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__f1(X) -> cons2(mark1(X), f1(g1(X)))
a__g1(0) -> s1(0)
a__g1(s1(X)) -> s1(s1(a__g1(mark1(X))))
a__sel2(0, cons2(X, Y)) -> mark1(X)
a__sel2(s1(X), cons2(Y, Z)) -> a__sel2(mark1(X), mark1(Z))
mark1(f1(X)) -> a__f1(mark1(X))
mark1(g1(X)) -> a__g1(mark1(X))
mark1(sel2(X1, X2)) -> a__sel2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__g1(X) -> g1(X)
a__sel2(X1, X2) -> sel2(X1, X2)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f1(X) -> cons2(mark1(X), f1(g1(X)))
a__g1(0) -> s1(0)
a__g1(s1(X)) -> s1(s1(a__g1(mark1(X))))
a__sel2(0, cons2(X, Y)) -> mark1(X)
a__sel2(s1(X), cons2(Y, Z)) -> a__sel2(mark1(X), mark1(Z))
mark1(f1(X)) -> a__f1(mark1(X))
mark1(g1(X)) -> a__g1(mark1(X))
mark1(sel2(X1, X2)) -> a__sel2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__g1(X) -> g1(X)
a__sel2(X1, X2) -> sel2(X1, X2)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
MARK1(g1(X)) -> A__G1(mark1(X))
A__SEL2(0, cons2(X, Y)) -> MARK1(X)
MARK1(f1(X)) -> A__F1(mark1(X))
MARK1(sel2(X1, X2)) -> A__SEL2(mark1(X1), mark1(X2))
A__F1(X) -> MARK1(X)
A__G1(s1(X)) -> MARK1(X)
A__SEL2(s1(X), cons2(Y, Z)) -> A__SEL2(mark1(X), mark1(Z))
MARK1(sel2(X1, X2)) -> MARK1(X2)
A__G1(s1(X)) -> A__G1(mark1(X))
A__SEL2(s1(X), cons2(Y, Z)) -> MARK1(Z)
MARK1(sel2(X1, X2)) -> MARK1(X1)
MARK1(s1(X)) -> MARK1(X)
MARK1(g1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__SEL2(s1(X), cons2(Y, Z)) -> MARK1(X)
MARK1(f1(X)) -> MARK1(X)
The TRS R consists of the following rules:
a__f1(X) -> cons2(mark1(X), f1(g1(X)))
a__g1(0) -> s1(0)
a__g1(s1(X)) -> s1(s1(a__g1(mark1(X))))
a__sel2(0, cons2(X, Y)) -> mark1(X)
a__sel2(s1(X), cons2(Y, Z)) -> a__sel2(mark1(X), mark1(Z))
mark1(f1(X)) -> a__f1(mark1(X))
mark1(g1(X)) -> a__g1(mark1(X))
mark1(sel2(X1, X2)) -> a__sel2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__g1(X) -> g1(X)
a__sel2(X1, X2) -> sel2(X1, X2)
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(g1(X)) -> A__G1(mark1(X))
A__SEL2(0, cons2(X, Y)) -> MARK1(X)
MARK1(f1(X)) -> A__F1(mark1(X))
MARK1(sel2(X1, X2)) -> A__SEL2(mark1(X1), mark1(X2))
A__F1(X) -> MARK1(X)
A__G1(s1(X)) -> MARK1(X)
A__SEL2(s1(X), cons2(Y, Z)) -> A__SEL2(mark1(X), mark1(Z))
MARK1(sel2(X1, X2)) -> MARK1(X2)
A__G1(s1(X)) -> A__G1(mark1(X))
A__SEL2(s1(X), cons2(Y, Z)) -> MARK1(Z)
MARK1(sel2(X1, X2)) -> MARK1(X1)
MARK1(s1(X)) -> MARK1(X)
MARK1(g1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__SEL2(s1(X), cons2(Y, Z)) -> MARK1(X)
MARK1(f1(X)) -> MARK1(X)
The TRS R consists of the following rules:
a__f1(X) -> cons2(mark1(X), f1(g1(X)))
a__g1(0) -> s1(0)
a__g1(s1(X)) -> s1(s1(a__g1(mark1(X))))
a__sel2(0, cons2(X, Y)) -> mark1(X)
a__sel2(s1(X), cons2(Y, Z)) -> a__sel2(mark1(X), mark1(Z))
mark1(f1(X)) -> a__f1(mark1(X))
mark1(g1(X)) -> a__g1(mark1(X))
mark1(sel2(X1, X2)) -> a__sel2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
a__f1(X) -> f1(X)
a__g1(X) -> g1(X)
a__sel2(X1, X2) -> sel2(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.