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__2nd1(cons12(X, cons2(Y, Z))) -> mark1(Y)
a__2nd1(cons2(X, X1)) -> a__2nd1(cons12(mark1(X), mark1(X1)))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(2nd1(X)) -> a__2nd1(mark1(X))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(cons12(X1, X2)) -> cons12(mark1(X1), mark1(X2))
a__2nd1(X) -> 2nd1(X)
a__from1(X) -> from1(X)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__2nd1(cons12(X, cons2(Y, Z))) -> mark1(Y)
a__2nd1(cons2(X, X1)) -> a__2nd1(cons12(mark1(X), mark1(X1)))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(2nd1(X)) -> a__2nd1(mark1(X))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(cons12(X1, X2)) -> cons12(mark1(X1), mark1(X2))
a__2nd1(X) -> 2nd1(X)
a__from1(X) -> from1(X)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
MARK1(2nd1(X)) -> MARK1(X)
A__2ND1(cons2(X, X1)) -> MARK1(X1)
A__2ND1(cons2(X, X1)) -> A__2ND1(cons12(mark1(X), mark1(X1)))
A__2ND1(cons12(X, cons2(Y, Z))) -> MARK1(Y)
A__2ND1(cons2(X, X1)) -> MARK1(X)
MARK1(s1(X)) -> MARK1(X)
MARK1(cons12(X1, X2)) -> MARK1(X1)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons12(X1, X2)) -> MARK1(X2)
A__FROM1(X) -> MARK1(X)
MARK1(2nd1(X)) -> A__2ND1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
The TRS R consists of the following rules:
a__2nd1(cons12(X, cons2(Y, Z))) -> mark1(Y)
a__2nd1(cons2(X, X1)) -> a__2nd1(cons12(mark1(X), mark1(X1)))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(2nd1(X)) -> a__2nd1(mark1(X))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(cons12(X1, X2)) -> cons12(mark1(X1), mark1(X2))
a__2nd1(X) -> 2nd1(X)
a__from1(X) -> from1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MARK1(2nd1(X)) -> MARK1(X)
A__2ND1(cons2(X, X1)) -> MARK1(X1)
A__2ND1(cons2(X, X1)) -> A__2ND1(cons12(mark1(X), mark1(X1)))
A__2ND1(cons12(X, cons2(Y, Z))) -> MARK1(Y)
A__2ND1(cons2(X, X1)) -> MARK1(X)
MARK1(s1(X)) -> MARK1(X)
MARK1(cons12(X1, X2)) -> MARK1(X1)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons12(X1, X2)) -> MARK1(X2)
A__FROM1(X) -> MARK1(X)
MARK1(2nd1(X)) -> A__2ND1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
The TRS R consists of the following rules:
a__2nd1(cons12(X, cons2(Y, Z))) -> mark1(Y)
a__2nd1(cons2(X, X1)) -> a__2nd1(cons12(mark1(X), mark1(X1)))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(2nd1(X)) -> a__2nd1(mark1(X))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(cons12(X1, X2)) -> cons12(mark1(X1), mark1(X2))
a__2nd1(X) -> 2nd1(X)
a__from1(X) -> from1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.