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:
c(c(x1)) → a(a(a(b(b(b(x1))))))
b(b(b(a(x1)))) → b(b(b(b(b(b(b(b(x1))))))))
b(b(c(c(x1)))) → c(c(c(a(a(a(a(x1)))))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
c(c(x1)) → a(a(a(b(b(b(x1))))))
b(b(b(a(x1)))) → b(b(b(b(b(b(b(b(x1))))))))
b(b(c(c(x1)))) → c(c(c(a(a(a(a(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(b(a(x1)))) → B(b(b(b(b(x1)))))
B(b(b(a(x1)))) → B(b(b(b(x1))))
C(c(x1)) → B(x1)
B(b(c(c(x1)))) → C(a(a(a(a(x1)))))
B(b(b(a(x1)))) → B(b(b(b(b(b(b(x1)))))))
B(b(b(a(x1)))) → B(b(b(b(b(b(b(b(x1))))))))
B(b(b(a(x1)))) → B(b(b(x1)))
B(b(b(a(x1)))) → B(x1)
B(b(b(a(x1)))) → B(b(b(b(b(b(x1))))))
B(b(c(c(x1)))) → C(c(a(a(a(a(x1))))))
C(c(x1)) → B(b(x1))
B(b(b(a(x1)))) → B(b(x1))
B(b(c(c(x1)))) → C(c(c(a(a(a(a(x1)))))))
C(c(x1)) → B(b(b(x1)))
The TRS R consists of the following rules:
c(c(x1)) → a(a(a(b(b(b(x1))))))
b(b(b(a(x1)))) → b(b(b(b(b(b(b(b(x1))))))))
b(b(c(c(x1)))) → c(c(c(a(a(a(a(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(b(b(a(x1)))) → B(b(b(b(b(x1)))))
B(b(b(a(x1)))) → B(b(b(b(x1))))
C(c(x1)) → B(x1)
B(b(c(c(x1)))) → C(a(a(a(a(x1)))))
B(b(b(a(x1)))) → B(b(b(b(b(b(b(x1)))))))
B(b(b(a(x1)))) → B(b(b(b(b(b(b(b(x1))))))))
B(b(b(a(x1)))) → B(b(b(x1)))
B(b(b(a(x1)))) → B(x1)
B(b(b(a(x1)))) → B(b(b(b(b(b(x1))))))
B(b(c(c(x1)))) → C(c(a(a(a(a(x1))))))
C(c(x1)) → B(b(x1))
B(b(b(a(x1)))) → B(b(x1))
B(b(c(c(x1)))) → C(c(c(a(a(a(a(x1)))))))
C(c(x1)) → B(b(b(x1)))
The TRS R consists of the following rules:
c(c(x1)) → a(a(a(b(b(b(x1))))))
b(b(b(a(x1)))) → b(b(b(b(b(b(b(b(x1))))))))
b(b(c(c(x1)))) → c(c(c(a(a(a(a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(b(b(a(x1)))) → B(b(b(b(b(x1)))))
B(b(b(a(x1)))) → B(b(b(b(x1))))
C(c(x1)) → B(x1)
B(b(b(a(x1)))) → B(b(b(b(b(b(b(x1)))))))
B(b(b(a(x1)))) → B(b(b(b(b(b(b(b(x1))))))))
B(b(b(a(x1)))) → B(b(b(x1)))
B(b(b(a(x1)))) → B(x1)
B(b(b(a(x1)))) → B(b(b(b(b(b(x1))))))
B(b(c(c(x1)))) → C(c(a(a(a(a(x1))))))
C(c(x1)) → B(b(x1))
B(b(b(a(x1)))) → B(b(x1))
B(b(c(c(x1)))) → C(c(c(a(a(a(a(x1)))))))
C(c(x1)) → B(b(b(x1)))
The TRS R consists of the following rules:
c(c(x1)) → a(a(a(b(b(b(x1))))))
b(b(b(a(x1)))) → b(b(b(b(b(b(b(b(x1))))))))
b(b(c(c(x1)))) → c(c(c(a(a(a(a(x1)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.