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:

  max/plus interpretations on natural numbers:
  
    a__oddNs#_A = 21
    a__incr#_A(x1) = max{0, x1 - 2}
    a__pairNs_A = 23
    cons_A(x1,x2) = max{23, x1 + 2, x2 - 17}
    mark#_A(x1) = max{1, x1 - 8}
    pair_A(x1,x2) = max{x1 + 10, x2 + 10}
    s_A(x1) = x1 + 6
    repItems_A(x1) = x1 + 40
    a__repItems#_A(x1) = max{32, x1 - 2}
    mark_A(x1) = x1 + 4
    tail_A(x1) = max{23, x1 + 22}
    a__tail#_A(x1) = x1 + 9
    zip_A(x1,x2) = max{x1 + 40, x2 + 41}
    a__zip#_A(x1,x2) = max{30, x1 + 19, x2 + 26}
    take_A(x1,x2) = max{x1 + 40, x2 + 40}
    a__take#_A(x1,x2) = max{x1 + 28, x2 + 27}
    oddNs_A = 30
    incr_A(x1) = x1 + 10
    a__oddNs_A = 34
    a__incr_A(x1) = x1 + 10
    a__take_A(x1,x2) = max{x1 + 40, x2 + 40}
    |0|_A = 4
    nil_A = 6
    a__zip_A(x1,x2) = max{x1 + 40, x2 + 41}
    a__tail_A(x1) = max{23, x1 + 22}
    a__repItems_A(x1) = x1 + 40
    pairNs_A = 19

The next rules are strictly ordered:

  p2, p3, p4, p6, p7, p9, p10, p11, p13, p14, p15, p16, p17, p18, p20, p21, p22, p24

We remove them from the problem.

-- SCC decomposition.

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

p1: a__oddNs#() -> a__incr#(a__pairNs())
p2: mark#(s(X)) -> mark#(X)
p3: mark#(repItems(X)) -> a__repItems#(mark(X))
p4: a__tail#(cons(X,XS)) -> mark#(XS)
p5: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p6: mark#(incr(X)) -> a__incr#(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 estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: mark#(s(X)) -> 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

  (no rules)

Take the monotone reduction pair:

  max/plus interpretations on natural numbers:
  
    mark#_A(x1) = x1
    s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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