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:
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(n__s1(X)))
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
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:
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(n__s1(X)))
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
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__first2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(activate1(X1), activate1(X2))
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
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(n__s1(X)))
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
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__first2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(activate1(X1), activate1(X2))
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
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(n__s1(X)))
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(activate1(X1), activate1(X2))
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
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(n__s1(X)))
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
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.