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(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(a(b(x1))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(a(b(x1))))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
B(b(a(x1))) → A(b(x1))
B(b(a(x1))) → B(x1)
A(a(a(b(x1)))) → B(a(b(a(x1))))
B(b(a(x1))) → A(a(a(b(x1))))
A(a(a(b(x1)))) → A(x1)
A(a(a(b(x1)))) → A(b(a(x1)))
B(b(a(x1))) → A(a(b(x1)))
A(a(a(b(x1)))) → B(a(x1))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(a(b(x1))))
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:
B(b(a(x1))) → A(b(x1))
B(b(a(x1))) → B(x1)
A(a(a(b(x1)))) → B(a(b(a(x1))))
B(b(a(x1))) → A(a(a(b(x1))))
A(a(a(b(x1)))) → A(x1)
A(a(a(b(x1)))) → A(b(a(x1)))
B(b(a(x1))) → A(a(b(x1)))
A(a(a(b(x1)))) → B(a(x1))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(a(b(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.