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:

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

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

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

first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
activate1(n__first2(X1, X2)) -> first2(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 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
The remaining pairs can at least be oriented weakly.

FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( FIRST2(x1, x2) ) = max{0, x2 - 1}


POL( s1(x1) ) = max{0, x1 - 1}


POL( cons2(x1, x2) ) = x1 + x2 + 1


POL( ACTIVATE1(x1) ) = x1


POL( n__first2(x1, x2) ) = x1 + x2 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
QDP
              ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
activate1(n__first2(X1, X2)) -> first2(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 0 SCCs with 1 less node.