> tail(X)
r30: a__repItems(X) -> repItems(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: mark#(pair(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__pairNs() -> cons(|0|(),incr(oddNs()))
r2: a__oddNs() -> a__incr(a__pairNs())
r3: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r4: a__take(|0|(),XS) -> nil()
r5: a__take(s(N),cons(X,XS)) -> cons(mark(X),take(N,XS))
r6: a__zip(nil(),XS) -> nil()
r7: a__zip(X,nil()) -> nil()
r8: a__zip(cons(X,XS),cons(Y,YS)) -> cons(pair(mark(X),mark(Y)),zip(XS,YS))
r9: a__tail(cons(X,XS)) -> mark(XS)
r10: a__repItems(nil()) -> nil()
r11: a__repItems(cons(X,XS)) -> cons(mark(X),cons(X,repItems(XS)))
r12: mark(pairNs()) -> a__pairNs()
r13: mark(incr(X)) -> a__incr(mark(X))
r14: mark(oddNs()) -> a__oddNs()
r15: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r16: mark(zip(X1,X2)) -> a__zip(mark(X1),mark(X2))
r17: mark(tail(X)) -> a__tail(mark(X))
r18: mark(repItems(X)) -> a__repItems(mark(X))
r19: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r20: mark(|0|()) -> |0|()
r21: mark(s(X)) -> s(mark(X))
r22: mark(nil()) -> nil()
r23: mark(pair(X1,X2)) -> pair(mark(X1),mark(X2))
r24: a__pairNs() -> pairNs()
r25: a__incr(X) -> incr(X)
r26: a__oddNs() -> oddNs()
r27: a__take(X1,X2) -> take(X1,X2)
r28: a__zip(X1,X2) -> zip(X1,X2)
r29: a__tail(X) -> tail(X)
r30: a__repItems(X) -> repItems(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{3, x1 + 2}
          pair_A(x1,x2) = max{1, x2}
    
      precedence:
      
        mark# = pair
      
      partial status:
    
        pi(mark#) = [1]
        pi(pair) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{3, x1 + 2}
          pair_A(x1,x2) = x2 + 1
    
      precedence:
      
        mark# = pair
      
      partial status:
    
        pi(mark#) = []
        pi(pair) = [2]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.