Termination w.r.t. Q proof of /tmp/tmpj0r_sf/Ex5_7

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:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

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

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

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

DBLS(cons(X, Y)) → DBLS(Y)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
DBLS(cons(X, Y)) → DBL(X)
QUOTE(dbl(X)) → DBL1(X)
DBL(s(X)) → DBL(X)
QUOTE(sel(X, Y)) → SEL1(X, Y)
DBL1(s(X)) → DBL1(X)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
FROM(X) → FROM(s(X))
QUOTE(s(X)) → QUOTE(X)
INDX(cons(X, Y), Z) → INDX(Y, Z)
INDX(cons(X, Y), Z) → SEL(X, Z)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

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:

DBLS(cons(X, Y)) → DBLS(Y)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
DBLS(cons(X, Y)) → DBL(X)
QUOTE(dbl(X)) → DBL1(X)
DBL(s(X)) → DBL(X)
QUOTE(sel(X, Y)) → SEL1(X, Y)
DBL1(s(X)) → DBL1(X)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
FROM(X) → FROM(s(X))
QUOTE(s(X)) → QUOTE(X)
INDX(cons(X, Y), Z) → INDX(Y, Z)
INDX(cons(X, Y), Z) → SEL(X, Z)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 8 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)

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

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

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.

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)

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:

SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DBL1(s(X)) → DBL1(X)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DBL1(s(X)) → DBL1(X)

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

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

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.

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DBL1(s(X)) → DBL1(X)

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

DBL1(s(X)) → DBL1(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

QUOTE(s(X)) → QUOTE(X)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

QUOTE(s(X)) → QUOTE(X)

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

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

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.

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

QUOTE(s(X)) → QUOTE(X)

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

QUOTE(s(X)) → QUOTE(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

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.

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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)))).

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

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

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

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.

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

INDX(cons(X, Y), Z) → INDX(Y, Z)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

INDX(cons(X, Y), Z) → INDX(Y, Z)

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

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

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.

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

INDX(cons(X, Y), Z) → INDX(Y, Z)

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

INDX(cons(X, Y), Z) → INDX(Y, Z)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

DBL(s(X)) → DBL(X)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

DBL(s(X)) → DBL(X)

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

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

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.

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

DBL(s(X)) → DBL(X)

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

DBL(s(X)) → DBL(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

DBLS(cons(X, Y)) → DBLS(Y)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

The set Q consists of the following terms:

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

DBLS(cons(X, Y)) → DBLS(Y)

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

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

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.

dbls(nil)
quote(dbl(x0))
dbls(cons(x0, x1))
sel1(0, cons(x0, x1))
sel(0, cons(x0, x1))
indx(nil, x0)
sel1(s(x0), cons(x1, x2))
quote(s(x0))
from(x0)
dbl(0)
dbl(s(x0))
dbl1(0)
indx(cons(x0, x1), x2)
sel(s(x0), cons(x1, x2))
quote(sel(x0, x1))
dbl1(s(x0))
quote(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

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

DBLS(cons(X, Y)) → DBLS(Y)

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

DBLS(cons(X, Y)) → DBLS(Y)
The graph contains the following edges 1 > 1