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:
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.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
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.