Termination w.r.t. Q proof of TRS/TRCSREx1_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(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)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(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)))
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)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(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:

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

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, 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)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(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:

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

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, 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)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(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 3 SCCs with 11 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, 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)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(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 oriented strictly and are deleted.

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PLUS₂(x₁, x₂)) = 3·x₁
POL(s₁(x₁)) = 1 + x₁

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, 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)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(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

            ↳ 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, 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)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(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 oriented strictly and are deleted.

TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(TIMES₂(x₁, x₂)) = 3·x₁
POL(s₁(x₁)) = 1 + x₁

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, 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)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(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

            ↳ QDPOrderProof

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(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)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(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 oriented strictly and are deleted.

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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(2NDSNEG₂(x₁, x₂)) = 3·x₁ + 3·x₂
POL(2NDSPOS₂(x₁, x₂)) = 2 + 3·x₁ + 3·x₂
POL(activate₁(x₁)) = 3 + x₁
POL(cons₂(x₁, x₂)) = 3 + 3·x₂
POL(from₁(x₁)) = 3
POL(n__cons₂(x₁, x₂)) = 1 + 3·x₂
POL(n__from₁(x₁)) = 0
POL(s₁(x₁)) = 3 + x₁

The following usable rules [14] were oriented:

activate₁(X) -> X
from₁(X) -> n__from₁(X)
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
activate₁(n__from₁(X)) -> from₁(X)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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)))
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)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(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.