Termination w.r.t. Q proof of TRS/TRCSREx3_2_Luc97

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:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(n__dbl₁(activate₁(X)), n__dbls₁(activate₁(Y)))
sel₂(0, cons₂(X, Y)) -> activate₁(X)
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(activate₁(X), activate₁(Z))
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(n__sel₂(activate₁(X), activate₁(Z)), n__indx₂(activate₁(Y), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
dbls₁(X) -> n__dbls₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
indx₂(X1, X2) -> n__indx₂(X1, X2)
from₁(X) -> n__from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__dbls₁(X)) -> dbls₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(n__indx₂(X1, X2)) -> indx₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(n__dbl₁(activate₁(X)), n__dbls₁(activate₁(Y)))
sel₂(0, cons₂(X, Y)) -> activate₁(X)
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(activate₁(X), activate₁(Z))
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(n__sel₂(activate₁(X), activate₁(Z)), n__indx₂(activate₁(Y), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
dbls₁(X) -> n__dbls₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
indx₂(X1, X2) -> n__indx₂(X1, X2)
from₁(X) -> n__from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__dbls₁(X)) -> dbls₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(n__indx₂(X1, X2)) -> indx₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(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:

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
INDX₂(cons₂(X, Y), Z) -> ACTIVATE₁(Y)
ACTIVATE₁(n__indx₂(X1, X2)) -> INDX₂(X1, X2)
INDX₂(cons₂(X, Y), Z) -> ACTIVATE₁(X)
DBLS₁(cons₂(X, Y)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> S₁(X)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
INDX₂(cons₂(X, Y), Z) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(activate₁(X), activate₁(Z))
ACTIVATE₁(n__dbls₁(X)) -> DBLS₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__sel₂(X1, X2)) -> SEL₂(X1, X2)
DBLS₁(cons₂(X, Y)) -> ACTIVATE₁(Y)
FROM₁(X) -> ACTIVATE₁(X)
SEL₂(0, cons₂(X, Y)) -> ACTIVATE₁(X)
DBL₁(s₁(X)) -> S₁(n__s₁(n__dbl₁(activate₁(X))))

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(n__dbl₁(activate₁(X)), n__dbls₁(activate₁(Y)))
sel₂(0, cons₂(X, Y)) -> activate₁(X)
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(activate₁(X), activate₁(Z))
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(n__sel₂(activate₁(X), activate₁(Z)), n__indx₂(activate₁(Y), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
dbls₁(X) -> n__dbls₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
indx₂(X1, X2) -> n__indx₂(X1, X2)
from₁(X) -> n__from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__dbls₁(X)) -> dbls₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(n__indx₂(X1, X2)) -> indx₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(X) -> X

Q is empty.
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:

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
INDX₂(cons₂(X, Y), Z) -> ACTIVATE₁(Y)
ACTIVATE₁(n__indx₂(X1, X2)) -> INDX₂(X1, X2)
INDX₂(cons₂(X, Y), Z) -> ACTIVATE₁(X)
DBLS₁(cons₂(X, Y)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> S₁(X)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
INDX₂(cons₂(X, Y), Z) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(activate₁(X), activate₁(Z))
ACTIVATE₁(n__dbls₁(X)) -> DBLS₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__sel₂(X1, X2)) -> SEL₂(X1, X2)
DBLS₁(cons₂(X, Y)) -> ACTIVATE₁(Y)
FROM₁(X) -> ACTIVATE₁(X)
SEL₂(0, cons₂(X, Y)) -> ACTIVATE₁(X)
DBL₁(s₁(X)) -> S₁(n__s₁(n__dbl₁(activate₁(X))))

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(n__dbl₁(activate₁(X)), n__dbls₁(activate₁(Y)))
sel₂(0, cons₂(X, Y)) -> activate₁(X)
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(activate₁(X), activate₁(Z))
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(n__sel₂(activate₁(X), activate₁(Z)), n__indx₂(activate₁(Y), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
dbls₁(X) -> n__dbls₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
indx₂(X1, X2) -> n__indx₂(X1, X2)
from₁(X) -> n__from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__dbls₁(X)) -> dbls₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(n__indx₂(X1, X2)) -> indx₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(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

      ↳ DependencyGraphProof

        ↳ QDP

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

ACTIVATE₁(n__dbl₁(X)) -> DBL₁(X)
INDX₂(cons₂(X, Y), Z) -> ACTIVATE₁(Y)
ACTIVATE₁(n__indx₂(X1, X2)) -> INDX₂(X1, X2)
INDX₂(cons₂(X, Y), Z) -> ACTIVATE₁(X)
DBLS₁(cons₂(X, Y)) -> ACTIVATE₁(X)
DBL₁(s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
INDX₂(cons₂(X, Y), Z) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(activate₁(X), activate₁(Z))
ACTIVATE₁(n__dbls₁(X)) -> DBLS₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__sel₂(X1, X2)) -> SEL₂(X1, X2)
DBLS₁(cons₂(X, Y)) -> ACTIVATE₁(Y)
FROM₁(X) -> ACTIVATE₁(X)
SEL₂(0, cons₂(X, Y)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(n__s₁(n__dbl₁(activate₁(X))))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(n__dbl₁(activate₁(X)), n__dbls₁(activate₁(Y)))
sel₂(0, cons₂(X, Y)) -> activate₁(X)
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(activate₁(X), activate₁(Z))
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(n__sel₂(activate₁(X), activate₁(Z)), n__indx₂(activate₁(Y), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
s₁(X) -> n__s₁(X)
dbl₁(X) -> n__dbl₁(X)
dbls₁(X) -> n__dbls₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
indx₂(X1, X2) -> n__indx₂(X1, X2)
from₁(X) -> n__from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__dbl₁(X)) -> dbl₁(X)
activate₁(n__dbls₁(X)) -> dbls₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(n__indx₂(X1, X2)) -> indx₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(X) -> X

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