YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__oddNs#_A() = (2,0)
        a__incr#_A(x1) = ((0,1),(0,0)) x1 + (1,0)
        a__pairNs_A() = (3,1)
        cons_A(x1,x2) = ((0,1),(0,1)) x1 + x2
        mark#_A(x1) = ((0,1),(0,0)) x1 + (1,0)
        pair_A(x1,x2) = ((0,1),(0,1)) x1 + x2 + (1,0)
        s_A(x1) = ((1,1),(0,1)) x1 + (1,0)
        repItems_A(x1) = ((1,1),(1,1)) x1 + (1,2)
        a__repItems#_A(x1) = x1 + (2,0)
        mark_A(x1) = x1
        tail_A(x1) = x1 + (1,2)
        a__tail#_A(x1) = ((0,1),(0,0)) x1 + (2,0)
        zip_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,1),(1,1)) x2 + (2,0)
        a__zip#_A(x1,x2) = ((0,1),(0,0)) x1 + x2 + (1,0)
        take_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(1,1)) x2 + (2,1)
        a__take#_A(x1,x2) = x1 + x2
        oddNs_A() = (2,1)
        incr_A(x1) = ((0,1),(0,1)) x1 + (1,0)
        a__oddNs_A() = (2,1)
        a__incr_A(x1) = ((0,1),(0,1)) x1 + (1,0)
        a__take_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(1,1)) x2 + (2,1)
        |0|_A() = (1,0)
        nil_A() = (1,0)
        a__zip_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,1),(1,1)) x2 + (2,0)
        a__tail_A(x1) = x1 + (1,2)
        a__repItems_A(x1) = ((1,1),(1,1)) x1 + (1,2)
        pairNs_A() = (3,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__oddNs#_A() = (7,1)
        a__incr#_A(x1) = (7,1)
        a__pairNs_A() = (2,4)
        cons_A(x1,x2) = (3,5)
        mark#_A(x1) = (7,1)
        pair_A(x1,x2) = (1,1)
        s_A(x1) = ((0,0),(1,0)) x1
        repItems_A(x1) = ((0,0),(1,1)) x1 + (3,1)
        a__repItems#_A(x1) = ((1,1),(1,1)) x1
        mark_A(x1) = ((0,0),(1,1)) x1 + (3,3)
        tail_A(x1) = (1,1)
        a__tail#_A(x1) = (6,0)
        zip_A(x1,x2) = ((0,0),(1,0)) x1 + (1,1)
        a__zip#_A(x1,x2) = x2 + (4,1)
        take_A(x1,x2) = (1,5)
        a__take#_A(x1,x2) = (7,2)
        oddNs_A() = (0,2)
        incr_A(x1) = (0,2)
        a__oddNs_A() = (3,5)
        a__incr_A(x1) = (3,5)
        a__take_A(x1,x2) = (3,5)
        |0|_A() = (1,1)
        nil_A() = (1,1)
        a__zip_A(x1,x2) = ((0,0),(1,0)) x1 + (3,2)
        a__tail_A(x1) = (3,3)
        a__repItems_A(x1) = ((0,0),(1,1)) x1 + (3,1)
        pairNs_A() = (1,1)
    

The next rules are strictly ordered:

  p7, p8, p9, p10, p11, p12, p17, p18, p19

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#(zip(X1,X2)) -> mark#(X2)
p8: mark#(zip(X1,X2)) -> mark#(X1)
p9: mark#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p10: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(Y)
p11: a__take#(s(N),cons(X,XS)) -> mark#(X)
p12: mark#(oddNs()) -> a__oddNs#()
p13: mark#(incr(X)) -> mark#(X)
p14: mark#(incr(X)) -> a__incr#(mark(X))
p15: 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, p8, p9, p10, p12, p13, p14, p15}


-- 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#(zip(X1,X2)) -> a__zip#(mark(X1),mark(X2))
p7: a__zip#(cons(X,XS),cons(Y,YS)) -> mark#(X)
p8: mark#(zip(X1,X2)) -> mark#(X1)
p9: mark#(zip(X1,X2)) -> mark#(X2)
p10: mark#(cons(X1,X2)) -> mark#(X1)
p11: mark#(s(X)) -> mark#(X)
p12: mark#(pair(X1,X2)) -> mark#(X1)
p13: mark#(pair(X1,X2)) -> mark#(X2)
p14: 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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__oddNs#_A() = (1,0)
        a__incr#_A(x1) = ((0,1),(0,0)) x1
        a__pairNs_A() = (1,1)
        cons_A(x1,x2) = x1 + ((0,1),(1,0)) x2
        mark#_A(x1) = ((0,1),(0,0)) x1
        incr_A(x1) = x1
        mark_A(x1) = x1
        oddNs_A() = (1,1)
        zip_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (2,2)
        a__zip#_A(x1,x2) = ((1,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (1,0)
        s_A(x1) = x1 + (1,0)
        pair_A(x1,x2) = x1 + x2 + (1,0)
        a__oddNs_A() = (1,1)
        a__incr_A(x1) = x1
        a__take_A(x1,x2) = x2 + (1,1)
        |0|_A() = (1,0)
        nil_A() = (1,0)
        take_A(x1,x2) = x2 + (1,1)
        a__zip_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (2,2)
        a__tail_A(x1) = ((0,1),(1,0)) x1 + (1,1)
        a__repItems_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        repItems_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        tail_A(x1) = ((0,1),(1,0)) x1 + (1,1)
        pairNs_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__oddNs#_A() = (5,0)
        a__incr#_A(x1) = (5,0)
        a__pairNs_A() = (4,1)
        cons_A(x1,x2) = (3,1)
        mark#_A(x1) = (5,0)
        incr_A(x1) = (4,10)
        mark_A(x1) = ((1,1),(1,1)) x1 + (1,5)
        oddNs_A() = (1,4)
        zip_A(x1,x2) = x2 + (1,1)
        a__zip#_A(x1,x2) = ((1,1),(0,0)) x1 + (0,1)
        s_A(x1) = (1,1)
        pair_A(x1,x2) = (1,1)
        a__oddNs_A() = (5,10)
        a__incr_A(x1) = (4,10)
        a__take_A(x1,x2) = ((0,1),(1,0)) x2 + (2,11)
        |0|_A() = (0,1)
        nil_A() = (2,2)
        take_A(x1,x2) = ((0,1),(1,0)) x2 + (1,6)
        a__zip_A(x1,x2) = x2 + (1,1)
        a__tail_A(x1) = (3,5)
        a__repItems_A(x1) = ((1,1),(1,1)) x1 + (1,5)
        repItems_A(x1) = ((1,1),(1,1)) x1 + (1,5)
        tail_A(x1) = (1,1)
        pairNs_A() = (3,1)
    

The next rules are strictly ordered:

  p6, p7, p8, p9, p14

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)) -> a__incr#(mark(X))
p4: mark#(incr(X)) -> mark#(X)
p5: mark#(oddNs()) -> a__oddNs#()
p6: mark#(cons(X1,X2)) -> mark#(X1)
p7: mark#(s(X)) -> mark#(X)
p8: mark#(pair(X1,X2)) -> mark#(X1)
p9: 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, p2, p3, p4, p5, p6, p7, p8, p9}


-- 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#(oddNs()) -> a__oddNs#()
p8: mark#(incr(X)) -> mark#(X)
p9: 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 set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__oddNs#_A() = (3,4)
        a__incr#_A(x1) = ((0,1),(1,1)) x1
        a__pairNs_A() = (2,2)
        cons_A(x1,x2) = ((0,1),(1,1)) x1 + ((0,1),(1,0)) x2
        mark#_A(x1) = ((1,1),(1,1)) x1
        pair_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(0,1)) x2
        s_A(x1) = ((1,1),(0,0)) x1
        oddNs_A() = (2,2)
        incr_A(x1) = x1
        mark_A(x1) = x1
        a__oddNs_A() = (2,2)
        a__incr_A(x1) = x1
        a__take_A(x1,x2) = ((1,1),(1,1)) x2 + (2,2)
        |0|_A() = (0,0)
        nil_A() = (1,1)
        take_A(x1,x2) = ((1,1),(1,1)) x2 + (2,2)
        a__zip_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        zip_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        a__tail_A(x1) = ((0,1),(1,1)) x1 + (1,1)
        a__repItems_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        repItems_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        tail_A(x1) = ((0,1),(1,1)) x1 + (1,1)
        pairNs_A() = (2,2)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__oddNs#_A() = (1,1)
        a__incr#_A(x1) = (0,0)
        a__pairNs_A() = (2,1)
        cons_A(x1,x2) = (2,1)
        mark#_A(x1) = (0,0)
        pair_A(x1,x2) = (1,1)
        s_A(x1) = (0,1)
        oddNs_A() = (4,1)
        incr_A(x1) = (1,1)
        mark_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        a__oddNs_A() = (4,6)
        a__incr_A(x1) = (3,3)
        a__take_A(x1,x2) = (3,2)
        |0|_A() = (1,1)
        nil_A() = (4,3)
        take_A(x1,x2) = (3,1)
        a__zip_A(x1,x2) = x2 + (2,2)
        zip_A(x1,x2) = x2 + (1,2)
        a__tail_A(x1) = (2,1)
        a__repItems_A(x1) = ((1,1),(1,1)) x1 + (1,2)
        repItems_A(x1) = ((1,1),(1,1)) x1 + (1,2)
        tail_A(x1) = (1,1)
        pairNs_A() = (1,1)
    

The next rules are strictly ordered:

  p1, p7

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#(incr(X)) -> mark#(X)
p7: 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:

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


-- Reduction pair.

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

p1: a__incr#(cons(X,XS)) -> mark#(X)
p2: mark#(incr(X)) -> a__incr#(mark(X))
p3: mark#(incr(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(s(X)) -> mark#(X)
p6: mark#(pair(X1,X2)) -> mark#(X1)
p7: mark#(pair(X1,X2)) -> mark#(X2)

and R consists of:

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

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__incr#_A(x1) = x1
        cons_A(x1,x2) = x1 + ((0,1),(1,0)) x2
        mark#_A(x1) = x1
        incr_A(x1) = x1
        mark_A(x1) = x1
        s_A(x1) = ((1,0),(1,1)) x1 + (0,1)
        pair_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (0,1)
        a__pairNs_A() = (0,0)
        |0|_A() = (0,1)
        oddNs_A() = (0,0)
        a__oddNs_A() = (0,0)
        a__incr_A(x1) = x1
        a__take_A(x1,x2) = ((1,1),(0,1)) x1 + x2 + (1,2)
        nil_A() = (1,1)
        take_A(x1,x2) = ((1,1),(0,1)) x1 + x2 + (1,2)
        a__zip_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,1)
        zip_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,1)
        a__tail_A(x1) = ((0,1),(1,0)) x1 + (1,1)
        a__repItems_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        repItems_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        pairNs_A() = (0,0)
        tail_A(x1) = ((0,1),(1,0)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__incr#_A(x1) = ((1,1),(0,0)) x1 + (2,0)
        cons_A(x1,x2) = x1 + (1,1)
        mark#_A(x1) = ((1,1),(0,0)) x1
        incr_A(x1) = ((0,1),(1,1)) x1 + (1,4)
        mark_A(x1) = ((0,1),(1,1)) x1 + (1,2)
        s_A(x1) = x1 + (1,1)
        pair_A(x1,x2) = x1 + x2 + (2,3)
        a__pairNs_A() = (4,4)
        |0|_A() = (2,1)
        oddNs_A() = (7,6)
        a__oddNs_A() = (7,12)
        a__incr_A(x1) = ((0,1),(1,1)) x1 + (2,4)
        a__take_A(x1,x2) = ((0,1),(1,1)) x2 + (3,5)
        nil_A() = (2,1)
        take_A(x1,x2) = ((0,1),(1,1)) x2 + (1,5)
        a__zip_A(x1,x2) = ((0,1),(1,1)) x1 + ((0,1),(1,1)) x2 + (5,9)
        zip_A(x1,x2) = ((0,1),(1,1)) x1 + ((0,1),(1,1)) x2 + (4,9)
        a__tail_A(x1) = (1,1)
        a__repItems_A(x1) = ((0,1),(1,1)) x1 + (2,4)
        repItems_A(x1) = ((0,1),(1,1)) x1 + (2,4)
        pairNs_A() = (1,4)
        tail_A(x1) = (0,1)
    

The next rules are strictly ordered:

  p1, p3, p4, p5, p6, p7

We remove them from the problem.

-- SCC decomposition.

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

p1: 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:

  (no SCCs)