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))) → b(b(a(b(a(b(a(b(x1))))))))
b(a(b(b(x1)))) → b(a(b(a(b(a(a(b(x1))))))))
b(b(a(b(a(b(a(a(b(x1))))))))) → b(b(b(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))) → b(b(a(b(a(b(a(b(x1))))))))
b(a(b(b(x1)))) → b(a(b(a(b(a(a(b(x1))))))))
b(b(a(b(a(b(a(a(b(x1))))))))) → b(b(b(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:
B(a(b(b(x1)))) → B(a(a(b(x1))))
B(a(b(b(x1)))) → B(a(b(a(b(a(a(b(x1))))))))
B(b(a(b(a(b(a(a(b(x1))))))))) → B(b(b(b(a(b(x1))))))
B(b(b(x1))) → B(a(b(a(b(x1)))))
B(b(a(b(a(b(a(a(b(x1))))))))) → B(b(b(a(b(x1)))))
B(a(b(b(x1)))) → B(a(b(a(a(b(x1))))))
B(b(b(x1))) → B(a(b(a(b(a(b(x1)))))))
B(b(b(x1))) → B(b(a(b(a(b(a(b(x1))))))))
B(b(a(b(a(b(a(a(b(x1))))))))) → B(b(a(b(x1))))
B(b(b(x1))) → B(a(b(x1)))
B(b(a(b(a(b(a(a(b(x1))))))))) → B(a(b(x1)))
The TRS R consists of the following rules:
b(b(b(x1))) → b(b(a(b(a(b(a(b(x1))))))))
b(a(b(b(x1)))) → b(a(b(a(b(a(a(b(x1))))))))
b(b(a(b(a(b(a(a(b(x1))))))))) → b(b(b(b(a(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
B(a(b(b(x1)))) → B(a(a(b(x1))))
B(a(b(b(x1)))) → B(a(b(a(b(a(a(b(x1))))))))
B(b(a(b(a(b(a(a(b(x1))))))))) → B(b(b(b(a(b(x1))))))
B(b(b(x1))) → B(a(b(a(b(x1)))))
B(b(a(b(a(b(a(a(b(x1))))))))) → B(b(b(a(b(x1)))))
B(a(b(b(x1)))) → B(a(b(a(a(b(x1))))))
B(b(b(x1))) → B(a(b(a(b(a(b(x1)))))))
B(b(b(x1))) → B(b(a(b(a(b(a(b(x1))))))))
B(b(a(b(a(b(a(a(b(x1))))))))) → B(b(a(b(x1))))
B(b(b(x1))) → B(a(b(x1)))
B(b(a(b(a(b(a(a(b(x1))))))))) → B(a(b(x1)))
The TRS R consists of the following rules:
b(b(b(x1))) → b(b(a(b(a(b(a(b(x1))))))))
b(a(b(b(x1)))) → b(a(b(a(b(a(a(b(x1))))))))
b(b(a(b(a(b(a(a(b(x1))))))))) → b(b(b(b(a(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 7 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(b(a(b(a(b(a(a(b(x1))))))))) → B(b(b(b(a(b(x1))))))
B(b(a(b(a(b(a(a(b(x1))))))))) → B(b(b(a(b(x1)))))
B(b(b(x1))) → B(b(a(b(a(b(a(b(x1))))))))
B(b(a(b(a(b(a(a(b(x1))))))))) → B(b(a(b(x1))))
The TRS R consists of the following rules:
b(b(b(x1))) → b(b(a(b(a(b(a(b(x1))))))))
b(a(b(b(x1)))) → b(a(b(a(b(a(a(b(x1))))))))
b(b(a(b(a(b(a(a(b(x1))))))))) → b(b(b(b(a(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.