Termination w.r.t. Q proof of TRS/TRCSRExAppendixB_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 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 Order [17,21] with Interpretation:

POL( s₁(x₁) ) = 2x₁ + 2

POL( PLUS₂(x₁, x₂) ) = max{0, 2x₁ + x₂ - 2}

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 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 Order [17,21] with Interpretation:

POL( s₁(x₁) ) = 2x₁ + 2

POL( TIMES₂(x₁, x₂) ) = max{0, 2x₁ + x₂ - 2}

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 oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from₁(x₁) ) = 1

POL( n__from₁(x₁) ) = max{0, -2}

POL( 2NDSNEG₂(x₁, x₂) ) = 2x₁ + x₂

POL( 2NDSPOS₂(x₁, x₂) ) = max{0, 2x₁ + x₂ - 1}

POL( s₁(x₁) ) = 2x₁ + 2

POL( cons2₂(x₁, x₂) ) = max{0, x₂ - 2}

POL( cons₂(x₁, x₂) ) = x₂ + 1

POL( activate₁(x₁) ) = x₁ + 2

The following usable rules [14] were oriented:

activate₁(n__from₁(X)) -> from₁(X)
activate₁(X) -> X
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
from₁(X) -> n__from₁(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

2NDSPOS₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSNEG₂(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, 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 1 less node.