Termination w.r.t. Q proof of /tmp/tmpPfSnWj/Ex5_DLMMU04

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 PisEmptyProof (⇔)

    ↳29 TRUE

  ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 PisEmptyProof (⇔)

    ↳40 TRUE

  ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 QDPOrderProof (⇔)

    ↳47 QDP

    ↳48 QDPOrderProof (⇔)

    ↳49 QDP

    ↳50 PisEmptyProof (⇔)

    ↳51 TRUE

  ↳52 QDP

    ↳53 QDPOrderProof (⇔)

    ↳54 QDP

    ↳55 QDPOrderProof (⇔)

    ↳56 QDP

    ↳57 PisEmptyProof (⇔)

    ↳58 TRUE

  ↳59 QDP

    ↳60 QDPOrderProof (⇔)

    ↳61 QDP

    ↳62 QDPOrderProof (⇔)

    ↳63 QDP

    ↳64 PisEmptyProof (⇔)

    ↳65 TRUE

  ↳66 QDP

    ↳67 QDPOrderProof (⇔)

    ↳68 QDP

    ↳69 QDPOrderProof (⇔)

    ↳70 QDP

    ↳71 QDPOrderProof (⇔)

    ↳72 QDP

    ↳73 QDPOrderProof (⇔)

    ↳74 QDP

    ↳75 PisEmptyProof (⇔)

    ↳76 TRUE

  ↳77 QDP

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(pairNs) → MARK(cons(0, incr(oddNs)))
ACTIVE(pairNs) → CONS(0, incr(oddNs))
ACTIVE(pairNs) → INCR(oddNs)
ACTIVE(oddNs) → MARK(incr(pairNs))
ACTIVE(oddNs) → INCR(pairNs)
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
ACTIVE(incr(cons(X, XS))) → CONS(s(X), incr(XS))
ACTIVE(incr(cons(X, XS))) → S(X)
ACTIVE(incr(cons(X, XS))) → INCR(XS)
ACTIVE(take(0, XS)) → MARK(nil)
ACTIVE(take(s(N), cons(X, XS))) → MARK(cons(X, take(N, XS)))
ACTIVE(take(s(N), cons(X, XS))) → CONS(X, take(N, XS))
ACTIVE(take(s(N), cons(X, XS))) → TAKE(N, XS)
ACTIVE(zip(nil, XS)) → MARK(nil)
ACTIVE(zip(X, nil)) → MARK(nil)
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → MARK(cons(pair(X, Y), zip(XS, YS)))
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → CONS(pair(X, Y), zip(XS, YS))
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → PAIR(X, Y)
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → ZIP(XS, YS)
ACTIVE(tail(cons(X, XS))) → MARK(XS)
ACTIVE(repItems(nil)) → MARK(nil)
ACTIVE(repItems(cons(X, XS))) → MARK(cons(X, cons(X, repItems(XS))))
ACTIVE(repItems(cons(X, XS))) → CONS(X, cons(X, repItems(XS)))
ACTIVE(repItems(cons(X, XS))) → CONS(X, repItems(XS))
ACTIVE(repItems(cons(X, XS))) → REPITEMS(XS)
MARK(pairNs) → ACTIVE(pairNs)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(0) → ACTIVE(0)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → ACTIVE(oddNs)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → TAKE(mark(X1), mark(X2))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(nil) → ACTIVE(nil)
MARK(zip(X1, X2)) → ACTIVE(zip(mark(X1), mark(X2)))
MARK(zip(X1, X2)) → ZIP(mark(X1), mark(X2))
MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → MARK(X2)
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → PAIR(mark(X1), mark(X2))
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → TAIL(mark(X))
MARK(tail(X)) → MARK(X)
MARK(repItems(X)) → ACTIVE(repItems(mark(X)))
MARK(repItems(X)) → REPITEMS(mark(X))
MARK(repItems(X)) → MARK(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
INCR(mark(X)) → INCR(X)
INCR(active(X)) → INCR(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)
ZIP(mark(X1), X2) → ZIP(X1, X2)
ZIP(X1, mark(X2)) → ZIP(X1, X2)
ZIP(active(X1), X2) → ZIP(X1, X2)
ZIP(X1, active(X2)) → ZIP(X1, X2)
PAIR(mark(X1), X2) → PAIR(X1, X2)
PAIR(X1, mark(X2)) → PAIR(X1, X2)
PAIR(active(X1), X2) → PAIR(X1, X2)
PAIR(X1, active(X2)) → PAIR(X1, X2)
TAIL(mark(X)) → TAIL(X)
TAIL(active(X)) → TAIL(X)
REPITEMS(mark(X)) → REPITEMS(X)
REPITEMS(active(X)) → REPITEMS(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 9 SCCs with 28 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

REPITEMS(active(X)) → REPITEMS(X)
REPITEMS(mark(X)) → REPITEMS(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

REPITEMS(mark(X)) → REPITEMS(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
REPITEMS(x1) = x1
active(x1) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

REPITEMS(active(X)) → REPITEMS(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

REPITEMS(active(X)) → REPITEMS(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
REPITEMS(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAIL(mark(X)) → TAIL(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAIL(x1) = x1
active(x1) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TAIL(active(X)) → TAIL(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAIL(active(X)) → TAIL(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAIL(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(17) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(18) TRUE

(19) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PAIR(X1, mark(X2)) → PAIR(X1, X2)
PAIR(mark(X1), X2) → PAIR(X1, X2)
PAIR(active(X1), X2) → PAIR(X1, X2)
PAIR(X1, active(X2)) → PAIR(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PAIR(X1, active(X2)) → PAIR(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PAIR(x1, x2) = x2
mark(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PAIR(X1, mark(X2)) → PAIR(X1, X2)
PAIR(mark(X1), X2) → PAIR(X1, X2)
PAIR(active(X1), X2) → PAIR(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PAIR(X1, mark(X2)) → PAIR(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PAIR(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PAIR(mark(X1), X2) → PAIR(X1, X2)
PAIR(active(X1), X2) → PAIR(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PAIR(active(X1), X2) → PAIR(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PAIR(x1, x2) = x1
mark(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PAIR(mark(X1), X2) → PAIR(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PAIR(mark(X1), X2) → PAIR(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PAIR(x1, x2) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(28) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(29) TRUE

(30) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ZIP(X1, mark(X2)) → ZIP(X1, X2)
ZIP(mark(X1), X2) → ZIP(X1, X2)
ZIP(active(X1), X2) → ZIP(X1, X2)
ZIP(X1, active(X2)) → ZIP(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ZIP(X1, active(X2)) → ZIP(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZIP(x1, x2) = x2
mark(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ZIP(X1, mark(X2)) → ZIP(X1, X2)
ZIP(mark(X1), X2) → ZIP(X1, X2)
ZIP(active(X1), X2) → ZIP(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ZIP(X1, mark(X2)) → ZIP(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZIP(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ZIP(mark(X1), X2) → ZIP(X1, X2)
ZIP(active(X1), X2) → ZIP(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ZIP(active(X1), X2) → ZIP(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZIP(x1, x2) = x1
mark(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(36) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ZIP(mark(X1), X2) → ZIP(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ZIP(mark(X1), X2) → ZIP(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZIP(x1, x2) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(38) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(39) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(40) TRUE

(41) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAKE(X1, active(X2)) → TAKE(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2) = x2
mark(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAKE(X1, mark(X2)) → TAKE(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(45) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(46) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAKE(active(X1), X2) → TAKE(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2) = x1
mark(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(47) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TAKE(mark(X1), X2) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(48) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAKE(mark(X1), X2) → TAKE(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(49) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(50) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(51) TRUE

(52) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(53) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = x1
active(x1) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(54) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(active(X)) → S(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(56) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(57) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(58) TRUE

(59) Obligation:

Q DP problem:
The TRS P consists of the following rules:

INCR(active(X)) → INCR(X)
INCR(mark(X)) → INCR(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INCR(mark(X)) → INCR(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INCR(x1) = x1
active(x1) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(61) Obligation:

Q DP problem:
The TRS P consists of the following rules:

INCR(active(X)) → INCR(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INCR(active(X)) → INCR(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INCR(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(63) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(64) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(65) TRUE

(66) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(67) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, active(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = x2
mark(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(68) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(69) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, mark(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = x2
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(70) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(71) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(active(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = x1
mark(x1) = x1
active(x1) = active(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(72) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(73) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(mark(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = x1
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(74) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(75) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(76) TRUE

(77) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(pairNs) → ACTIVE(pairNs)
ACTIVE(pairNs) → MARK(cons(0, incr(oddNs)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(take(s(N), cons(X, XS))) → MARK(cons(X, take(N, XS)))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → ACTIVE(oddNs)
ACTIVE(oddNs) → MARK(incr(pairNs))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → MARK(cons(pair(X, Y), zip(XS, YS)))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
ACTIVE(tail(cons(X, XS))) → MARK(XS)
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(zip(X1, X2)) → ACTIVE(zip(mark(X1), mark(X2)))
ACTIVE(repItems(cons(X, XS))) → MARK(cons(X, cons(X, repItems(XS))))
MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → MARK(X2)
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → MARK(X)
MARK(repItems(X)) → ACTIVE(repItems(mark(X)))
MARK(repItems(X)) → MARK(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.