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:
a1(d1(x)) -> d1(c1(b1(a1(x))))
b1(c1(x)) -> c1(d1(a1(b1(x))))
a1(c1(x)) -> x
b1(d1(x)) -> x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a1(d1(x)) -> d1(c1(b1(a1(x))))
b1(c1(x)) -> c1(d1(a1(b1(x))))
a1(c1(x)) -> x
b1(d1(x)) -> 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:
A1(d1(x)) -> B1(a1(x))
B1(c1(x)) -> A1(b1(x))
B1(c1(x)) -> B1(x)
A1(d1(x)) -> A1(x)
The TRS R consists of the following rules:
a1(d1(x)) -> d1(c1(b1(a1(x))))
b1(c1(x)) -> c1(d1(a1(b1(x))))
a1(c1(x)) -> x
b1(d1(x)) -> 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:
A1(d1(x)) -> B1(a1(x))
B1(c1(x)) -> A1(b1(x))
B1(c1(x)) -> B1(x)
A1(d1(x)) -> A1(x)
The TRS R consists of the following rules:
a1(d1(x)) -> d1(c1(b1(a1(x))))
b1(c1(x)) -> c1(d1(a1(b1(x))))
a1(c1(x)) -> x
b1(d1(x)) -> x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.