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)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
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)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
activate1(X) -> X
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__take2(X1, X2)) -> TAKE2(X1, X2)
TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
SEL2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
2ND1(cons2(X, XS)) -> HEAD1(activate1(XS))
ACTIVATE1(n__from1(X)) -> FROM1(X)
2ND1(cons2(X, XS)) -> ACTIVATE1(XS)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
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__take2(X1, X2)) -> TAKE2(X1, X2)
TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
SEL2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
2ND1(cons2(X, XS)) -> HEAD1(activate1(XS))
ACTIVATE1(n__from1(X)) -> FROM1(X)
2ND1(cons2(X, XS)) -> ACTIVATE1(XS)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__take2(X1, X2)) -> TAKE2(X1, X2)
TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
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.
TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
Used argument filtering: ACTIVATE1(x1) = x1
n__take2(x1, x2) = x2
TAKE2(x1, x2) = x2
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__take2(X1, X2)) -> TAKE2(X1, X2)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
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(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
Used argument filtering: SEL2(x1, x2) = x1
s1(x1) = s1(x1)
activate1(x1) = x1
n__from1(x1) = n__from
from1(x1) = from
n__take2(x1, x2) = n__take
take2(x1, x2) = take
0 = 0
nil = nil
cons2(x1, x2) = cons
Used ordering: Quasi Precedence:
[n__from, from, n__take, take, cons] > nil
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
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.