Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/TRCSRSLASHEx7_BLR02

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(X) -> X

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(X1, X2)
TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
2ND₁(cons₂(X, XS)) -> HEAD₁(activate₁(XS))
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
2ND₁(cons₂(X, XS)) -> ACTIVATE₁(XS)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(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:

ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(X1, X2)
TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
2ND₁(cons₂(X, XS)) -> HEAD₁(activate₁(XS))
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
2ND₁(cons₂(X, XS)) -> ACTIVATE₁(XS)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(X1, X2)
TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(X1, X2)

The remaining pairs can at least by weakly be oriented.

TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)

Used ordering: Combined order from the following AFS and order.
ACTIVATE₁(x1) = ACTIVATE₁(x1)
n__take₂(x1, x2) = n__take₂(x1, x2)
TAKE₂(x1, x2) = TAKE₁(x2)
s₁(x1) = s
cons₂(x1, x2) = x2

Lexicographic Path Order [19].
Precedence:

s > [ACTIVATE₁, n_{_take₂}, TAKE_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SEL₂(x1, x2) = x1
s₁(x1) = s₁(x1)
cons₂(x1, x2) = x1
activate₁(x1) = activate₁(x1)
n__from₁(x1) = n__from₁(x1)
from₁(x1) = from₁(x1)
n__take₂(x1, x2) = n__take₁(x2)
take₂(x1, x2) = take₁(x2)
0 = 0
nil = nil

Lexicographic Path Order [19].
Precedence:

s₁ > [n_{_from₁}, from_1]
[n_{_take₁}, take_1] > activate₁ > [n_{_from₁}, from_1]
[n_{_take₁}, take_1] > [0, nil] > [n_{_from₁}, from_1]

The following usable rules [14] were oriented:

activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(X) -> X
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
take₂(X1, X2) -> n__take₂(X1, X2)
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
from₁(X) -> n__from₁(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.