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:
b(b(b(x1))) → a(a(b(x1)))
b(a(a(x1))) → a(a(b(x1)))
a(a(x1)) → b(a(b(x1)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(b(x1)))
b(a(a(x1))) → a(a(b(x1)))
a(a(x1)) → b(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:
A(a(x1)) → A(b(x1))
B(a(a(x1))) → A(b(x1))
B(b(b(x1))) → A(b(x1))
A(a(x1)) → B(a(b(x1)))
B(a(a(x1))) → B(x1)
B(a(a(x1))) → A(a(b(x1)))
B(b(b(x1))) → A(a(b(x1)))
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(b(x1)))
b(a(a(x1))) → a(a(b(x1)))
a(a(x1)) → b(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:
A(a(x1)) → A(b(x1))
B(a(a(x1))) → A(b(x1))
B(b(b(x1))) → A(b(x1))
A(a(x1)) → B(a(b(x1)))
B(a(a(x1))) → B(x1)
B(a(a(x1))) → A(a(b(x1)))
B(b(b(x1))) → A(a(b(x1)))
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(b(x1)))
b(a(a(x1))) → a(a(b(x1)))
a(a(x1)) → b(a(b(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.