NO Termination w.r.t. Q proof of Transformed_CSR_04_Ex5_DLMMU04_FR.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 73 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 QDP
↳9 MRRProof (⇔, 13 ms)
↳10 QDP
↳11 DependencyGraphProof (⇔, 0 ms)
↳12 QDP
↳13 TransformationProof (⇔, 0 ms)
↳14 QDP
↳15 QDPOrderProof (⇔, 79 ms)
↳16 QDP
↳17 TransformationProof (⇔, 0 ms)
↳18 QDP
↳19 MRRProof (⇔, 0 ms)
↳20 QDP
↳21 MRRProof (⇔, 10 ms)
↳22 QDP
↳23 MRRProof (⇔, 0 ms)
↳24 QDP
↳25 NonTerminationLoopProof (⇒, 0 ms)
↳26 NO

(0) Obligation:

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 1 + 2·x1   
POL(n__take(x1, x2)) = x1 + x2   
POL(n__zip(x1, x2)) = 1 + x1 + x2   
POL(nil) = 0   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 1 + 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + 2·x1   
POL(take(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = 1 + x1 + x2   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

zip(nil, XS) → nil
zip(X, nil) → nil
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil


(2) Obligation:

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2 + 2·x1   
POL(n__take(x1, x2)) = 2 + 2·x1 + x2   
POL(n__zip(x1, x2)) = 2 + 2·x1 + x2   
POL(nil) = 1   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2 + 2·x1   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = 2 + 2·x1 + x2   
POL(zip(x1, x2)) = 2 + 2·x1 + x2   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

take(0, XS) → nil


(4) Obligation:

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

PAIRNS → CONS(0, n__incr(n__oddNs))
ODDNS → INCR(pairNs)
ODDNS → PAIRNS
INCR(cons(X, XS)) → CONS(s(X), n__incr(activate(XS)))
INCR(cons(X, XS)) → ACTIVATE(XS)
TAKE(s(N), cons(X, XS)) → CONS(X, n__take(N, activate(XS)))
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ZIP(cons(X, XS), cons(Y, YS)) → CONS(pair(X, Y), n__zip(activate(XS), activate(YS)))
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
REPITEMS(cons(X, XS)) → CONS(X, n__cons(X, n__repItems(activate(XS))))
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__zip(X1, X2)) → ZIP(activate(X1), activate(X2))
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__repItems(X)) → REPITEMS(activate(X))
ACTIVATE(n__repItems(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 7 less nodes.

(8) Obligation:

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

ODDNS → INCR(pairNs)
INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__zip(X1, X2)) → ZIP(activate(X1), activate(X2))
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__repItems(X)) → REPITEMS(activate(X))
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__repItems(X)) → ACTIVATE(X)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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

(9) 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:

ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__zip(X1, X2)) → ZIP(activate(X1), activate(X2))
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X2)
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__repItems(X)) → ACTIVATE(X)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVATE(x1)) = x1   
POL(INCR(x1)) = x1   
POL(ODDNS) = 0   
POL(REPITEMS(x1)) = 2 + x1   
POL(TAKE(x1, x2)) = 1 + 2·x1 + x2   
POL(ZIP(x1, x2)) = 1 + x1 + 2·x2   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2 + 2·x1   
POL(n__take(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(n__zip(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(oddNs) = 0   
POL(pair(x1, x2)) = 2·x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2 + 2·x1   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(zip(x1, x2)) = 2 + 2·x1 + 2·x2   

(10) Obligation:

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

ODDNS → INCR(pairNs)
INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__repItems(X)) → REPITEMS(activate(X))

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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 1 less node.

(12) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ODDNS → INCR(pairNs)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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

(13) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ODDNS → INCR(pairNs) at position [0] we obtained the following new rules [LPAR04]:

ODDNS → INCR(cons(0, n__incr(n__oddNs))) → ODDNS → INCR(cons(0, n__incr(n__oddNs)))

(14) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ODDNS → INCR(cons(0, n__incr(n__oddNs)))

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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,JAR06].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(INCR(x1)) = 2A + 3A·x1

POL(cons(x1, x2)) = 0A + 1A·x1 + 0A·x2

POL(ACTIVATE(x1)) = 2A + 3A·x1

POL(n__incr(x1)) = -I + 0A·x1

POL(activate(x1)) = -I + 0A·x1

POL(n__oddNs) = 1A

POL(ODDNS) = 4A

POL(n__cons(x1, x2)) = 0A + 1A·x1 + 0A·x2

POL(0) = 0A

POL(incr(x1)) = -I + 0A·x1

POL(oddNs) = 1A

POL(n__take(x1, x2)) = 4A + -I·x1 + 1A·x2

POL(take(x1, x2)) = 4A + -I·x1 + 1A·x2

POL(n__zip(x1, x2)) = -I + -I·x1 + 1A·x2

POL(zip(x1, x2)) = -I + -I·x1 + 1A·x2

POL(n__repItems(x1)) = 1A + 1A·x1

POL(repItems(x1)) = 1A + 1A·x1

POL(s(x1)) = -I + 0A·x1

POL(pairNs) = 1A

POL(pair(x1, x2)) = 0A + -I·x1 + -I·x2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X
cons(X1, X2) → n__cons(X1, X2)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
oddNs → n__oddNs
oddNs → incr(pairNs)
pairNs → cons(0, n__incr(n__oddNs))
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
repItems(X) → n__repItems(X)
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))

(16) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ODDNS → INCR(cons(0, n__incr(n__oddNs)))

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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

(17) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ACTIVATE(n__incr(X)) → INCR(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0))) → ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs) → ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs)
ACTIVATE(n__incr(n__take(x0, x1))) → INCR(take(activate(x0), activate(x1))) → ACTIVATE(n__incr(n__take(x0, x1))) → INCR(take(activate(x0), activate(x1)))
ACTIVATE(n__incr(n__zip(x0, x1))) → INCR(zip(activate(x0), activate(x1))) → ACTIVATE(n__incr(n__zip(x0, x1))) → INCR(zip(activate(x0), activate(x1)))
ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1)) → ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1))
ACTIVATE(n__incr(n__repItems(x0))) → INCR(repItems(activate(x0))) → ACTIVATE(n__incr(n__repItems(x0))) → INCR(repItems(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0) → ACTIVATE(n__incr(x0)) → INCR(x0)

(18) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ODDNS → INCR(cons(0, n__incr(n__oddNs)))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs)
ACTIVATE(n__incr(n__take(x0, x1))) → INCR(take(activate(x0), activate(x1)))
ACTIVATE(n__incr(n__zip(x0, x1))) → INCR(zip(activate(x0), activate(x1)))
ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1))
ACTIVATE(n__incr(n__repItems(x0))) → INCR(repItems(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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

(19) 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:

ACTIVATE(n__incr(n__zip(x0, x1))) → INCR(zip(activate(x0), activate(x1)))


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVATE(x1)) = 2 + 2·x1   
POL(INCR(x1)) = 2 + 2·x1   
POL(ODDNS) = 2   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2·x1   
POL(n__take(x1, x2)) = x1 + 2·x2   
POL(n__zip(x1, x2)) = 2 + 2·x1 + x2   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = x1 + 2·x2   
POL(zip(x1, x2)) = 2 + 2·x1 + x2   

(20) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ODDNS → INCR(cons(0, n__incr(n__oddNs)))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs)
ACTIVATE(n__incr(n__take(x0, x1))) → INCR(take(activate(x0), activate(x1)))
ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1))
ACTIVATE(n__incr(n__repItems(x0))) → INCR(repItems(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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

(21) 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:

ACTIVATE(n__incr(n__take(x0, x1))) → INCR(take(activate(x0), activate(x1)))


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVATE(x1)) = x1   
POL(INCR(x1)) = x1   
POL(ODDNS) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2·x1   
POL(n__take(x1, x2)) = 2 + 2·x1 + x2   
POL(n__zip(x1, x2)) = 2 + x1 + x2   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = 2 + 2·x1 + x2   
POL(zip(x1, x2)) = 2 + x1 + x2   

(22) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ODDNS → INCR(cons(0, n__incr(n__oddNs)))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs)
ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1))
ACTIVATE(n__incr(n__repItems(x0))) → INCR(repItems(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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

(23) 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:

ACTIVATE(n__incr(n__repItems(x0))) → INCR(repItems(activate(x0)))


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVATE(x1)) = 1 + x1   
POL(INCR(x1)) = 1 + x1   
POL(ODDNS) = 1   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 1 + 2·x1   
POL(n__take(x1, x2)) = 2·x1 + 2·x2   
POL(n__zip(x1, x2)) = 2 + 2·x1 + x2   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 1 + 2·x1   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = 2·x1 + 2·x2   
POL(zip(x1, x2)) = 2 + 2·x1 + x2   

(24) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ODDNS → INCR(cons(0, n__incr(n__oddNs)))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs)
ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1))
ACTIVATE(n__incr(x0)) → INCR(x0)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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

(25) NonTerminationLoopProof (COMPLETE transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = ACTIVATE(n__incr(n__oddNs)) evaluates to t =ACTIVATE(n__incr(n__oddNs))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

ACTIVATE(n__incr(n__oddNs)) → ACTIVATE(n__oddNs)
with rule ACTIVATE(n__incr(X)) → ACTIVATE(X) at position [] and matcher [X / n__oddNs]

ACTIVATE(n__oddNs) → ODDNS
with rule ACTIVATE(n__oddNs) → ODDNS at position [] and matcher [ ]

ODDNS → INCR(cons(0, n__incr(n__oddNs)))
with rule ODDNS → INCR(cons(0, n__incr(n__oddNs))) at position [] and matcher [ ]

INCR(cons(0, n__incr(n__oddNs))) → ACTIVATE(n__incr(n__oddNs))
with rule INCR(cons(X, XS)) → ACTIVATE(XS)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(26) NO