Termination w.r.t. Q proof of TRS/TRCSREx3_2_Luc97_Z.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:

dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(n__dbl(activate(X)), n__dbls(activate(Y)))
sel(0, cons(X, Y)) → activate(X)
sel(s(X), cons(Y, Z)) → sel(activate(X), activate(Z))
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
s(X) → n__s(X)
dbl(X) → n__dbl(X)
dbls(X) → n__dbls(X)
sel(X1, X2) → n__sel(X1, X2)
indx(X1, X2) → n__indx(X1, X2)
from(X) → n__from(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__dbls(X)) → dbls(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(n__indx(X1, X2)) → indx(X1, X2)
activate(n__from(X)) → from(X)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(n__dbl(activate(X)), n__dbls(activate(Y)))
sel(0, cons(X, Y)) → activate(X)
sel(s(X), cons(Y, Z)) → sel(activate(X), activate(Z))
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
s(X) → n__s(X)
dbl(X) → n__dbl(X)
dbls(X) → n__dbls(X)
sel(X1, X2) → n__sel(X1, X2)
indx(X1, X2) → n__indx(X1, X2)
from(X) → n__from(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__dbls(X)) → dbls(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(n__indx(X1, X2)) → indx(X1, X2)
activate(n__from(X)) → from(X)
activate(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:

ACTIVATE(n__dbl(X)) → DBL(X)
INDX(cons(X, Y), Z) → ACTIVATE(Y)
ACTIVATE(n__indx(X1, X2)) → INDX(X1, X2)
INDX(cons(X, Y), Z) → ACTIVATE(X)
DBLS(cons(X, Y)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(X)
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → FROM(X)
INDX(cons(X, Y), Z) → ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) → SEL(activate(X), activate(Z))
ACTIVATE(n__dbls(X)) → DBLS(X)
SEL(s(X), cons(Y, Z)) → ACTIVATE(X)
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
DBLS(cons(X, Y)) → ACTIVATE(Y)
FROM(X) → ACTIVATE(X)
SEL(0, cons(X, Y)) → ACTIVATE(X)
DBL(s(X)) → S(n__s(n__dbl(activate(X))))

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(n__dbl(activate(X)), n__dbls(activate(Y)))
sel(0, cons(X, Y)) → activate(X)
sel(s(X), cons(Y, Z)) → sel(activate(X), activate(Z))
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
s(X) → n__s(X)
dbl(X) → n__dbl(X)
dbls(X) → n__dbls(X)
sel(X1, X2) → n__sel(X1, X2)
indx(X1, X2) → n__indx(X1, X2)
from(X) → n__from(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__dbls(X)) → dbls(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(n__indx(X1, X2)) → indx(X1, X2)
activate(n__from(X)) → from(X)
activate(X) → 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:

ACTIVATE(n__dbl(X)) → DBL(X)
INDX(cons(X, Y), Z) → ACTIVATE(Y)
ACTIVATE(n__indx(X1, X2)) → INDX(X1, X2)
INDX(cons(X, Y), Z) → ACTIVATE(X)
DBLS(cons(X, Y)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(X)
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → FROM(X)
INDX(cons(X, Y), Z) → ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) → SEL(activate(X), activate(Z))
ACTIVATE(n__dbls(X)) → DBLS(X)
SEL(s(X), cons(Y, Z)) → ACTIVATE(X)
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
DBLS(cons(X, Y)) → ACTIVATE(Y)
FROM(X) → ACTIVATE(X)
SEL(0, cons(X, Y)) → ACTIVATE(X)
DBL(s(X)) → S(n__s(n__dbl(activate(X))))

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(n__dbl(activate(X)), n__dbls(activate(Y)))
sel(0, cons(X, Y)) → activate(X)
sel(s(X), cons(Y, Z)) → sel(activate(X), activate(Z))
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
s(X) → n__s(X)
dbl(X) → n__dbl(X)
dbls(X) → n__dbls(X)
sel(X1, X2) → n__sel(X1, X2)
indx(X1, X2) → n__indx(X1, X2)
from(X) → n__from(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__dbls(X)) → dbls(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(n__indx(X1, X2)) → indx(X1, X2)
activate(n__from(X)) → from(X)
activate(X) → 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:

ACTIVATE(n__dbl(X)) → DBL(X)
INDX(cons(X, Y), Z) → ACTIVATE(Y)
INDX(cons(X, Y), Z) → ACTIVATE(X)
ACTIVATE(n__indx(X1, X2)) → INDX(X1, X2)
DBLS(cons(X, Y)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(X)
DBL(s(X)) → ACTIVATE(X)
INDX(cons(X, Y), Z) → ACTIVATE(Z)
ACTIVATE(n__from(X)) → FROM(X)
SEL(s(X), cons(Y, Z)) → SEL(activate(X), activate(Z))
ACTIVATE(n__dbls(X)) → DBLS(X)
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
FROM(X) → ACTIVATE(X)
DBLS(cons(X, Y)) → ACTIVATE(Y)
SEL(0, cons(X, Y)) → ACTIVATE(X)
DBL(s(X)) → S(n__s(n__dbl(activate(X))))

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(n__dbl(activate(X)), n__dbls(activate(Y)))
sel(0, cons(X, Y)) → activate(X)
sel(s(X), cons(Y, Z)) → sel(activate(X), activate(Z))
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
s(X) → n__s(X)
dbl(X) → n__dbl(X)
dbls(X) → n__dbls(X)
sel(X1, X2) → n__sel(X1, X2)
indx(X1, X2) → n__indx(X1, X2)
from(X) → n__from(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__dbls(X)) → dbls(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(n__indx(X1, X2)) → indx(X1, X2)
activate(n__from(X)) → from(X)
activate(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 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

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

ACTIVATE(n__dbl(X)) → DBL(X)
INDX(cons(X, Y), Z) → ACTIVATE(Y)
ACTIVATE(n__indx(X1, X2)) → INDX(X1, X2)
INDX(cons(X, Y), Z) → ACTIVATE(X)
DBLS(cons(X, Y)) → ACTIVATE(X)
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → FROM(X)
INDX(cons(X, Y), Z) → ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) → SEL(activate(X), activate(Z))
ACTIVATE(n__dbls(X)) → DBLS(X)
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
DBLS(cons(X, Y)) → ACTIVATE(Y)
FROM(X) → ACTIVATE(X)
SEL(0, cons(X, Y)) → ACTIVATE(X)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(n__dbl(activate(X)), n__dbls(activate(Y)))
sel(0, cons(X, Y)) → activate(X)
sel(s(X), cons(Y, Z)) → sel(activate(X), activate(Z))
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
s(X) → n__s(X)
dbl(X) → n__dbl(X)
dbls(X) → n__dbls(X)
sel(X1, X2) → n__sel(X1, X2)
indx(X1, X2) → n__indx(X1, X2)
from(X) → n__from(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__dbls(X)) → dbls(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(n__indx(X1, X2)) → indx(X1, X2)
activate(n__from(X)) → from(X)
activate(X) → X

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