Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

from1(X) -> cons2(X, n__from1(s1(X)))
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
sel2(0, cons2(X, Z)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
from1(X) -> n__from1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

from1(X) -> cons2(X, n__from1(s1(X)))
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
sel2(0, cons2(X, Z)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
from1(X) -> n__from1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X

Q is empty.

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

SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
ACTIVATE1(n__from1(X)) -> FROM1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(s1(X)))
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
sel2(0, cons2(X, Z)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
from1(X) -> n__from1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
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:

SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
ACTIVATE1(n__from1(X)) -> FROM1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(s1(X)))
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
sel2(0, cons2(X, Z)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
from1(X) -> n__from1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(s1(X)))
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
sel2(0, cons2(X, Z)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
from1(X) -> n__from1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
Used argument filtering: ACTIVATE1(x1)  =  x1
n__first2(x1, x2)  =  x2
FIRST2(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ DependencyGraphProof
          ↳ QDP

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

ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(s1(X)))
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
sel2(0, cons2(X, Z)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
from1(X) -> n__from1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof

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

SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(s1(X)))
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
sel2(0, cons2(X, Z)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
from1(X) -> n__from1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
Used argument filtering: SEL2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(s1(X)))
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
sel2(0, cons2(X, Z)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
from1(X) -> n__from1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.