Termination proof of ../tpdb/TRS/CSR/Ex1_2

YES Termination proof of ../tpdb/TRS/CSR/Ex1_2_AEL03.trs
Termination of the following Term Rewriting System could be proven:

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 replacement map contains the following entries:

from: {1}
cons: {1}
s: {1}
2ndspos: {1, 2}
0: empty set
rnil: empty set
rcons: {1, 2}
posrecip: {1}
2ndsneg: {1, 2}
negrecip: {1}
pi: {1}
plus: {1, 2}
times: {1, 2}
square: {1}

↳ CSR

  ↳ Zantema-Transformation

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 replacement map contains the following entries:

from: {1}
cons: {1}
s: {1}
2ndspos: {1, 2}
0: empty set
rnil: empty set
rcons: {1, 2}
posrecip: {1}
2ndsneg: {1, 2}
negrecip: {1}
pi: {1}
plus: {1, 2}
times: {1, 2}
square: {1}

We applied the Zantema [34] to transform the context-sensitive TRS to an usual TRS.

↳ CSR

  ↳ Zantema-Transformation

    ↳ QTRS

      ↳ DependencyPairsProof

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

from(X) → cons(X, fromInact(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, consInact(Y, Z))) → rcons(posrecip(a(Y)), 2ndsneg(N, a(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, consInact(Y, Z))) → rcons(negrecip(a(Y)), 2ndspos(N, a(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)
a(x) → x
from(x1) → fromInact(x1)
a(fromInact(x1)) → from(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

Q is empty.

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

2NDSNEG(s(N), cons(X, consInact(Y, Z))) → A(Z)
2NDSPOS(s(N), cons(X, consInact(Y, Z))) → A(Z)
2NDSPOS(s(N), cons(X, consInact(Y, Z))) → A(Y)
2NDSPOS(s(N), cons(X, consInact(Y, Z))) → 2NDSNEG(N, a(Z))
PI(X) → FROM(0)
SQUARE(X) → TIMES(X, X)
PI(X) → 2NDSPOS(X, from(0))
PLUS(s(X), Y) → PLUS(X, Y)
TIMES(s(X), Y) → PLUS(Y, times(X, Y))
A(fromInact(x1)) → FROM(x1)
2NDSNEG(s(N), cons(X, consInact(Y, Z))) → A(Y)
2NDSNEG(s(N), cons(X, consInact(Y, Z))) → 2NDSPOS(N, a(Z))
A(consInact(x1, x2)) → CONS(x1, x2)
TIMES(s(X), Y) → TIMES(X, Y)
FROM(X) → CONS(X, fromInact(s(X)))

The TRS R consists of the following rules:

from(X) → cons(X, fromInact(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, consInact(Y, Z))) → rcons(posrecip(a(Y)), 2ndsneg(N, a(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, consInact(Y, Z))) → rcons(negrecip(a(Y)), 2ndspos(N, a(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)
a(x) → x
from(x1) → fromInact(x1)
a(fromInact(x1)) → from(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

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

↳ CSR

  ↳ Zantema-Transformation

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

2NDSNEG(s(N), cons(X, consInact(Y, Z))) → A(Z)
2NDSPOS(s(N), cons(X, consInact(Y, Z))) → A(Z)
2NDSPOS(s(N), cons(X, consInact(Y, Z))) → A(Y)
2NDSPOS(s(N), cons(X, consInact(Y, Z))) → 2NDSNEG(N, a(Z))
PI(X) → FROM(0)
SQUARE(X) → TIMES(X, X)
PI(X) → 2NDSPOS(X, from(0))
PLUS(s(X), Y) → PLUS(X, Y)
TIMES(s(X), Y) → PLUS(Y, times(X, Y))
A(fromInact(x1)) → FROM(x1)
2NDSNEG(s(N), cons(X, consInact(Y, Z))) → A(Y)
2NDSNEG(s(N), cons(X, consInact(Y, Z))) → 2NDSPOS(N, a(Z))
A(consInact(x1, x2)) → CONS(x1, x2)
TIMES(s(X), Y) → TIMES(X, Y)
FROM(X) → CONS(X, fromInact(s(X)))

The TRS R consists of the following rules:

from(X) → cons(X, fromInact(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, consInact(Y, Z))) → rcons(posrecip(a(Y)), 2ndsneg(N, a(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, consInact(Y, Z))) → rcons(negrecip(a(Y)), 2ndspos(N, a(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)
a(x) → x
from(x1) → fromInact(x1)
a(fromInact(x1)) → from(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 11 less nodes.

↳ CSR

  ↳ Zantema-Transformation

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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, fromInact(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, consInact(Y, Z))) → rcons(posrecip(a(Y)), 2ndsneg(N, a(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, consInact(Y, Z))) → rcons(negrecip(a(Y)), 2ndspos(N, a(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)
a(x) → x
from(x1) → fromInact(x1)
a(fromInact(x1)) → from(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ CSR

  ↳ Zantema-Transformation

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

              ↳ 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

↳ CSR

  ↳ Zantema-Transformation

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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, fromInact(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, consInact(Y, Z))) → rcons(posrecip(a(Y)), 2ndsneg(N, a(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, consInact(Y, Z))) → rcons(negrecip(a(Y)), 2ndspos(N, a(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)
a(x) → x
from(x1) → fromInact(x1)
a(fromInact(x1)) → from(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

↳ CSR

  ↳ Zantema-Transformation

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

              ↳ 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

↳ CSR

  ↳ Zantema-Transformation

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPSizeChangeProof

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

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

The TRS R consists of the following rules:

from(X) → cons(X, fromInact(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, consInact(Y, Z))) → rcons(posrecip(a(Y)), 2ndsneg(N, a(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, consInact(Y, Z))) → rcons(negrecip(a(Y)), 2ndspos(N, a(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)
a(x) → x
from(x1) → fromInact(x1)
a(fromInact(x1)) → from(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

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:

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