Termination w.r.t. Q proof of /tmp/tmpegREE6/Ex5_DLMMU04

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS

↳1 QTRSRRRProof (⇔)

↳2 QTRS

↳3 QTRSRRRProof (⇔)

↳4 QTRS

↳5 QTRSRRRProof (⇔)

↳6 QTRS

↳7 QTRSRRRProof (⇔)

↳8 QTRS

↳9 DependencyPairsProof (⇔)

↳10 QDP

↳11 DependencyGraphProof (⇔)

↳12 QDP

↳13 MRRProof (⇔)

↳14 QDP

↳15 DependencyGraphProof (⇔)

↳16 QDP

↳17 MRRProof (⇔)

↳18 QDP

↳19 MRRProof (⇔)

↳20 QDP

↳21 DependencyGraphProof (⇔)

↳22 QDP

↳23 QDPOrderProof (⇔)

↳24 QDP

↳25 QDPOrderProof (⇔)

↳26 QDP

↳27 DependencyGraphProof (⇔)

↳28 QDP

↳29 UsableRulesProof (⇔)

↳30 QDP

↳31 QDPSizeChangeProof (⇔)

↳32 TRUE

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(a__incr(x₁)) = x₁
POL(a__oddNs) = 0
POL(a__pairNs) = 0
POL(a__repItems(x₁)) = 2·x₁
POL(a__tail(x₁)) = x₁
POL(a__take(x₁, x₂)) = 2·x₁ + x₂
POL(a__zip(x₁, x₂)) = 2 + x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(oddNs) = 0
POL(pair(x₁, x₂)) = x₁ + x₂
POL(pairNs) = 0
POL(repItems(x₁)) = 2·x₁
POL(s(x₁)) = x₁
POL(tail(x₁)) = x₁
POL(take(x₁, x₂)) = 2·x₁ + x₂
POL(zip(x₁, x₂)) = 2 + x₁ + x₂

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__zip(nil, XS) → nil
a__zip(X, nil) → nil

(2) Obligation:

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

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(a__incr(x₁)) = 2·x₁
POL(a__oddNs) = 0
POL(a__pairNs) = 0
POL(a__repItems(x₁)) = 2·x₁
POL(a__tail(x₁)) = x₁
POL(a__take(x₁, x₂)) = 2 + x₁ + x₂
POL(a__zip(x₁, x₂)) = x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(oddNs) = 0
POL(pair(x₁, x₂)) = x₁ + x₂
POL(pairNs) = 0
POL(repItems(x₁)) = 2·x₁
POL(s(x₁)) = x₁
POL(tail(x₁)) = x₁
POL(take(x₁, x₂)) = 2 + x₁ + x₂
POL(zip(x₁, x₂)) = x₁ + x₂

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__take(0, XS) → nil

(4) Obligation:

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

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

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(a__incr(x₁)) = x₁
POL(a__oddNs) = 0
POL(a__pairNs) = 0
POL(a__repItems(x₁)) = 2·x₁
POL(a__tail(x₁)) = 2·x₁
POL(a__take(x₁, x₂)) = x₁ + x₂
POL(a__zip(x₁, x₂)) = x₁ + x₂
POL(cons(x₁, x₂)) = x₁ + x₂
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 2
POL(oddNs) = 0
POL(pair(x₁, x₂)) = x₁ + x₂
POL(pairNs) = 0
POL(repItems(x₁)) = 2·x₁
POL(s(x₁)) = x₁
POL(tail(x₁)) = 2·x₁
POL(take(x₁, x₂)) = x₁ + x₂
POL(zip(x₁, x₂)) = x₁ + x₂

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__repItems(nil) → nil

(6) Obligation:

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

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

Q is empty.

(7) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(a__incr(x₁)) = 2·x₁
POL(a__oddNs) = 0
POL(a__pairNs) = 0
POL(a__repItems(x₁)) = 2·x₁
POL(a__tail(x₁)) = 1 + x₁
POL(a__take(x₁, x₂)) = x₁ + x₂
POL(a__zip(x₁, x₂)) = x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(oddNs) = 0
POL(pair(x₁, x₂)) = x₁ + x₂
POL(pairNs) = 0
POL(repItems(x₁)) = 2·x₁
POL(s(x₁)) = 2·x₁
POL(tail(x₁)) = 1 + x₁
POL(take(x₁, x₂)) = x₁ + x₂
POL(zip(x₁, x₂)) = x₁ + x₂

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__tail(cons(X, XS)) → mark(XS)

(8) Obligation:

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

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

Q is empty.

(9) DependencyPairsProof (EQUIVALENT transformation)

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

(10) Obligation:

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

A__ODDNS → A__INCR(a__pairNs)
A__ODDNS → A__PAIRNS
A__INCR(cons(X, XS)) → MARK(X)
A__TAKE(s(N), cons(X, XS)) → MARK(X)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(pairNs) → A__PAIRNS
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → A__ODDNS
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → MARK(X2)
MARK(tail(X)) → A__TAIL(mark(X))
MARK(tail(X)) → MARK(X)
MARK(repItems(X)) → A__REPITEMS(mark(X))
MARK(repItems(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

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

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(12) Obligation:

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

A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__ODDNS → A__INCR(a__pairNs)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
A__TAKE(s(N), cons(X, XS)) → MARK(X)
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → MARK(X2)
MARK(tail(X)) → MARK(X)
MARK(repItems(X)) → A__REPITEMS(mark(X))
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(repItems(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)

The TRS R consists of the following rules:

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

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

(13) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
A__TAKE(s(N), cons(X, XS)) → MARK(X)
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → MARK(X2)
MARK(tail(X)) → MARK(X)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(A__INCR(x₁)) = x₁
POL(A__ODDNS) = 0
POL(A__REPITEMS(x₁)) = x₁
POL(A__TAKE(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(A__ZIP(x₁, x₂)) = x₁ + x₂
POL(MARK(x₁)) = 2·x₁
POL(a__incr(x₁)) = 2·x₁
POL(a__oddNs) = 0
POL(a__pairNs) = 0
POL(a__repItems(x₁)) = 2·x₁
POL(a__tail(x₁)) = 1 + x₁
POL(a__take(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(a__zip(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(oddNs) = 0
POL(pair(x₁, x₂)) = x₁ + 2·x₂
POL(pairNs) = 0
POL(repItems(x₁)) = 2·x₁
POL(s(x₁)) = x₁
POL(tail(x₁)) = 1 + x₁
POL(take(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(zip(x₁, x₂)) = 2 + x₁ + 2·x₂

(14) Obligation:

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

A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__ODDNS → A__INCR(a__pairNs)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(repItems(X)) → A__REPITEMS(mark(X))
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(repItems(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)

The TRS R consists of the following rules:

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

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(16) Obligation:

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

MARK(incr(X)) → A__INCR(mark(X))
A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__ODDNS → A__INCR(a__pairNs)
MARK(repItems(X)) → A__REPITEMS(mark(X))
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(repItems(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

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

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

(17) MRRProof (EQUIVALENT transformation)

MARK(repItems(X)) → A__REPITEMS(mark(X))
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(repItems(X)) → MARK(X)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(A__INCR(x₁)) = x₁
POL(A__ODDNS) = 0
POL(A__REPITEMS(x₁)) = 1 + 2·x₁
POL(MARK(x₁)) = x₁
POL(a__incr(x₁)) = 2·x₁
POL(a__oddNs) = 0
POL(a__pairNs) = 0
POL(a__repItems(x₁)) = 2 + 2·x₁
POL(a__tail(x₁)) = 2·x₁
POL(a__take(x₁, x₂)) = 2·x₁ + 2·x₂
POL(a__zip(x₁, x₂)) = x₁ + 2·x₂
POL(cons(x₁, x₂)) = x₁ + x₂
POL(incr(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(oddNs) = 0
POL(pair(x₁, x₂)) = x₁ + x₂
POL(pairNs) = 0
POL(repItems(x₁)) = 2 + 2·x₁
POL(s(x₁)) = 2·x₁
POL(tail(x₁)) = 2·x₁
POL(take(x₁, x₂)) = 2·x₁ + 2·x₂
POL(zip(x₁, x₂)) = x₁ + 2·x₂

(18) Obligation:

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

MARK(incr(X)) → A__INCR(mark(X))
A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__ODDNS → A__INCR(a__pairNs)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

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

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

(19) MRRProof (EQUIVALENT transformation)

A__ODDNS → A__INCR(a__pairNs)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(A__INCR(x₁)) = x₁
POL(A__ODDNS) = 2
POL(MARK(x₁)) = 2·x₁
POL(a__incr(x₁)) = x₁
POL(a__oddNs) = 1
POL(a__pairNs) = 1
POL(a__repItems(x₁)) = 2·x₁
POL(a__tail(x₁)) = x₁
POL(a__take(x₁, x₂)) = 2·x₁ + x₂
POL(a__zip(x₁, x₂)) = x₁ + 2·x₂
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(oddNs) = 1
POL(pair(x₁, x₂)) = x₁ + 2·x₂
POL(pairNs) = 1
POL(repItems(x₁)) = 2·x₁
POL(s(x₁)) = x₁
POL(tail(x₁)) = x₁
POL(take(x₁, x₂)) = 2·x₁ + x₂
POL(zip(x₁, x₂)) = x₁ + 2·x₂

(20) Obligation:

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

MARK(incr(X)) → A__INCR(mark(X))
A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(oddNs) → A__ODDNS
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

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

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

(21) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(22) Obligation:

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

A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

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

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(A__INCR(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(a__incr(x₁)) = x₁
POL(a__oddNs) = 1
POL(a__pairNs) = 1
POL(a__repItems(x₁)) = x₁
POL(a__tail(x₁)) = 0
POL(a__take(x₁, x₂)) = x₁ + x₂
POL(a__zip(x₁, x₂)) = 1 + x₁ + x₂
POL(cons(x₁, x₂)) = x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(oddNs) = 1
POL(pair(x₁, x₂)) = 1 + x₁ + x₂
POL(pairNs) = 1
POL(repItems(x₁)) = x₁
POL(s(x₁)) = x₁
POL(tail(x₁)) = 0
POL(take(x₁, x₂)) = x₁ + x₂
POL(zip(x₁, x₂)) = 1 + x₁ + x₂

The following usable rules [FROCOS05] were oriented:

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

(24) Obligation:

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

A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

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

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

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(A__INCR(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(a__incr(x₁)) = 1 + x₁
POL(a__oddNs) = 1
POL(a__pairNs) = 0
POL(a__repItems(x₁)) = x₁
POL(a__tail(x₁)) = x₁
POL(a__take(x₁, x₂)) = x₂
POL(a__zip(x₁, x₂)) = x₁
POL(cons(x₁, x₂)) = x₁
POL(incr(x₁)) = 1 + x₁
POL(mark(x₁)) = x₁
POL(nil) = 1
POL(oddNs) = 1
POL(pair(x₁, x₂)) = 0
POL(pairNs) = 0
POL(repItems(x₁)) = x₁
POL(s(x₁)) = 1 + x₁
POL(tail(x₁)) = x₁
POL(take(x₁, x₂)) = x₂
POL(zip(x₁, x₂)) = x₁

The following usable rules [FROCOS05] were oriented:

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

(26) Obligation:

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

A__INCR(cons(X, XS)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

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

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

(27) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(28) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

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

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

(29) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(30) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)

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

(31) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

MARK(cons(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1

(0) Obligation:

(1) QTRSRRRProof (EQUIVALENT transformation)

(2) Obligation:

(3) QTRSRRRProof (EQUIVALENT transformation)

(4) Obligation:

(5) QTRSRRRProof (EQUIVALENT transformation)

(6) Obligation:

(7) QTRSRRRProof (EQUIVALENT transformation)

(8) Obligation:

(9) DependencyPairsProof (EQUIVALENT transformation)

(10) Obligation:

(11) DependencyGraphProof (EQUIVALENT transformation)

(12) Obligation:

(13) MRRProof (EQUIVALENT transformation)

(14) Obligation:

(15) DependencyGraphProof (EQUIVALENT transformation)

(16) Obligation:

(17) MRRProof (EQUIVALENT transformation)

(18) Obligation:

(19) MRRProof (EQUIVALENT transformation)

(20) Obligation:

(21) DependencyGraphProof (EQUIVALENT transformation)

(22) Obligation:

(23) QDPOrderProof (EQUIVALENT transformation)

(24) Obligation:

(25) QDPOrderProof (EQUIVALENT transformation)

(26) Obligation:

(27) DependencyGraphProof (EQUIVALENT transformation)

(28) Obligation:

(29) UsableRulesProof (EQUIVALENT transformation)

(30) Obligation:

(31) QDPSizeChangeProof (EQUIVALENT transformation)

(32) TRUE