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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__oddNs#_A = 48
        a__incr#_A(x1) = x1 + 18
        a__pairNs_A = 30
        cons_A(x1,x2) = max{x1, x2 - 12}
        mark#_A(x1) = x1 + 18
        pair_A(x1,x2) = max{12, x1 + 6, x2 + 9}
        s_A(x1) = max{4, x1}
        repItems_A(x1) = max{25, x1 + 19}
        a__repItems#_A(x1) = x1 + 20
        mark_A(x1) = max{2, x1}
        tail_A(x1) = max{21, x1 + 19}
        a__tail#_A(x1) = x1 + 37
        zip_A(x1,x2) = max{22, x1 + 11, x2 + 9}
        a__zip#_A(x1,x2) = max{23, x1 + 19, x2 + 19}
        take_A(x1,x2) = max{27, x1 + 24, x2 + 22}
        a__take#_A(x1,x2) = max{x1 + 6, x2 + 19}
        oddNs_A = 42
        incr_A(x1) = max{4, x1}
        a__oddNs_A = 42
        a__incr_A(x1) = max{4, x1}
        a__take_A(x1,x2) = max{27, x1 + 24, x2 + 22}
        |0|_A = 29
        nil_A = 28
        a__zip_A(x1,x2) = max{22, x1 + 11, x2 + 9}
        a__tail_A(x1) = max{21, x1 + 19}
        a__repItems_A(x1) = max{25, x1 + 19}
        pairNs_A = 30
  
    precedence:
    
      take = a__take > a__take# > a__oddNs# = tail = a__tail > a__incr# = mark# = s = mark = a__zip# = oddNs = a__oddNs = a__zip > a__repItems# = a__tail# > pair = incr = a__incr > a__pairNs = cons = repItems = nil = a__repItems = pairNs > |0| > zip
    
    partial status:
  
      pi(a__oddNs#) = []
      pi(a__incr#) = []
      pi(a__pairNs) = []
      pi(cons) = []
      pi(mark#) = []
      pi(pair) = [1, 2]
      pi(s) = [1]
      pi(repItems) = []
      pi(a__repItems#) = []
      pi(mark) = [1]
      pi(tail) = []
      pi(a__tail#) = [1]
      pi(zip) = [1, 2]
      pi(a__zip#) = []
      pi(take) = []
      pi(a__take#) = [1, 2]
      pi(oddNs) = []
      pi(incr) = [1]
      pi(a__oddNs) = []
      pi(a__incr) = [1]
      pi(a__take) = []
      pi(|0|) = []
      pi(nil) = []
      pi(a__zip) = []
      pi(a__tail) = []
      pi(a__repItems) = [1]
      pi(pairNs) = []

The next rules are strictly ordered:

  p8

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: 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: a__repItems#(cons(X,XS)) -> mark#(X)
p9: mark#(tail(X)) -> mark#(X)
p10: mark#(tail(X)) -> a__tail#(mark(X))
p11: a__tail#(cons(X,XS)) -> mark#(XS)
p12: mark#(zip(X1,X2)) -> mark#(X2)
p13: mark#(zip(X1,X2)) -> mark#(X1)
p14: mark#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p15: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(Y)
p16: mark#(take(X1,X2)) -> mark#(X2)
p17: mark#(take(X1,X2)) -> mark#(X1)
p18: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p19: a__take#(s(N),cons(X,XS)) -> mark#(X)
p20: mark#(oddNs()) -> a__oddNs#()
p21: mark#(incr(X)) -> mark#(X)
p22: mark#(incr(X)) -> a__incr#(mark(X))
p23: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21, p22, p23}


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

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__oddNs#_A = 80
        a__incr#_A(x1) = max{29, x1 + 10}
        a__pairNs_A = 69
        cons_A(x1,x2) = max{x1 + 6, x2 - 43}
        mark#_A(x1) = x1 + 6
        incr_A(x1) = x1 + 29
        mark_A(x1) = x1 + 18
        oddNs_A = 82
        take_A(x1,x2) = max{117, x1 + 58, x2 + 67}
        a__take#_A(x1,x2) = max{57, x1 + 1, x2 + 49}
        s_A(x1) = x1 + 10
        zip_A(x1,x2) = max{x1 + 74, x2 + 83}
        a__zip#_A(x1,x2) = max{73, x1 + 55, x2 + 65}
        tail_A(x1) = x1 + 78
        a__tail#_A(x1) = x1 + 50
        repItems_A(x1) = x1 + 18
        pair_A(x1,x2) = max{x1 + 18, x2 + 7}
        a__oddNs_A = 99
        a__incr_A(x1) = x1 + 29
        a__take_A(x1,x2) = max{118, x1 + 58, x2 + 67}
        |0|_A = 62
        nil_A = 119
        a__zip_A(x1,x2) = max{x1 + 74, x2 + 83}
        a__tail_A(x1) = x1 + 78
        a__repItems_A(x1) = x1 + 18
        pairNs_A = 52
  
    precedence:
    
      a__oddNs > oddNs > mark = s = a__incr > incr > |0| > pair > repItems = a__repItems > a__take > take > nil > a__tail > tail > a__zip > zip > pairNs > a__pairNs = cons > a__oddNs# = a__incr# = mark# = a__take# = a__zip# = a__tail#
    
    partial status:
  
      pi(a__oddNs#) = []
      pi(a__incr#) = [1]
      pi(a__pairNs) = []
      pi(cons) = []
      pi(mark#) = [1]
      pi(incr) = []
      pi(mark) = []
      pi(oddNs) = []
      pi(take) = [2]
      pi(a__take#) = [1, 2]
      pi(s) = []
      pi(zip) = []
      pi(a__zip#) = [1, 2]
      pi(tail) = [1]
      pi(a__tail#) = [1]
      pi(repItems) = []
      pi(pair) = []
      pi(a__oddNs) = []
      pi(a__incr) = []
      pi(a__take) = [2]
      pi(|0|) = []
      pi(nil) = []
      pi(a__zip) = []
      pi(a__tail) = [1]
      pi(a__repItems) = []
      pi(pairNs) = []

The next rules are strictly ordered:

  p3

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

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, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21}


-- 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#(tail(X)) -> mark#(X)
p9: mark#(tail(X)) -> a__tail#(mark(X))
p10: a__tail#(cons(X,XS)) -> mark#(XS)
p11: mark#(zip(X1,X2)) -> mark#(X2)
p12: mark#(zip(X1,X2)) -> mark#(X1)
p13: mark#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p14: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(Y)
p15: mark#(take(X1,X2)) -> mark#(X2)
p16: mark#(take(X1,X2)) -> mark#(X1)
p17: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p18: a__take#(s(N),cons(X,XS)) -> mark#(X)
p19: mark#(oddNs()) -> a__oddNs#()
p20: mark#(incr(X)) -> mark#(X)
p21: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__oddNs#_A = 23
        a__incr#_A(x1) = x1 + 4
        a__pairNs_A = 18
        cons_A(x1,x2) = max{17, x1 + 7, x2}
        mark#_A(x1) = x1 + 6
        pair_A(x1,x2) = max{x1 + 2, x2 + 2}
        s_A(x1) = x1
        repItems_A(x1) = x1 + 9
        tail_A(x1) = x1 + 5
        a__tail#_A(x1) = x1 + 7
        mark_A(x1) = x1
        zip_A(x1,x2) = max{x1 + 7, x2 + 26}
        a__zip#_A(x1,x2) = max{3, x1 - 1, x2 - 1}
        take_A(x1,x2) = max{x1 + 17, x2}
        a__take#_A(x1,x2) = max{3, x2 - 1}
        oddNs_A = 18
        incr_A(x1) = x1
        a__oddNs_A = 18
        a__incr_A(x1) = x1
        a__take_A(x1,x2) = max{x1 + 17, x2}
        |0|_A = 1
        nil_A = 9
        a__zip_A(x1,x2) = max{x1 + 7, x2 + 26}
        a__tail_A(x1) = x1 + 5
        a__repItems_A(x1) = x1 + 9
        pairNs_A = 18
  
    precedence:
    
      pair > zip = a__zip > a__oddNs# = mark# = a__tail# = a__zip# = a__take# > a__incr# > s = take = oddNs = a__oddNs = a__take > mark = incr = a__incr = |0| = nil > a__pairNs = cons = repItems = a__repItems = pairNs > tail = a__tail
    
    partial status:
  
      pi(a__oddNs#) = []
      pi(a__incr#) = [1]
      pi(a__pairNs) = []
      pi(cons) = []
      pi(mark#) = []
      pi(pair) = []
      pi(s) = []
      pi(repItems) = []
      pi(tail) = []
      pi(a__tail#) = []
      pi(mark) = [1]
      pi(zip) = []
      pi(a__zip#) = []
      pi(take) = []
      pi(a__take#) = []
      pi(oddNs) = []
      pi(incr) = [1]
      pi(a__oddNs) = []
      pi(a__incr) = [1]
      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

We remove them from the problem.

-- SCC decomposition.

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

p1: a__incr#(cons(X,XS)) -> mark#(X)
p2: mark#(pair(X1,X2)) -> mark#(X2)
p3: mark#(pair(X1,X2)) -> mark#(X1)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(repItems(X)) -> mark#(X)
p7: mark#(tail(X)) -> mark#(X)
p8: mark#(tail(X)) -> a__tail#(mark(X))
p9: a__tail#(cons(X,XS)) -> mark#(XS)
p10: mark#(zip(X1,X2)) -> mark#(X2)
p11: mark#(zip(X1,X2)) -> mark#(X1)
p12: mark#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p13: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(Y)
p14: mark#(take(X1,X2)) -> mark#(X2)
p15: mark#(take(X1,X2)) -> mark#(X1)
p16: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p17: a__take#(s(N),cons(X,XS)) -> mark#(X)
p18: mark#(oddNs()) -> a__oddNs#()
p19: mark#(incr(X)) -> mark#(X)
p20: 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 estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p19, p20}


-- Reduction pair.

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

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

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 7
        pair_A(x1,x2) = max{42, x1 + 33, x2 + 21}
        incr_A(x1) = x1 + 27
        take_A(x1,x2) = max{x1 + 51, x2 + 60}
        a__take#_A(x1,x2) = max{6, x1, x2}
        mark_A(x1) = x1 + 20
        s_A(x1) = x1 + 7
        cons_A(x1,x2) = max{x1 + 7, x2 - 37}
        zip_A(x1,x2) = max{x1 + 95, x2 + 87}
        a__zip#_A(x1,x2) = max{88, x1 + 74, x2 + 66}
        tail_A(x1) = x1 + 65
        a__tail#_A(x1) = x1 + 45
        repItems_A(x1) = x1 + 20
        a__pairNs_A = 18
        |0|_A = 10
        oddNs_A = 27
        a__oddNs_A = 46
        a__incr_A(x1) = x1 + 27
        a__take_A(x1,x2) = max{x1 + 51, x2 + 60}
        nil_A = 11
        a__zip_A(x1,x2) = max{96, x1 + 95, x2 + 87}
        a__tail_A(x1) = max{83, x1 + 65}
        a__repItems_A(x1) = x1 + 20
        pairNs_A = 0
  
    precedence:
    
      incr = take = mark = s = cons = repItems = oddNs = a__incr = a__take = nil = a__zip = a__repItems > pair = tail = a__tail > |0| > a__pairNs = a__oddNs > a__take# = pairNs > mark# > a__zip# > a__tail# > zip
    
    partial status:
  
      pi(mark#) = [1]
      pi(pair) = [2]
      pi(incr) = []
      pi(take) = []
      pi(a__take#) = [1, 2]
      pi(mark) = []
      pi(s) = []
      pi(cons) = []
      pi(zip) = []
      pi(a__zip#) = [1]
      pi(tail) = [1]
      pi(a__tail#) = [1]
      pi(repItems) = []
      pi(a__pairNs) = []
      pi(|0|) = []
      pi(oddNs) = []
      pi(a__oddNs) = []
      pi(a__incr) = []
      pi(a__take) = []
      pi(nil) = []
      pi(a__zip) = []
      pi(a__tail) = [1]
      pi(a__repItems) = []
      pi(pairNs) = []

The next rules are strictly ordered:

  p18

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)
p2: mark#(incr(X)) -> mark#(X)
p3: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p4: a__take#(s(N),cons(X,XS)) -> mark#(X)
p5: mark#(take(X1,X2)) -> mark#(X1)
p6: mark#(take(X1,X2)) -> mark#(X2)
p7: mark#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p8: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(X)
p9: mark#(zip(X1,X2)) -> mark#(X1)
p10: mark#(zip(X1,X2)) -> mark#(X2)
p11: mark#(tail(X)) -> a__tail#(mark(X))
p12: a__tail#(cons(X,XS)) -> mark#(XS)
p13: mark#(tail(X)) -> mark#(X)
p14: mark#(repItems(X)) -> mark#(X)
p15: mark#(cons(X1,X2)) -> mark#(X1)
p16: mark#(s(X)) -> mark#(X)
p17: mark#(pair(X1,X2)) -> mark#(X1)

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, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17}


-- Reduction pair.

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

p1: mark#(pair(X1,X2)) -> mark#(X2)
p2: mark#(pair(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(repItems(X)) -> mark#(X)
p6: mark#(tail(X)) -> mark#(X)
p7: mark#(tail(X)) -> a__tail#(mark(X))
p8: a__tail#(cons(X,XS)) -> mark#(XS)
p9: mark#(zip(X1,X2)) -> mark#(X2)
p10: mark#(zip(X1,X2)) -> mark#(X1)
p11: mark#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p12: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(X)
p13: mark#(take(X1,X2)) -> mark#(X2)
p14: mark#(take(X1,X2)) -> mark#(X1)
p15: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p16: a__take#(s(N),cons(X,XS)) -> mark#(X)
p17: mark#(incr(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

  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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 5
        pair_A(x1,x2) = max{x1 + 19, x2 + 40}
        s_A(x1) = x1 + 3
        cons_A(x1,x2) = max{x1 + 6, x2 - 58}
        repItems_A(x1) = x1 + 87
        tail_A(x1) = max{93, x1 + 91}
        a__tail#_A(x1) = max{92, x1 + 64}
        mark_A(x1) = x1 + 26
        zip_A(x1,x2) = max{x1 + 87, x2 + 66}
        a__zip#_A(x1,x2) = max{70, x1 + 60, x2}
        take_A(x1,x2) = max{91, x1 + 90, x2 + 88}
        a__take#_A(x1,x2) = max{89, x1 + 63, x2 + 61}
        incr_A(x1) = max{34, x1 + 29}
        a__pairNs_A = 96
        |0|_A = 89
        oddNs_A = 125
        a__oddNs_A = 125
        a__incr_A(x1) = max{35, x1 + 29}
        a__take_A(x1,x2) = max{92, x1 + 90, x2 + 88}
        nil_A = 88
        a__zip_A(x1,x2) = max{x1 + 87, x2 + 66}
        a__tail_A(x1) = max{119, x1 + 91}
        a__repItems_A(x1) = max{89, x1 + 87}
        pairNs_A = 96
  
    precedence:
    
      oddNs = a__oddNs > mark# = tail = a__tail# = mark = a__zip# = a__take# = a__pairNs = |0| = a__zip = a__tail = a__repItems > repItems = zip > s = a__take > take > incr = a__incr > pair = cons = nil > pairNs
    
    partial status:
  
      pi(mark#) = [1]
      pi(pair) = [1]
      pi(s) = [1]
      pi(cons) = []
      pi(repItems) = [1]
      pi(tail) = []
      pi(a__tail#) = [1]
      pi(mark) = [1]
      pi(zip) = []
      pi(a__zip#) = [1, 2]
      pi(take) = []
      pi(a__take#) = [1, 2]
      pi(incr) = []
      pi(a__pairNs) = []
      pi(|0|) = []
      pi(oddNs) = []
      pi(a__oddNs) = []
      pi(a__incr) = [1]
      pi(a__take) = []
      pi(nil) = []
      pi(a__zip) = []
      pi(a__tail) = []
      pi(a__repItems) = []
      pi(pairNs) = []

The next rules are strictly ordered:

  p16

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)
p2: mark#(pair(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(repItems(X)) -> mark#(X)
p6: mark#(tail(X)) -> mark#(X)
p7: mark#(tail(X)) -> a__tail#(mark(X))
p8: a__tail#(cons(X,XS)) -> mark#(XS)
p9: mark#(zip(X1,X2)) -> mark#(X2)
p10: mark#(zip(X1,X2)) -> mark#(X1)
p11: mark#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p12: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(X)
p13: mark#(take(X1,X2)) -> mark#(X2)
p14: mark#(take(X1,X2)) -> mark#(X1)
p15: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p16: mark#(incr(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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p16}


-- Reduction pair.

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

p1: mark#(pair(X1,X2)) -> mark#(X2)
p2: mark#(incr(X)) -> mark#(X)
p3: mark#(take(X1,X2)) -> mark#(X1)
p4: mark#(take(X1,X2)) -> mark#(X2)
p5: mark#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p6: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(X)
p7: mark#(zip(X1,X2)) -> mark#(X1)
p8: mark#(zip(X1,X2)) -> mark#(X2)
p9: mark#(tail(X)) -> a__tail#(mark(X))
p10: a__tail#(cons(X,XS)) -> mark#(XS)
p11: mark#(tail(X)) -> mark#(X)
p12: mark#(repItems(X)) -> mark#(X)
p13: mark#(cons(X1,X2)) -> mark#(X1)
p14: mark#(s(X)) -> mark#(X)
p15: mark#(pair(X1,X2)) -> mark#(X1)

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{27, x1 - 17}
        pair_A(x1,x2) = max{87, x1 + 9, x2 + 15}
        incr_A(x1) = max{33, x1 + 32}
        take_A(x1,x2) = max{x1 + 11, x2 + 13}
        zip_A(x1,x2) = max{87, x1 + 33, x2 + 29}
        a__zip#_A(x1,x2) = max{x1, x2 - 2}
        mark_A(x1) = x1 + 13
        cons_A(x1,x2) = max{27, x1, x2 - 53}
        tail_A(x1) = x1 + 112
        a__tail#_A(x1) = x1 + 81
        repItems_A(x1) = x1 + 18
        s_A(x1) = x1 + 18
        a__pairNs_A = 34
        |0|_A = 34
        oddNs_A = 55
        a__oddNs_A = 67
        a__incr_A(x1) = max{46, x1 + 32}
        a__take_A(x1,x2) = max{23, x1 + 11, x2 + 13}
        nil_A = 22
        a__zip_A(x1,x2) = max{87, x1 + 33, x2 + 29}
        a__tail_A(x1) = max{123, x1 + 112}
        a__repItems_A(x1) = max{19, x1 + 18}
        pairNs_A = 33
  
    precedence:
    
      |0| = nil > pair = mark = tail = oddNs = a__oddNs = a__incr = a__tail > zip = cons = repItems = a__pairNs = a__take = a__zip = a__repItems = pairNs > a__tail# > mark# > s > incr > take = a__zip#
    
    partial status:
  
      pi(mark#) = []
      pi(pair) = []
      pi(incr) = [1]
      pi(take) = []
      pi(zip) = []
      pi(a__zip#) = [1]
      pi(mark) = []
      pi(cons) = []
      pi(tail) = []
      pi(a__tail#) = [1]
      pi(repItems) = []
      pi(s) = [1]
      pi(a__pairNs) = []
      pi(|0|) = []
      pi(oddNs) = []
      pi(a__oddNs) = []
      pi(a__incr) = []
      pi(a__take) = []
      pi(nil) = []
      pi(a__zip) = []
      pi(a__tail) = []
      pi(a__repItems) = []
      pi(pairNs) = []

The next rules are strictly ordered:

  p6

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)
p2: mark#(incr(X)) -> mark#(X)
p3: mark#(take(X1,X2)) -> mark#(X1)
p4: mark#(take(X1,X2)) -> mark#(X2)
p5: mark#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p6: mark#(zip(X1,X2)) -> mark#(X1)
p7: mark#(zip(X1,X2)) -> mark#(X2)
p8: mark#(tail(X)) -> a__tail#(mark(X))
p9: a__tail#(cons(X,XS)) -> mark#(XS)
p10: mark#(tail(X)) -> mark#(X)
p11: mark#(repItems(X)) -> mark#(X)
p12: mark#(cons(X1,X2)) -> mark#(X1)
p13: mark#(s(X)) -> mark#(X)
p14: mark#(pair(X1,X2)) -> mark#(X1)

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, p2, p3, p4, p6, p7, p8, p9, p10, p11, p12, p13, p14}


-- Reduction pair.

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

p1: mark#(pair(X1,X2)) -> mark#(X2)
p2: mark#(pair(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(repItems(X)) -> mark#(X)
p6: mark#(tail(X)) -> mark#(X)
p7: mark#(tail(X)) -> a__tail#(mark(X))
p8: a__tail#(cons(X,XS)) -> mark#(XS)
p9: mark#(zip(X1,X2)) -> mark#(X2)
p10: mark#(zip(X1,X2)) -> mark#(X1)
p11: mark#(take(X1,X2)) -> mark#(X2)
p12: mark#(take(X1,X2)) -> mark#(X1)
p13: mark#(incr(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

  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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{12, x1}
        pair_A(x1,x2) = max{18, x1 + 15, x2 + 13}
        s_A(x1) = max{3, x1}
        cons_A(x1,x2) = max{12, x1 + 8, x2}
        repItems_A(x1) = max{16, x1 + 11}
        tail_A(x1) = max{2, x1}
        a__tail#_A(x1) = max{12, x1}
        mark_A(x1) = max{1, x1}
        zip_A(x1,x2) = max{25, x1 + 21, x2 + 19}
        take_A(x1,x2) = max{17, x1 + 15, x2 + 13}
        incr_A(x1) = max{12, x1}
        a__pairNs_A = 26
        |0|_A = 18
        oddNs_A = 26
        a__oddNs_A = 26
        a__incr_A(x1) = max{12, x1}
        a__take_A(x1,x2) = max{17, x1 + 15, x2 + 13}
        nil_A = 17
        a__zip_A(x1,x2) = max{25, x1 + 21, x2 + 19}
        a__tail_A(x1) = max{2, x1}
        a__repItems_A(x1) = max{16, x1 + 11}
        pairNs_A = 26
  
    precedence:
    
      tail = mark = a__zip = a__tail > a__take > nil = a__repItems > zip > s = repItems = take = a__oddNs > a__pairNs = a__incr > pair = cons = oddNs > incr > mark# > a__tail# = |0| = pairNs
    
    partial status:
  
      pi(mark#) = [1]
      pi(pair) = []
      pi(s) = [1]
      pi(cons) = [2]
      pi(repItems) = []
      pi(tail) = [1]
      pi(a__tail#) = [1]
      pi(mark) = [1]
      pi(zip) = [1]
      pi(take) = []
      pi(incr) = [1]
      pi(a__pairNs) = []
      pi(|0|) = []
      pi(oddNs) = []
      pi(a__oddNs) = []
      pi(a__incr) = [1]
      pi(a__take) = [1]
      pi(nil) = []
      pi(a__zip) = []
      pi(a__tail) = [1]
      pi(a__repItems) = []
      pi(pairNs) = []

The next rules are strictly ordered:

  p7

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)
p2: mark#(pair(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(repItems(X)) -> mark#(X)
p6: mark#(tail(X)) -> mark#(X)
p7: a__tail#(cons(X,XS)) -> mark#(XS)
p8: mark#(zip(X1,X2)) -> mark#(X2)
p9: mark#(zip(X1,X2)) -> mark#(X1)
p10: mark#(take(X1,X2)) -> mark#(X2)
p11: mark#(take(X1,X2)) -> mark#(X1)
p12: mark#(incr(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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p8, p9, p10, p11, p12}


-- Reduction pair.

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

p1: mark#(pair(X1,X2)) -> mark#(X2)
p2: mark#(incr(X)) -> mark#(X)
p3: mark#(take(X1,X2)) -> mark#(X1)
p4: mark#(take(X1,X2)) -> mark#(X2)
p5: mark#(zip(X1,X2)) -> mark#(X1)
p6: mark#(zip(X1,X2)) -> mark#(X2)
p7: mark#(tail(X)) -> mark#(X)
p8: mark#(repItems(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(s(X)) -> mark#(X)
p11: mark#(pair(X1,X2)) -> mark#(X1)

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{4, x1 + 3}
        pair_A(x1,x2) = max{x1 + 2, x2 + 2}
        incr_A(x1) = x1
        take_A(x1,x2) = max{x1, x2}
        zip_A(x1,x2) = max{x1, x2}
        tail_A(x1) = x1
        repItems_A(x1) = x1
        cons_A(x1,x2) = max{x1, x2 + 1}
        s_A(x1) = x1
  
    precedence:
    
      mark# = pair = incr = take = zip = tail = repItems = cons = s
    
    partial status:
  
      pi(mark#) = []
      pi(pair) = [2]
      pi(incr) = [1]
      pi(take) = [2]
      pi(zip) = [2]
      pi(tail) = [1]
      pi(repItems) = [1]
      pi(cons) = [2]
      pi(s) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(incr(X)) -> mark#(X)
p2: mark#(take(X1,X2)) -> mark#(X1)
p3: mark#(take(X1,X2)) -> mark#(X2)
p4: mark#(zip(X1,X2)) -> mark#(X1)
p5: mark#(zip(X1,X2)) -> mark#(X2)
p6: mark#(tail(X)) -> mark#(X)
p7: mark#(repItems(X)) -> mark#(X)
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(s(X)) -> mark#(X)
p10: mark#(pair(X1,X2)) -> mark#(X1)

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, p2, p3, p4, p5, p6, p7, p8, p9, p10}


-- Reduction pair.

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

p1: mark#(incr(X)) -> mark#(X)
p2: mark#(pair(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(repItems(X)) -> mark#(X)
p6: mark#(tail(X)) -> mark#(X)
p7: mark#(zip(X1,X2)) -> mark#(X2)
p8: mark#(zip(X1,X2)) -> mark#(X1)
p9: mark#(take(X1,X2)) -> mark#(X2)
p10: mark#(take(X1,X2)) -> mark#(X1)

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 3
        incr_A(x1) = x1 + 2
        pair_A(x1,x2) = max{x1, x2}
        s_A(x1) = x1
        cons_A(x1,x2) = max{x1, x2}
        repItems_A(x1) = x1
        tail_A(x1) = x1
        zip_A(x1,x2) = max{x1, x2}
        take_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      mark# = incr = pair = s = cons = repItems = tail = zip = take
    
    partial status:
  
      pi(mark#) = [1]
      pi(incr) = [1]
      pi(pair) = [1, 2]
      pi(s) = [1]
      pi(cons) = [1, 2]
      pi(repItems) = [1]
      pi(tail) = [1]
      pi(zip) = [1, 2]
      pi(take) = [1, 2]

The next rules are strictly ordered:

  p1

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#(X1)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(repItems(X)) -> mark#(X)
p5: mark#(tail(X)) -> mark#(X)
p6: mark#(zip(X1,X2)) -> mark#(X2)
p7: mark#(zip(X1,X2)) -> mark#(X1)
p8: mark#(take(X1,X2)) -> mark#(X2)
p9: mark#(take(X1,X2)) -> mark#(X1)

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, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

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

p1: mark#(pair(X1,X2)) -> mark#(X1)
p2: mark#(take(X1,X2)) -> mark#(X1)
p3: mark#(take(X1,X2)) -> mark#(X2)
p4: mark#(zip(X1,X2)) -> mark#(X1)
p5: mark#(zip(X1,X2)) -> mark#(X2)
p6: mark#(tail(X)) -> mark#(X)
p7: mark#(repItems(X)) -> mark#(X)
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 1
        pair_A(x1,x2) = max{x1, x2}
        take_A(x1,x2) = max{x1, x2}
        zip_A(x1,x2) = max{x1, x2}
        tail_A(x1) = x1
        repItems_A(x1) = x1
        cons_A(x1,x2) = max{x1, x2}
        s_A(x1) = x1
  
    precedence:
    
      mark# = pair = take = zip = tail = repItems = cons = s
    
    partial status:
  
      pi(mark#) = [1]
      pi(pair) = [1, 2]
      pi(take) = [1, 2]
      pi(zip) = [1, 2]
      pi(tail) = [1]
      pi(repItems) = [1]
      pi(cons) = [1, 2]
      pi(s) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(take(X1,X2)) -> mark#(X1)
p2: mark#(take(X1,X2)) -> mark#(X2)
p3: mark#(zip(X1,X2)) -> mark#(X1)
p4: mark#(zip(X1,X2)) -> mark#(X2)
p5: mark#(tail(X)) -> mark#(X)
p6: mark#(repItems(X)) -> mark#(X)
p7: mark#(cons(X1,X2)) -> mark#(X1)
p8: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

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

p1: mark#(take(X1,X2)) -> mark#(X1)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(repItems(X)) -> mark#(X)
p5: mark#(tail(X)) -> mark#(X)
p6: mark#(zip(X1,X2)) -> mark#(X2)
p7: mark#(zip(X1,X2)) -> mark#(X1)
p8: mark#(take(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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 3
        take_A(x1,x2) = max{x1 + 1, x2}
        s_A(x1) = x1 + 3
        cons_A(x1,x2) = max{x1, x2}
        repItems_A(x1) = x1
        tail_A(x1) = x1
        zip_A(x1,x2) = max{x1, x2 + 2}
  
    precedence:
    
      mark# = take = s = cons = repItems = tail = zip
    
    partial status:
  
      pi(mark#) = []
      pi(take) = [2]
      pi(s) = [1]
      pi(cons) = [2]
      pi(repItems) = [1]
      pi(tail) = [1]
      pi(zip) = [2]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(take(X1,X2)) -> mark#(X1)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(repItems(X)) -> mark#(X)
p4: mark#(tail(X)) -> mark#(X)
p5: mark#(zip(X1,X2)) -> mark#(X2)
p6: mark#(zip(X1,X2)) -> mark#(X1)
p7: mark#(take(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, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

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

p1: mark#(take(X1,X2)) -> mark#(X1)
p2: mark#(take(X1,X2)) -> mark#(X2)
p3: mark#(zip(X1,X2)) -> mark#(X1)
p4: mark#(zip(X1,X2)) -> mark#(X2)
p5: mark#(tail(X)) -> mark#(X)
p6: mark#(repItems(X)) -> mark#(X)
p7: mark#(cons(X1,X2)) -> mark#(X1)

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 2
        take_A(x1,x2) = max{x1, x2}
        zip_A(x1,x2) = max{x1 + 1, x2 + 1}
        tail_A(x1) = x1
        repItems_A(x1) = x1
        cons_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      mark# = take = zip = tail = repItems = cons
    
    partial status:
  
      pi(mark#) = [1]
      pi(take) = [1, 2]
      pi(zip) = [1, 2]
      pi(tail) = [1]
      pi(repItems) = [1]
      pi(cons) = [1, 2]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(take(X1,X2)) -> mark#(X1)
p2: mark#(zip(X1,X2)) -> mark#(X1)
p3: mark#(zip(X1,X2)) -> mark#(X2)
p4: mark#(tail(X)) -> mark#(X)
p5: mark#(repItems(X)) -> mark#(X)
p6: mark#(cons(X1,X2)) -> mark#(X1)

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, p2, p3, p4, p5, p6}


-- Reduction pair.

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

p1: mark#(take(X1,X2)) -> mark#(X1)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(repItems(X)) -> mark#(X)
p4: mark#(tail(X)) -> mark#(X)
p5: mark#(zip(X1,X2)) -> mark#(X2)
p6: mark#(zip(X1,X2)) -> mark#(X1)

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{2, x1 - 4}
        take_A(x1,x2) = max{x1 + 2, x2 + 2}
        cons_A(x1,x2) = max{7, x1 + 1, x2}
        repItems_A(x1) = x1 + 1
        tail_A(x1) = x1
        zip_A(x1,x2) = max{x1 + 5, x2 + 3}
  
    precedence:
    
      mark# = take = cons = repItems = tail = zip
    
    partial status:
  
      pi(mark#) = []
      pi(take) = [2]
      pi(cons) = [2]
      pi(repItems) = [1]
      pi(tail) = [1]
      pi(zip) = [1]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(take(X1,X2)) -> mark#(X1)
p2: mark#(repItems(X)) -> mark#(X)
p3: mark#(tail(X)) -> mark#(X)
p4: mark#(zip(X1,X2)) -> mark#(X2)
p5: mark#(zip(X1,X2)) -> mark#(X1)

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, p2, p3, p4, p5}


-- Reduction pair.

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

p1: mark#(take(X1,X2)) -> mark#(X1)
p2: mark#(zip(X1,X2)) -> mark#(X1)
p3: mark#(zip(X1,X2)) -> mark#(X2)
p4: mark#(tail(X)) -> mark#(X)
p5: mark#(repItems(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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 1
        take_A(x1,x2) = max{x1, x2}
        zip_A(x1,x2) = max{x1, x2}
        tail_A(x1) = x1
        repItems_A(x1) = x1
  
    precedence:
    
      mark# = take = zip = tail = repItems
    
    partial status:
  
      pi(mark#) = [1]
      pi(take) = [1, 2]
      pi(zip) = [1, 2]
      pi(tail) = [1]
      pi(repItems) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(zip(X1,X2)) -> mark#(X1)
p2: mark#(zip(X1,X2)) -> mark#(X2)
p3: mark#(tail(X)) -> mark#(X)
p4: mark#(repItems(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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: mark#(zip(X1,X2)) -> mark#(X1)
p2: mark#(repItems(X)) -> mark#(X)
p3: mark#(tail(X)) -> mark#(X)
p4: mark#(zip(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 monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 1
        zip_A(x1,x2) = max{x1, x2}
        repItems_A(x1) = x1
        tail_A(x1) = x1
  
    precedence:
    
      mark# = zip = repItems = tail
    
    partial status:
  
      pi(mark#) = [1]
      pi(zip) = [1, 2]
      pi(repItems) = [1]
      pi(tail) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(repItems(X)) -> mark#(X)
p2: mark#(tail(X)) -> mark#(X)
p3: mark#(zip(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, p2, p3}


-- Reduction pair.

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

p1: mark#(repItems(X)) -> mark#(X)
p2: mark#(zip(X1,X2)) -> mark#(X2)
p3: mark#(tail(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 reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 2
        repItems_A(x1) = x1
        zip_A(x1,x2) = x2
        tail_A(x1) = x1
  
    precedence:
    
      mark# = repItems = zip = tail
    
    partial status:
  
      pi(mark#) = [1]
      pi(repItems) = [1]
      pi(zip) = [2]
      pi(tail) = [1]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(repItems(X)) -> mark#(X)
p2: mark#(tail(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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: mark#(repItems(X)) -> mark#(X)
p2: mark#(tail(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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{3, x1 + 2}
        repItems_A(x1) = x1 + 3
        tail_A(x1) = max{1, x1}
  
    precedence:
    
      mark# = repItems = tail
    
    partial status:
  
      pi(mark#) = [1]
      pi(repItems) = [1]
      pi(tail) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(tail(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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: mark#(tail(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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 2
        tail_A(x1) = x1 + 2
  
    precedence:
    
      mark# = tail
    
    partial status:
  
      pi(mark#) = [1]
      pi(tail) = [1]

The next rules are strictly ordered:

  p1

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