Termination w.r.t. Q proof of /tmp/tmp1Whwfn/Ex5_DLMMU04

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.

↳ QTRS

  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

TAKE(s(N), cons(X, XS)) → CONS(X, n__take(N, activate(XS)))
INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ODDNS → INCR(pairNs)
INCR(cons(X, XS)) → CONS(s(X), n__incr(activate(XS)))
ACTIVATE(n__repItems(X)) → REPITEMS(X)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
PAIRNS → ODDNS
PAIRNS → CONS(0, n__incr(oddNs))
ODDNS → PAIRNS
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
ZIP(cons(X, XS), cons(Y, YS)) → CONS(pair(X, Y), n__zip(activate(XS), activate(YS)))
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__zip(X1, X2)) → ZIP(X1, X2)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
REPITEMS(cons(X, XS)) → CONS(X, n__cons(X, n__repItems(activate(XS))))
TAIL(cons(X, XS)) → ACTIVATE(XS)
ZIP(cons(X, XS), cons(Y, YS)) → 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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 8 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ACTIVATE(n__zip(X1, X2)) → ZIP(X1, X2)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
ACTIVATE(n__repItems(X)) → REPITEMS(X)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
TAKE(s(N), 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.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)

Used ordering: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 1/4 + (3)x_2
POL(n__incr(x₁)) = 4 + (11/4)x_1
POL(TAKE(x₁, x₂)) = (3/4)x_1 + (2)x_2
POL(n__take(x₁, x₂)) = x_1 + (2)x_2
POL(ZIP(x₁, x₂)) = 1 + (7/4)x_1 + x_2
POL(n__repItems(x₁)) = 2 + (3/4)x_1
POL(s(x₁)) = 1/4
POL(n__zip(x₁, x₂)) = 3 + x_1 + (1/2)x_2
POL(REPITEMS(x₁)) = 7/2 + (3/2)x_1
POL(INCR(x₁)) = 4 + (3/4)x_1
POL(ACTIVATE(x₁)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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.