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:
s11(s11(s01(s01(x)))) -> s01(s01(s01(s11(s11(s11(x))))))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
s11(s11(s01(s01(x)))) -> s01(s01(s01(s11(s11(s11(x))))))
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
s11(s11(s01(s01(x)))) -> s01(s01(s01(s11(s11(s11(x))))))
The set Q consists of the following terms:
s11(s11(s01(s01(x0))))
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
S11(s11(s01(s01(x)))) -> S11(x)
S11(s11(s01(s01(x)))) -> S11(s11(x))
S11(s11(s01(s01(x)))) -> S11(s11(s11(x)))
The TRS R consists of the following rules:
s11(s11(s01(s01(x)))) -> s01(s01(s01(s11(s11(s11(x))))))
The set Q consists of the following terms:
s11(s11(s01(s01(x0))))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
S11(s11(s01(s01(x)))) -> S11(x)
S11(s11(s01(s01(x)))) -> S11(s11(x))
S11(s11(s01(s01(x)))) -> S11(s11(s11(x)))
The TRS R consists of the following rules:
s11(s11(s01(s01(x)))) -> s01(s01(s01(s11(s11(s11(x))))))
The set Q consists of the following terms:
s11(s11(s01(s01(x0))))
We have to consider all minimal (P,Q,R)-chains.