Termination w.r.t. Q proof of /tmp/tmpAE3diC/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 DependencyGraphProof (⇔)

    ↳9 QDP

  ↳10 QDP

(0) Obligation:

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

pairNs → cons(0, n__incr(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)
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(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → 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:

PAIRNS → CONS(0, n__incr(oddNs))
PAIRNS → 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)
TAIL(cons(X, XS)) → ACTIVATE(XS)
REPITEMS(cons(X, XS)) → CONS(X, n__cons(X, n__repItems(activate(XS))))
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
ACTIVATE(n__zip(X1, X2)) → ZIP(X1, X2)
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__repItems(X)) → REPITEMS(X)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(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)
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(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → 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 2 SCCs with 8 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(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)
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(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → 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.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INCR(x1) = x1
cons(x1, x2) = x2
ACTIVATE(x1) = x1
n__incr(x1) = x1
n__take(x1, x2) = n__take(x1, x2)
TAKE(x1, x2) = TAKE(x1, x2)
s(x1) = s(x1)
n__zip(x1, x2) = n__zip(x1, x2)
ZIP(x1, x2) = ZIP(x1, x2)
n__repItems(x1) = n__repItems(x1)
REPITEMS(x1) = REPITEMS(x1)
pairNs = pairNs
0 = 0
oddNs = oddNs
incr(x1) = x1
activate(x1) = x1
take(x1, x2) = take(x1, x2)
nil = nil
zip(x1, x2) = zip(x1, x2)
pair(x1, x2) = pair(x2)
tail(x1) = tail(x1)
repItems(x1) = repItems(x1)
n__cons(x1, x2) = x2

Lexicographic Path Order [LPO].
Precedence:

[n_{_take₂}, TAKE₂, take₂] > [0, nil]
s₁ > [0, nil]
[n_{_zip₂}, zip₂] > ZIP₂ > [0, nil]
[n_{_zip₂}, zip₂] > pair₁ > [0, nil]
[n_{_repItems₁}, repItems₁] > REPITEMS₁ > [0, nil]
[pairNs, oddNs] > [0, nil]
tail₁ > [0, nil]

The following usable rules [FROCOS05] were oriented:

pairNs → cons(0, n__incr(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)
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(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

(7) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(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)
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(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

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

(8) DependencyGraphProof (EQUIVALENT transformation)

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

(9) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(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)
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(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

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

(10) Obligation:

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

PAIRNS → ODDNS
ODDNS → PAIRNS

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(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)
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(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

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