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__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(X)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(X)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
MARK1(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
MARK1(zeros) -> A__ZEROS
MARK1(cons2(X1, X2)) -> MARK1(X1)
The TRS R consists of the following rules:
a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(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(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
MARK1(zeros) -> A__ZEROS
MARK1(cons2(X1, X2)) -> MARK1(X1)
The TRS R consists of the following rules:
a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(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
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
MARK1(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
The TRS R consists of the following rules:
a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(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__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
Used argument filtering: MARK1(x1) = x1
tail1(x1) = tail1(x1)
A__TAIL1(x1) = x1
mark1(x1) = mark1(x1)
cons2(x1, x2) = cons2(x1, x2)
zeros = zeros
a__zeros = a__zeros
a__tail1(x1) = a__tail1(x1)
0 = 0
Used ordering: Quasi Precedence:
[tail_1, mark_1, a__zeros, a__tail_1] > cons_2
[tail_1, mark_1, a__zeros, a__tail_1] > zeros
[tail_1, mark_1, a__zeros, a__tail_1] > 0
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MARK1(tail1(X)) -> A__TAIL1(mark1(X))
The TRS R consists of the following rules:
a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(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.