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:
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(X)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(X)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
MARK1(s1(X)) -> MARK1(X)
A__LENGTH1(cons2(X, Y)) -> A__LENGTH11(Y)
MARK1(length1(X)) -> A__LENGTH1(X)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__LENGTH11(X) -> A__LENGTH1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(length11(X)) -> A__LENGTH11(X)
A__FROM1(X) -> MARK1(X)
The TRS R consists of the following rules:
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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:
MARK1(s1(X)) -> MARK1(X)
A__LENGTH1(cons2(X, Y)) -> A__LENGTH11(Y)
MARK1(length1(X)) -> A__LENGTH1(X)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__LENGTH11(X) -> A__LENGTH1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(length11(X)) -> A__LENGTH11(X)
A__FROM1(X) -> MARK1(X)
The TRS R consists of the following rules:
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A__LENGTH1(cons2(X, Y)) -> A__LENGTH11(Y)
A__LENGTH11(X) -> A__LENGTH1(X)
The TRS R consists of the following rules:
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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.
A__LENGTH1(cons2(X, Y)) -> A__LENGTH11(Y)
Used argument filtering: A__LENGTH1(x1) = x1
cons2(x1, x2) = cons1(x2)
A__LENGTH11(x1) = x1
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:
A__LENGTH11(X) -> A__LENGTH1(X)
The TRS R consists of the following rules:
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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:
MARK1(s1(X)) -> MARK1(X)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FROM1(X) -> MARK1(X)
The TRS R consists of the following rules:
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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.
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
Used argument filtering: MARK1(x1) = x1
s1(x1) = x1
from1(x1) = from1(x1)
A__FROM1(x1) = x1
mark1(x1) = x1
cons2(x1, x2) = x1
a__from1(x1) = a__from1(x1)
length1(x1) = length
a__length1(x1) = a__length
length11(x1) = length1
a__length11(x1) = a__length1
nil = nil
0 = 0
Used ordering: Quasi Precedence:
[from_1, a__from_1]
[length, a__length, length1, a__length1] > 0
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MARK1(s1(X)) -> MARK1(X)
A__FROM1(X) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
The TRS R consists of the following rules:
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
MARK1(s1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
The TRS R consists of the following rules:
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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.
MARK1(cons2(X1, X2)) -> MARK1(X1)
Used argument filtering: MARK1(x1) = x1
s1(x1) = x1
cons2(x1, x2) = cons1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
MARK1(s1(X)) -> MARK1(X)
The TRS R consists of the following rules:
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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.
MARK1(s1(X)) -> MARK1(X)
Used argument filtering: MARK1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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.