Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/TRCSRSLASHExAppendixB_AEL03

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)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, activate₁(Z)))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(X)
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)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, activate₁(Z)))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(X)
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:

2NDSNEG₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSPOS₂(N, activate₁(Z))
2NDSPOS₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> ACTIVATE₁(Z)
2NDSNEG₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> ACTIVATE₁(Z)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
SQUARE₁(X) -> TIMES₂(X, X)
PI₁(X) -> FROM₁(0)
2NDSPOS₂(s₁(N), cons₂(X, Z)) -> 2NDSPOS₂(s₁(N), cons2₂(X, activate₁(Z)))
PI₁(X) -> 2NDSPOS₂(X, from₁(0))
PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)
TIMES₂(s₁(X), Y) -> PLUS₂(Y, times₂(X, Y))
TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)
2NDSPOS₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSNEG₂(N, activate₁(Z))
2NDSNEG₂(s₁(N), cons₂(X, Z)) -> 2NDSNEG₂(s₁(N), cons2₂(X, activate₁(Z)))
2NDSNEG₂(s₁(N), cons₂(X, Z)) -> ACTIVATE₁(Z)
2NDSPOS₂(s₁(N), cons₂(X, Z)) -> ACTIVATE₁(Z)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, activate₁(Z)))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(X)
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:

2NDSNEG₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSPOS₂(N, activate₁(Z))
2NDSPOS₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> ACTIVATE₁(Z)
2NDSNEG₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> ACTIVATE₁(Z)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
SQUARE₁(X) -> TIMES₂(X, X)
PI₁(X) -> FROM₁(0)
2NDSPOS₂(s₁(N), cons₂(X, Z)) -> 2NDSPOS₂(s₁(N), cons2₂(X, activate₁(Z)))
PI₁(X) -> 2NDSPOS₂(X, from₁(0))
PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)
TIMES₂(s₁(X), Y) -> PLUS₂(Y, times₂(X, Y))
TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)
2NDSPOS₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSNEG₂(N, activate₁(Z))
2NDSNEG₂(s₁(N), cons₂(X, Z)) -> 2NDSNEG₂(s₁(N), cons2₂(X, activate₁(Z)))
2NDSNEG₂(s₁(N), cons₂(X, Z)) -> ACTIVATE₁(Z)
2NDSPOS₂(s₁(N), cons₂(X, Z)) -> ACTIVATE₁(Z)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, activate₁(Z)))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(X)
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 3 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, activate₁(Z)))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(X)
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.

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PLUS₂(x1, x2) = PLUS₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[PLUS₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, activate₁(Z)))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(X)
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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, activate₁(Z)))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(X)
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.

TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TIMES₂(x1, x2) = TIMES₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[TIMES₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, activate₁(Z)))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(X)
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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

2NDSNEG₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSPOS₂(N, activate₁(Z))
2NDSNEG₂(s₁(N), cons₂(X, Z)) -> 2NDSNEG₂(s₁(N), cons2₂(X, activate₁(Z)))
2NDSPOS₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSNEG₂(N, activate₁(Z))
2NDSPOS₂(s₁(N), cons₂(X, Z)) -> 2NDSPOS₂(s₁(N), cons2₂(X, activate₁(Z)))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, activate₁(Z)))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(X)
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.

2NDSNEG₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSPOS₂(N, activate₁(Z))
2NDSPOS₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSNEG₂(N, activate₁(Z))

The remaining pairs can at least by weakly be oriented.

2NDSNEG₂(s₁(N), cons₂(X, Z)) -> 2NDSNEG₂(s₁(N), cons2₂(X, activate₁(Z)))
2NDSPOS₂(s₁(N), cons₂(X, Z)) -> 2NDSPOS₂(s₁(N), cons2₂(X, activate₁(Z)))

Used ordering: Combined order from the following AFS and order.
2NDSNEG₂(x1, x2) = 2NDSNEG₁(x1)
s₁(x1) = s₁(x1)
cons2₂(x1, x2) = cons2₁(x1)
cons₂(x1, x2) = cons
2NDSPOS₂(x1, x2) = 2NDSPOS₁(x1)
activate₁(x1) = activate
from₁(x1) = x1
n__from₁(x1) = n__from₁(x1)

Lexicographic Path Order [19].
Precedence:

[2NDSNEG₁, 2NDSPOS_1] > [s₁, cons2₁, cons, activate]
n_{_from₁} > [s₁, cons2₁, cons, activate]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

2NDSNEG₂(s₁(N), cons₂(X, Z)) -> 2NDSNEG₂(s₁(N), cons2₂(X, activate₁(Z)))
2NDSPOS₂(s₁(N), cons₂(X, Z)) -> 2NDSPOS₂(s₁(N), cons2₂(X, activate₁(Z)))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, activate₁(Z)))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, activate₁(Z)))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(X)
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 2 less nodes.