Termination w.r.t. Q proof of /tmp/tmpC1khZ8/Ex1_2

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

↳ QTRS

  ↳ DependencyPairsProof

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

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

TIMES(s(X), Y) → PLUS(Y, times(X, Y))
PI(X) → 2NDSPOS(X, from(0))
PI(X) → FROM(0)
2NDSNEG(s(N), cons(X, cons(Y, Z))) → 2NDSPOS(N, Z)
TIMES(s(X), Y) → TIMES(X, Y)
SQUARE(X) → TIMES(X, X)
2NDSPOS(s(N), cons(X, cons(Y, Z))) → 2NDSNEG(N, Z)
FROM(X) → FROM(s(X))
PLUS(s(X), Y) → PLUS(X, Y)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

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:

TIMES(s(X), Y) → PLUS(Y, times(X, Y))
PI(X) → 2NDSPOS(X, from(0))
PI(X) → FROM(0)
2NDSNEG(s(N), cons(X, cons(Y, Z))) → 2NDSPOS(N, Z)
TIMES(s(X), Y) → TIMES(X, Y)
SQUARE(X) → TIMES(X, X)
2NDSPOS(s(N), cons(X, cons(Y, Z))) → 2NDSNEG(N, Z)
FROM(X) → FROM(s(X))
PLUS(s(X), Y) → PLUS(X, Y)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(s(X), Y) → PLUS(X, Y)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(s(X), Y) → PLUS(X, Y)

R is empty.
The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(s(X), Y) → PLUS(X, Y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

PLUS(s(X), Y) → PLUS(X, Y)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

TIMES(s(X), Y) → TIMES(X, Y)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

TIMES(s(X), Y) → TIMES(X, Y)

R is empty.
The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

TIMES(s(X), Y) → TIMES(X, Y)

From the DPs we obtained the following set of size-change graphs:

TIMES(s(X), Y) → TIMES(X, Y)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

2NDSNEG(s(N), cons(X, cons(Y, Z))) → 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, cons(Y, Z))) → 2NDSNEG(N, Z)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

2NDSNEG(s(N), cons(X, cons(Y, Z))) → 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, cons(Y, Z))) → 2NDSNEG(N, Z)

R is empty.
The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

2NDSNEG(s(N), cons(X, cons(Y, Z))) → 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, cons(Y, Z))) → 2NDSNEG(N, Z)

From the DPs we obtained the following set of size-change graphs:

2NDSPOS(s(N), cons(X, cons(Y, Z))) → 2NDSNEG(N, Z)
The graph contains the following edges 1 > 1, 2 > 2
2NDSNEG(s(N), cons(X, cons(Y, Z))) → 2NDSPOS(N, Z)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)

The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

FROM(X) → FROM(s(X))

R is empty.
The set Q consists of the following terms:

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

2ndspos(0, x0)
from(x0)
plus(s(x0), x1)
times(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
times(s(x0), x1)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
square(x0)
plus(0, x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

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

FROM(X) → FROM(s(X))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule FROM(X) → FROM(s(X)) we obtained the following new rules:

FROM(s(z0)) → FROM(s(s(z0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

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

FROM(s(z0)) → FROM(s(s(z0)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule FROM(s(z0)) → FROM(s(s(z0))) we obtained the following new rules:

FROM(s(s(z0))) → FROM(s(s(s(z0))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

                          ↳ QDP

                            ↳ NonTerminationProof

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

FROM(s(s(z0))) → FROM(s(s(s(z0))))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

FROM(s(s(z0))) → FROM(s(s(s(z0))))

The TRS R consists of the following rules:none

s = FROM(s(s(z0))) evaluates to t =FROM(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [z0 / s(z0)]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from FROM(s(s(z0))) to FROM(s(s(s(z0)))).