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(n__s1(X)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(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(n__s1(X)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(X) -> X

Q is empty.

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

ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
2ND1(cons2(X, X1)) -> 2ND1(cons12(X, activate1(X1)))
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
2ND1(cons2(X, X1)) -> ACTIVATE1(X1)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

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(n__s1(X)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(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__from1(X)) -> FROM1(activate1(X))
2ND1(cons2(X, X1)) -> 2ND1(cons12(X, activate1(X1)))
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
2ND1(cons2(X, X1)) -> ACTIVATE1(X1)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

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(n__s1(X)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(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 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPAfsSolverProof

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

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

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(n__s1(X)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
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.

ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
Used argument filtering: ACTIVATE1(x1)  =  x1
n__from1(x1)  =  x1
n__s1(x1)  =  n__s1(x1)
Used ordering: Quasi Precedence: trivial


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

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

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)

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(n__s1(X)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
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.

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
Used argument filtering: ACTIVATE1(x1)  =  x1
n__from1(x1)  =  n__from1(x1)
Used ordering: Quasi Precedence: trivial


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

Q DP problem:
P is empty.
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(n__s1(X)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
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.