Termination w.r.t. Q proof of TRS/nontermin/CSRExAppendixB

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, 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)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, 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)

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:

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

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, 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)

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:

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

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, 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)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, 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)

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)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, 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)

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

          ↳ 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, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, 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)

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)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, 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)

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

          ↳ QDP

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

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

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, 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)

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(2NDSNEG₂(x₁, x₂)) = 2 + 2·x₁
POL(2NDSPOS₂(x₁, x₂)) = 1 + 2·x₁
POL(cons₂(x₁, x₂)) = 2·x₁
POL(cons2₂(x₁, x₂)) = 1 + 2·x₁
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, 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)

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.