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:
eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
EQ2(s1(X), s1(Y)) -> EQ2(X, Y)
TAKE2(s1(X), cons2(Y, L)) -> TAKE2(X, L)
LENGTH1(cons2(X, L)) -> LENGTH1(L)
INF1(X) -> INF1(s1(X))
The TRS R consists of the following rules:
eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))
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:
EQ2(s1(X), s1(Y)) -> EQ2(X, Y)
TAKE2(s1(X), cons2(Y, L)) -> TAKE2(X, L)
LENGTH1(cons2(X, L)) -> LENGTH1(L)
INF1(X) -> INF1(s1(X))
The TRS R consists of the following rules:
eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LENGTH1(cons2(X, L)) -> LENGTH1(L)
The TRS R consists of the following rules:
eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))
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.
LENGTH1(cons2(X, L)) -> LENGTH1(L)
Used argument filtering: LENGTH1(x1) = x1
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))
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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TAKE2(s1(X), cons2(Y, L)) -> TAKE2(X, L)
The TRS R consists of the following rules:
eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))
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(X), cons2(Y, L)) -> TAKE2(X, L)
Used argument filtering: TAKE2(x1, x2) = x2
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))
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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INF1(X) -> INF1(s1(X))
The TRS R consists of the following rules:
eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
EQ2(s1(X), s1(Y)) -> EQ2(X, Y)
The TRS R consists of the following rules:
eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))
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.
EQ2(s1(X), s1(Y)) -> EQ2(X, Y)
Used argument filtering: EQ2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
eq2(0, 0) -> true
eq2(s1(X), s1(Y)) -> eq2(X, Y)
eq2(X, Y) -> false
inf1(X) -> cons2(X, inf1(s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(Y, take2(X, L))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(length1(L))
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.