Termination w.r.t. Q proof of TRS/TRCSREx3_2_Luc97_GM.trs

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:

a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__dbls(nil) → nil
a__dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
a__indx(nil, X) → nil
a__indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
a__from(X) → cons(X, from(s(X)))
mark(dbl(X)) → a__dbl(mark(X))
mark(dbls(X)) → a__dbls(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(indx(X1, X2)) → a__indx(mark(X1), X2)
mark(from(X)) → a__from(X)
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__dbl(X) → dbl(X)
a__dbls(X) → dbls(X)
a__sel(X1, X2) → sel(X1, X2)
a__indx(X1, X2) → indx(X1, X2)
a__from(X) → from(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__dbls(nil) → nil
a__dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
a__indx(nil, X) → nil
a__indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
a__from(X) → cons(X, from(s(X)))
mark(dbl(X)) → a__dbl(mark(X))
mark(dbls(X)) → a__dbls(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(indx(X1, X2)) → a__indx(mark(X1), X2)
mark(from(X)) → a__from(X)
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__dbl(X) → dbl(X)
a__dbls(X) → dbls(X)
a__sel(X1, X2) → sel(X1, X2)
a__indx(X1, X2) → indx(X1, X2)
a__from(X) → from(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:

MARK(dbl(X)) → MARK(X)
A__SEL(0, cons(X, Y)) → MARK(X)
MARK(indx(X1, X2)) → MARK(X1)
MARK(dbl(X)) → A__DBL(mark(X))
MARK(indx(X1, X2)) → A__INDX(mark(X1), X2)
MARK(from(X)) → A__FROM(X)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(dbls(X)) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(dbls(X)) → A__DBLS(mark(X))
MARK(sel(X1, X2)) → MARK(X1)
A__SEL(s(X), cons(Y, Z)) → MARK(X)

The TRS R consists of the following rules:

a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__dbls(nil) → nil
a__dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
a__indx(nil, X) → nil
a__indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
a__from(X) → cons(X, from(s(X)))
mark(dbl(X)) → a__dbl(mark(X))
mark(dbls(X)) → a__dbls(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(indx(X1, X2)) → a__indx(mark(X1), X2)
mark(from(X)) → a__from(X)
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__dbl(X) → dbl(X)
a__dbls(X) → dbls(X)
a__sel(X1, X2) → sel(X1, X2)
a__indx(X1, X2) → indx(X1, X2)
a__from(X) → from(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

MARK(dbl(X)) → MARK(X)
A__SEL(0, cons(X, Y)) → MARK(X)
MARK(indx(X1, X2)) → MARK(X1)
MARK(dbl(X)) → A__DBL(mark(X))
MARK(indx(X1, X2)) → A__INDX(mark(X1), X2)
MARK(from(X)) → A__FROM(X)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(dbls(X)) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(dbls(X)) → A__DBLS(mark(X))
MARK(sel(X1, X2)) → MARK(X1)
A__SEL(s(X), cons(Y, Z)) → MARK(X)

The TRS R consists of the following rules:

a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__dbls(nil) → nil
a__dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
a__indx(nil, X) → nil
a__indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
a__from(X) → cons(X, from(s(X)))
mark(dbl(X)) → a__dbl(mark(X))
mark(dbls(X)) → a__dbls(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(indx(X1, X2)) → a__indx(mark(X1), X2)
mark(from(X)) → a__from(X)
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__dbl(X) → dbl(X)
a__dbls(X) → dbls(X)
a__sel(X1, X2) → sel(X1, X2)
a__indx(X1, X2) → indx(X1, X2)
a__from(X) → from(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK(dbl(X)) → MARK(X)
A__SEL(0, cons(X, Y)) → MARK(X)
MARK(indx(X1, X2)) → MARK(X1)
MARK(dbl(X)) → A__DBL(mark(X))
MARK(indx(X1, X2)) → A__INDX(mark(X1), X2)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
MARK(from(X)) → A__FROM(X)
MARK(dbls(X)) → MARK(X)
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
MARK(sel(X1, X2)) → MARK(X1)
MARK(dbls(X)) → A__DBLS(mark(X))
A__SEL(s(X), cons(Y, Z)) → MARK(X)

The TRS R consists of the following rules:

a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__dbls(nil) → nil
a__dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
a__indx(nil, X) → nil
a__indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
a__from(X) → cons(X, from(s(X)))
mark(dbl(X)) → a__dbl(mark(X))
mark(dbls(X)) → a__dbls(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(indx(X1, X2)) → a__indx(mark(X1), X2)
mark(from(X)) → a__from(X)
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__dbl(X) → dbl(X)
a__dbls(X) → dbls(X)
a__sel(X1, X2) → sel(X1, X2)
a__indx(X1, X2) → indx(X1, X2)
a__from(X) → from(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

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

MARK(dbl(X)) → MARK(X)
A__SEL(0, cons(X, Y)) → MARK(X)
MARK(indx(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → A__SEL(mark(X1), mark(X2))
A__SEL(s(X), cons(Y, Z)) → A__SEL(mark(X), mark(Z))
MARK(dbls(X)) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
A__SEL(s(X), cons(Y, Z)) → MARK(Z)
A__SEL(s(X), cons(Y, Z)) → MARK(X)
MARK(sel(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__dbls(nil) → nil
a__dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
a__indx(nil, X) → nil
a__indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
a__from(X) → cons(X, from(s(X)))
mark(dbl(X)) → a__dbl(mark(X))
mark(dbls(X)) → a__dbls(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(indx(X1, X2)) → a__indx(mark(X1), X2)
mark(from(X)) → a__from(X)
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__dbl(X) → dbl(X)
a__dbls(X) → dbls(X)
a__sel(X1, X2) → sel(X1, X2)
a__indx(X1, X2) → indx(X1, X2)
a__from(X) → from(X)

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