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)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
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:
from1(X) -> cons2(X, n__from1(s1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
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:
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
ACTIVATE1(n__from1(X)) -> FROM1(X)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
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:
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
ACTIVATE1(n__from1(X)) -> FROM1(X)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
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 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
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.
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)
activate1(x1) = x1
n__from1(x1) = n__from
from1(x1) = from
cons2(x1, x2) = cons
Used ordering: Quasi Precedence:
[n__from, from, cons]
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
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 TRS P is empty. Hence, there is no (P,Q,R) chain.