Termination w.r.t. Q proof of /tmp/tmp9QNPn0/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 PisEmptyProof (⇔)

    ↳27 TRUE

  ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 PisEmptyProof (⇔)

    ↳36 TRUE

  ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 QDPOrderProof (⇔)

    ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 PisEmptyProof (⇔)

    ↳45 TRUE

  ↳46 QDP

    ↳47 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 PisEmptyProof (⇔)

    ↳52 TRUE

  ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 PisEmptyProof (⇔)

    ↳59 TRUE

  ↳60 QDP

    ↳61 QDPOrderProof (⇔)

    ↳62 QDP

    ↳63 QDPOrderProof (⇔)

    ↳64 QDP

    ↳65 PisEmptyProof (⇔)

    ↳66 TRUE

  ↳67 QDP

    ↳68 QDPOrderProof (⇔)

    ↳69 QDP

    ↳70 QDPOrderProof (⇔)

    ↳71 QDP

    ↳72 QDPOrderProof (⇔)

    ↳73 QDP

    ↳74 QDPOrderProof (⇔)

    ↳75 QDP

    ↳76 PisEmptyProof (⇔)

    ↳77 TRUE

  ↳78 QDP

    ↳79 QDPOrderProof (⇔)

    ↳80 QDP

    ↳81 QDPOrderProof (⇔)

    ↳82 QDP

    ↳83 QDPOrderProof (⇔)

    ↳84 QDP

    ↳85 QDPOrderProof (⇔)

    ↳86 QDP

    ↳87 QDPOrderProof (⇔)

    ↳88 QDP

    ↳89 PisEmptyProof (⇔)

    ↳90 TRUE

  ↳91 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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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) → CONS(0, incr(oddNs))
ACTIVE(pairNs) → INCR(oddNs)
ACTIVE(oddNs) → INCR(pairNs)
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(s(N), cons(X, XS))) → CONS(X, take(N, XS))
ACTIVE(take(s(N), cons(X, XS))) → TAKE(N, XS)
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(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)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(incr(X)) → INCR(active(X))
ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → TAKE(active(X1), X2)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → TAKE(X1, active(X2))
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(zip(X1, X2)) → ZIP(active(X1), X2)
ACTIVE(zip(X1, X2)) → ACTIVE(X1)
ACTIVE(zip(X1, X2)) → ZIP(X1, active(X2))
ACTIVE(zip(X1, X2)) → ACTIVE(X2)
ACTIVE(pair(X1, X2)) → PAIR(active(X1), X2)
ACTIVE(pair(X1, X2)) → ACTIVE(X1)
ACTIVE(pair(X1, X2)) → PAIR(X1, active(X2))
ACTIVE(pair(X1, X2)) → ACTIVE(X2)
ACTIVE(tail(X)) → TAIL(active(X))
ACTIVE(tail(X)) → ACTIVE(X)
ACTIVE(repItems(X)) → REPITEMS(active(X))
ACTIVE(repItems(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
INCR(mark(X)) → INCR(X)
S(mark(X)) → S(X)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
ZIP(mark(X1), X2) → ZIP(X1, X2)
ZIP(X1, mark(X2)) → ZIP(X1, X2)
PAIR(mark(X1), X2) → PAIR(X1, X2)
PAIR(X1, mark(X2)) → PAIR(X1, X2)
TAIL(mark(X)) → TAIL(X)
REPITEMS(mark(X)) → REPITEMS(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(incr(X)) → INCR(proper(X))
PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(take(X1, X2)) → TAKE(proper(X1), proper(X2))
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(zip(X1, X2)) → ZIP(proper(X1), proper(X2))
PROPER(zip(X1, X2)) → PROPER(X1)
PROPER(zip(X1, X2)) → PROPER(X2)
PROPER(pair(X1, X2)) → PAIR(proper(X1), proper(X2))
PROPER(pair(X1, X2)) → PROPER(X1)
PROPER(pair(X1, X2)) → PROPER(X2)
PROPER(tail(X)) → TAIL(proper(X))
PROPER(tail(X)) → PROPER(X)
PROPER(repItems(X)) → REPITEMS(proper(X))
PROPER(repItems(X)) → PROPER(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
INCR(ok(X)) → INCR(X)
S(ok(X)) → S(X)
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)
ZIP(ok(X1), ok(X2)) → ZIP(X1, X2)
PAIR(ok(X1), ok(X2)) → PAIR(X1, X2)
TAIL(ok(X)) → TAIL(X)
REPITEMS(ok(X)) → REPITEMS(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 11 SCCs with 35 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

REPITEMS(ok(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(X)) → REPITEMS(X)

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

Lexicographic path order with status [LPO].
Precedence:

ok₁ > REPITEMS₁

Status:

ok₁: [1]
REPITEMS₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(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
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(X)) → TAIL(X)

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

Lexicographic path order with status [LPO].
Precedence:

ok₁ > TAIL₁

Status:

ok₁: [1]
TAIL₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(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
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(X1), ok(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(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) = PAIR(x1)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

mark₁ > PAIR₁

Status:

mark₁: [1]
PAIR₁: [1]

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(ok(X1), ok(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(X1), ok(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) = PAIR(x1)
mark(x1) = x1
ok(x1) = ok(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

ok₁: [1]
PAIR₁: [1]

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

PAIR(X1, mark(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(X1, mark(X2)) → PAIR(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > PAIR₂

Status:

PAIR₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(25) 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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE

(28) 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(ok(X1), ok(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(29) 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) = ZIP(x1)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

mark₁ > ZIP₁

Status:

mark₁: [1]
ZIP₁: [1]

The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

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

ZIP(X1, mark(X2)) → ZIP(X1, X2)
ZIP(ok(X1), ok(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(X1), ok(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) = ZIP(x1)
mark(x1) = x1
ok(x1) = ok(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

ok₁: [1]
ZIP₁: [1]

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)

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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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: Lexicographic path order with status [LPO].
Precedence:

mark₁ > ZIP₂

Status:

mark₁: [1]
ZIP₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(34) 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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) 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(ok(X1), ok(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(38) 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) = TAKE(x1)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

mark₁ > TAKE₁

Status:

TAKE₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(ok(X1), ok(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAKE(ok(X1), ok(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) = TAKE(x1)
mark(x1) = x1
ok(x1) = ok(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

TAKE₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(41) Obligation:

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

TAKE(X1, mark(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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, mark(X2)) → TAKE(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > TAKE₂

Status:

TAKE₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(43) 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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(44) PisEmptyProof (EQUIVALENT transformation)

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

(45) TRUE

(46) Obligation:

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

S(ok(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(ok(X)) → S(X)

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

Lexicographic path order with status [LPO].
Precedence:

ok₁ > S₁

Status:

ok₁: [1]
S₁: [1]

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(49) 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
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(50) 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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) TRUE

(53) Obligation:

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

INCR(ok(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INCR(ok(X)) → INCR(X)

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

Lexicographic path order with status [LPO].
Precedence:

ok₁ > INCR₁

Status:

ok₁: [1]
INCR₁: [1]

The following usable rules [FROCOS05] were oriented: none

(55) Obligation:

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

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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(56) 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
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(57) 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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(58) PisEmptyProof (EQUIVALENT transformation)

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

(59) TRUE

(60) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)
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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(61) 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) = CONS(x1)
ok(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

mark₁ > CONS₁

Status:

CONS₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(62) Obligation:

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

CONS(ok(X1), ok(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(ok(X1), ok(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) = CONS(x1)
ok(x1) = ok(x1)

Lexicographic path order with status [LPO].
Precedence:

ok₁ > CONS₁

Status:

CONS₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(64) 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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(65) PisEmptyProof (EQUIVALENT transformation)

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

(66) TRUE

(67) Obligation:

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

PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(zip(X1, X2)) → PROPER(X1)
PROPER(zip(X1, X2)) → PROPER(X2)
PROPER(pair(X1, X2)) → PROPER(X1)
PROPER(pair(X1, X2)) → PROPER(X2)
PROPER(tail(X)) → PROPER(X)
PROPER(repItems(X)) → PROPER(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(68) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(zip(X1, X2)) → PROPER(X1)
PROPER(zip(X1, X2)) → PROPER(X2)
PROPER(pair(X1, X2)) → PROPER(X1)
PROPER(pair(X1, X2)) → PROPER(X2)
PROPER(repItems(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
cons(x1, x2) = cons(x1, x2)
incr(x1) = x1
s(x1) = x1
take(x1, x2) = take(x1, x2)
zip(x1, x2) = zip(x1, x2)
pair(x1, x2) = pair(x1, x2)
tail(x1) = x1
repItems(x1) = repItems(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

zip₂: [1,2]
PROPER₁: [1]
cons₂: [1,2]
pair₂: [1,2]
take₂: [1,2]
repItems₁: [1]

The following usable rules [FROCOS05] were oriented: none

(69) Obligation:

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

PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(tail(X)) → PROPER(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(70) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(tail(X)) → PROPER(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

PROPER₁: [1]
tail₁: [1]

The following usable rules [FROCOS05] were oriented: none

(71) Obligation:

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

PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → PROPER(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(72) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(incr(X)) → PROPER(X)

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

Lexicographic path order with status [LPO].
Precedence:

incr₁ > PROPER₁

Status:

PROPER₁: [1]
incr₁: [1]

The following usable rules [FROCOS05] were oriented: none

(73) Obligation:

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

PROPER(s(X)) → PROPER(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(74) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(s(X)) → PROPER(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(75) 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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(76) PisEmptyProof (EQUIVALENT transformation)

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

(77) TRUE

(78) Obligation:

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

ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(zip(X1, X2)) → ACTIVE(X1)
ACTIVE(zip(X1, X2)) → ACTIVE(X2)
ACTIVE(pair(X1, X2)) → ACTIVE(X1)
ACTIVE(pair(X1, X2)) → ACTIVE(X2)
ACTIVE(tail(X)) → ACTIVE(X)
ACTIVE(repItems(X)) → ACTIVE(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(79) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(zip(X1, X2)) → ACTIVE(X1)
ACTIVE(zip(X1, X2)) → ACTIVE(X2)
ACTIVE(pair(X1, X2)) → ACTIVE(X1)
ACTIVE(pair(X1, X2)) → ACTIVE(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
incr(x1) = x1
cons(x1, x2) = x1
s(x1) = s(x1)
take(x1, x2) = take(x1, x2)
zip(x1, x2) = zip(x1, x2)
pair(x1, x2) = pair(x1, x2)
tail(x1) = x1
repItems(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

zip₂: [1,2]
pair₂: [1,2]
s₁: [1]
take₂: [1,2]
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented: none

(80) Obligation:

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

ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(tail(X)) → ACTIVE(X)
ACTIVE(repItems(X)) → ACTIVE(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(81) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(cons(X1, X2)) → ACTIVE(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
incr(x1) = x1
cons(x1, x2) = cons(x1, x2)
tail(x1) = x1
repItems(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

cons₂ > ACTIVE₁

Status:

cons₂: [1,2]
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented: none

(82) Obligation:

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

ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(tail(X)) → ACTIVE(X)
ACTIVE(repItems(X)) → ACTIVE(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(83) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(repItems(X)) → ACTIVE(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

repItems₁: [1]
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented: none

(84) Obligation:

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

ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(tail(X)) → ACTIVE(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(85) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(incr(X)) → ACTIVE(X)

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

Lexicographic path order with status [LPO].
Precedence:

incr₁ > ACTIVE₁

Status:

incr₁: [1]
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented: none

(86) Obligation:

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

ACTIVE(tail(X)) → ACTIVE(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(87) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(tail(X)) → ACTIVE(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

tail₁: [1]

The following usable rules [FROCOS05] were oriented: none

(88) 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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(89) PisEmptyProof (EQUIVALENT transformation)

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

(90) TRUE

(91) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(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))))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(zip(X1, X2)) → zip(active(X1), X2)
active(zip(X1, X2)) → zip(X1, active(X2))
active(pair(X1, X2)) → pair(active(X1), X2)
active(pair(X1, X2)) → pair(X1, active(X2))
active(tail(X)) → tail(active(X))
active(repItems(X)) → repItems(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
zip(mark(X1), X2) → mark(zip(X1, X2))
zip(X1, mark(X2)) → mark(zip(X1, X2))
pair(mark(X1), X2) → mark(pair(X1, X2))
pair(X1, mark(X2)) → mark(pair(X1, X2))
tail(mark(X)) → mark(tail(X))
repItems(mark(X)) → mark(repItems(X))
proper(pairNs) → ok(pairNs)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(oddNs) → ok(oddNs)
proper(s(X)) → s(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(zip(X1, X2)) → zip(proper(X1), proper(X2))
proper(pair(X1, X2)) → pair(proper(X1), proper(X2))
proper(tail(X)) → tail(proper(X))
proper(repItems(X)) → repItems(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
zip(ok(X1), ok(X2)) → ok(zip(X1, X2))
pair(ok(X1), ok(X2)) → ok(pair(X1, X2))
tail(ok(X)) → ok(tail(X))
repItems(ok(X)) → ok(repItems(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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