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(activate1(X))
activate1(n__dbls1(X)) -> dbls1(activate1(X))
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(X2))
activate1(n__indx2(X1, X2)) -> indx2(activate1(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(activate1(X))
activate1(n__dbls1(X)) -> dbls1(activate1(X))
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(X2))
activate1(n__indx2(X1, X2)) -> indx2(activate1(X1), X2)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> X

Q is empty.

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

ACTIVATE1(n__indx2(X1, X2)) -> ACTIVATE1(X1)
INDX2(cons2(X, Y), Z) -> ACTIVATE1(Y)
ACTIVATE1(n__sel2(X1, X2)) -> SEL2(activate1(X1), activate1(X2))
INDX2(cons2(X, Y), Z) -> ACTIVATE1(X)
ACTIVATE1(n__dbls1(X)) -> DBLS1(activate1(X))
ACTIVATE1(n__indx2(X1, X2)) -> INDX2(activate1(X1), X2)
DBLS1(cons2(X, Y)) -> ACTIVATE1(X)
ACTIVATE1(n__dbls1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> S1(X)
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X1)
DBL1(s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__dbl1(X)) -> DBL1(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))
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__dbl1(X)) -> ACTIVATE1(X)
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(activate1(X))
activate1(n__dbls1(X)) -> dbls1(activate1(X))
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(X2))
activate1(n__indx2(X1, X2)) -> indx2(activate1(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__indx2(X1, X2)) -> ACTIVATE1(X1)
INDX2(cons2(X, Y), Z) -> ACTIVATE1(Y)
ACTIVATE1(n__sel2(X1, X2)) -> SEL2(activate1(X1), activate1(X2))
INDX2(cons2(X, Y), Z) -> ACTIVATE1(X)
ACTIVATE1(n__dbls1(X)) -> DBLS1(activate1(X))
ACTIVATE1(n__indx2(X1, X2)) -> INDX2(activate1(X1), X2)
DBLS1(cons2(X, Y)) -> ACTIVATE1(X)
ACTIVATE1(n__dbls1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> S1(X)
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X1)
DBL1(s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__dbl1(X)) -> DBL1(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))
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__dbl1(X)) -> ACTIVATE1(X)
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(activate1(X))
activate1(n__dbls1(X)) -> dbls1(activate1(X))
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(X2))
activate1(n__indx2(X1, X2)) -> indx2(activate1(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 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__indx2(X1, X2)) -> ACTIVATE1(X1)
INDX2(cons2(X, Y), Z) -> ACTIVATE1(Y)
ACTIVATE1(n__sel2(X1, X2)) -> SEL2(activate1(X1), activate1(X2))
INDX2(cons2(X, Y), Z) -> ACTIVATE1(X)
ACTIVATE1(n__dbls1(X)) -> DBLS1(activate1(X))
ACTIVATE1(n__indx2(X1, X2)) -> INDX2(activate1(X1), X2)
DBLS1(cons2(X, Y)) -> ACTIVATE1(X)
ACTIVATE1(n__dbls1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X1)
DBL1(s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__dbl1(X)) -> DBL1(activate1(X))
INDX2(cons2(X, Y), Z) -> ACTIVATE1(Z)
ACTIVATE1(n__from1(X)) -> FROM1(X)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(activate1(X), activate1(Z))
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__dbl1(X)) -> ACTIVATE1(X)
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(activate1(X))
activate1(n__dbls1(X)) -> dbls1(activate1(X))
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(X2))
activate1(n__indx2(X1, X2)) -> indx2(activate1(X1), X2)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> X

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