Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/TRCSRSLASHEx1_2_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(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(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(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(X) -> X

Q is empty.

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

ACTIVATE₁(n__cons₂(X1, X2)) -> ACTIVATE₁(X1)
2NDSPOS₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> ACTIVATE₁(Z)
PLUS₂(s₁(X), Y) -> S₁(plus₂(X, Y))
2NDSNEG₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> ACTIVATE₁(Y)
SQUARE₁(X) -> TIMES₂(X, X)
PI₁(X) -> FROM₁(0)
2NDSNEG₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> ACTIVATE₁(Z)
2NDSPOS₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> ACTIVATE₁(Y)
ACTIVATE₁(n__from₁(X)) -> FROM₁(activate₁(X))
PI₁(X) -> 2NDSPOS₂(X, from₁(0))
PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)
TIMES₂(s₁(X), Y) -> PLUS₂(Y, times₂(X, Y))
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
2NDSPOS₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> 2NDSNEG₂(N, activate₁(Z))
TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)
FROM₁(X) -> CONS₂(X, n__from₁(n__s₁(X)))
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
2NDSNEG₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> 2NDSPOS₂(N, activate₁(Z))
ACTIVATE₁(n__cons₂(X1, X2)) -> CONS₂(activate₁(X1), X2)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(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__cons₂(X1, X2)) -> ACTIVATE₁(X1)
2NDSPOS₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> ACTIVATE₁(Z)
PLUS₂(s₁(X), Y) -> S₁(plus₂(X, Y))
2NDSNEG₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> ACTIVATE₁(Y)
SQUARE₁(X) -> TIMES₂(X, X)
PI₁(X) -> FROM₁(0)
2NDSNEG₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> ACTIVATE₁(Z)
2NDSPOS₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> ACTIVATE₁(Y)
ACTIVATE₁(n__from₁(X)) -> FROM₁(activate₁(X))
PI₁(X) -> 2NDSPOS₂(X, from₁(0))
PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)
TIMES₂(s₁(X), Y) -> PLUS₂(Y, times₂(X, Y))
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
2NDSPOS₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> 2NDSNEG₂(N, activate₁(Z))
TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)
FROM₁(X) -> CONS₂(X, n__from₁(n__s₁(X)))
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
2NDSNEG₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> 2NDSPOS₂(N, activate₁(Z))
ACTIVATE₁(n__cons₂(X1, X2)) -> CONS₂(activate₁(X1), X2)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(X) -> X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ 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(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)

Used argument filtering: PLUS₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ 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(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)

Used argument filtering: TIMES₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVATE₁(n__cons₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

Used argument filtering: ACTIVATE₁(x1) = x1
n__cons₂(x1, x2) = x1
n__from₁(x1) = x1
n__s₁(x1) = n__s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVATE₁(n__cons₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)

Used argument filtering: ACTIVATE₁(x1) = x1
n__cons₂(x1, x2) = x1
n__from₁(x1) = n__from₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVATE₁(n__cons₂(X1, X2)) -> ACTIVATE₁(X1)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE₁(n__cons₂(X1, X2)) -> ACTIVATE₁(X1)

Used argument filtering: ACTIVATE₁(x1) = x1
n__cons₂(x1, x2) = n__cons₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

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

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

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

2NDSPOS₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> 2NDSNEG₂(N, activate₁(Z))
2NDSNEG₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> 2NDSPOS₂(N, activate₁(Z))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

2NDSPOS₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> 2NDSNEG₂(N, activate₁(Z))
2NDSNEG₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> 2NDSPOS₂(N, activate₁(Z))

Used argument filtering: 2NDSPOS₂(x1, x2) = x1
s₁(x1) = s₁(x1)
2NDSNEG₂(x1, x2) = x1
activate₁(x1) = x1
n__from₁(x1) = n__from
from₁(x1) = from
n__s₁(x1) = n__s₁(x1)
n__cons₂(x1, x2) = n__cons
cons₂(x1, x2) = cons
Used ordering: Quasi Precedence: [s_1, n__s_1] [n__from, from] > [n__cons, cons]

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

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

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(posrecip₁(activate₁(Y)), 2ndsneg₂(N, activate₁(Z)))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, n__cons₂(Y, Z))) -> rcons₂(negrecip₁(activate₁(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)
s₁(X) -> n__s₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(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.