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:

2nd1(cons12(X, cons2(Y, Z))) -> Y
2nd1(cons2(X, X1)) -> 2nd1(cons12(X, activate1(X1)))
from1(X) -> cons2(X, n__from1(s1(X)))
from1(X) -> n__from1(X)
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:

2nd1(cons12(X, cons2(Y, Z))) -> Y
2nd1(cons2(X, X1)) -> 2nd1(cons12(X, activate1(X1)))
from1(X) -> cons2(X, n__from1(s1(X)))
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> X

Q is empty.

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

2ND1(cons2(X, X1)) -> 2ND1(cons12(X, activate1(X1)))
ACTIVATE1(n__from1(X)) -> FROM1(X)
2ND1(cons2(X, X1)) -> ACTIVATE1(X1)

The TRS R consists of the following rules:

2nd1(cons12(X, cons2(Y, Z))) -> Y
2nd1(cons2(X, X1)) -> 2nd1(cons12(X, activate1(X1)))
from1(X) -> cons2(X, n__from1(s1(X)))
from1(X) -> n__from1(X)
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:

2ND1(cons2(X, X1)) -> 2ND1(cons12(X, activate1(X1)))
ACTIVATE1(n__from1(X)) -> FROM1(X)
2ND1(cons2(X, X1)) -> ACTIVATE1(X1)

The TRS R consists of the following rules:

2nd1(cons12(X, cons2(Y, Z))) -> Y
2nd1(cons2(X, X1)) -> 2nd1(cons12(X, activate1(X1)))
from1(X) -> cons2(X, n__from1(s1(X)))
from1(X) -> n__from1(X)
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 0 SCCs with 3 less nodes.