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(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → 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:
B1(a(x1)) → B1(x1)
B1(a(x1)) → A1(B(x1))
A1(b(x1)) → B1(A(x1))
A1(b(x1)) → A1(x1)
The TRS R consists of the following rules:
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → 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:
B1(a(x1)) → B1(x1)
B1(a(x1)) → A1(B(x1))
A1(b(x1)) → B1(A(x1))
A1(b(x1)) → A1(x1)
The TRS R consists of the following rules:
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.