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(a(b(x1)))) → b(a(b(b(b(b(x1))))))
b(a(b(b(x1)))) → b(b(b(x1)))
b(b(b(x1))) → b(b(a(a(b(a(b(x1)))))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(b(x1)))) → b(a(b(b(b(b(x1))))))
b(a(b(b(x1)))) → b(b(b(x1)))
b(b(b(x1))) → b(b(a(a(b(a(b(x1)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(a(b(x1)))) → b(a(b(b(b(b(x1))))))
b(a(b(b(x1)))) → b(b(b(x1)))
b(b(b(x1))) → b(b(a(a(b(a(b(x1)))))))
The set Q is empty.
We have obtained the following QTRS:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
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(b(x1)))) → B(b(b(b(x1))))
B(a(b(b(x1)))) → B(b(b(x1)))
B(b(a(b(x1)))) → B(a(b(b(b(b(x1))))))
B(b(a(b(x1)))) → B(b(x1))
B(b(a(b(x1)))) → B(b(b(x1)))
B(b(b(x1))) → B(a(a(b(a(b(x1))))))
B(b(b(x1))) → B(a(b(x1)))
B(b(b(x1))) → B(b(a(a(b(a(b(x1)))))))
The TRS R consists of the following rules:
b(b(a(b(x1)))) → b(a(b(b(b(b(x1))))))
b(a(b(b(x1)))) → b(b(b(x1)))
b(b(b(x1))) → b(b(a(a(b(a(b(x1)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(b(x1)))) → B(b(b(b(x1))))
B(a(b(b(x1)))) → B(b(b(x1)))
B(b(a(b(x1)))) → B(a(b(b(b(b(x1))))))
B(b(a(b(x1)))) → B(b(x1))
B(b(a(b(x1)))) → B(b(b(x1)))
B(b(b(x1))) → B(a(a(b(a(b(x1))))))
B(b(b(x1))) → B(a(b(x1)))
B(b(b(x1))) → B(b(a(a(b(a(b(x1)))))))
The TRS R consists of the following rules:
b(b(a(b(x1)))) → b(a(b(b(b(b(x1))))))
b(a(b(b(x1)))) → b(b(b(x1)))
b(b(b(x1))) → b(b(a(a(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 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(b(x1)))) → B(b(b(b(x1))))
B(b(a(b(x1)))) → B(a(b(b(b(b(x1))))))
B(a(b(b(x1)))) → B(b(b(x1)))
B(b(a(b(x1)))) → B(b(x1))
B(b(a(b(x1)))) → B(b(b(x1)))
B(b(b(x1))) → B(a(b(x1)))
The TRS R consists of the following rules:
b(b(a(b(x1)))) → b(a(b(b(b(b(x1))))))
b(a(b(b(x1)))) → b(b(b(x1)))
b(b(b(x1))) → b(b(a(a(b(a(b(x1)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(b(x1)))) → b(a(b(b(b(b(x1))))))
b(a(b(b(x1)))) → b(b(b(x1)))
b(b(b(x1))) → b(b(a(a(b(a(b(x1)))))))
B(b(a(b(x1)))) → B(b(b(b(x1))))
B(b(a(b(x1)))) → B(a(b(b(b(b(x1))))))
B(a(b(b(x1)))) → B(b(b(x1)))
B(b(a(b(x1)))) → B(b(x1))
B(b(a(b(x1)))) → B(b(b(x1)))
B(b(b(x1))) → B(a(b(x1)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(a(b(x1)))) → b(a(b(b(b(b(x1))))))
b(a(b(b(x1)))) → b(b(b(x1)))
b(b(b(x1))) → b(b(a(a(b(a(b(x1)))))))
B(b(a(b(x1)))) → B(b(b(b(x1))))
B(b(a(b(x1)))) → B(a(b(b(b(b(x1))))))
B(a(b(b(x1)))) → B(b(b(x1)))
B(b(a(b(x1)))) → B(b(x1))
B(b(a(b(x1)))) → B(b(b(x1)))
B(b(b(x1))) → B(a(b(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
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(b(b(x))) → B1(a(a(b(b(x)))))
B1(b(B(x))) → B1(a(B(x)))
B1(a(b(B(x)))) → B1(b(a(B(x))))
B1(a(b(B(x)))) → B1(b(b(a(B(x)))))
B1(a(b(b(x)))) → B1(b(b(a(b(x)))))
B1(a(b(B(x)))) → B1(a(B(x)))
B1(a(b(b(x)))) → B1(b(b(b(a(b(x))))))
B1(b(a(B(x)))) → B1(b(B(x)))
B1(b(a(b(x)))) → B1(b(x))
B1(b(b(x))) → B1(a(b(a(a(b(b(x)))))))
B1(b(a(B(x)))) → B1(B(x))
B1(a(b(B(x)))) → B1(b(b(B(x))))
B1(a(b(b(x)))) → B1(a(b(x)))
B1(a(b(b(x)))) → B1(b(a(b(x))))
B1(b(a(b(x)))) → B1(b(b(x)))
B1(a(b(B(x)))) → B1(b(B(x)))
B1(a(b(B(x)))) → B1(b(b(b(a(B(x))))))
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(x))) → B1(a(a(b(b(x)))))
B1(b(B(x))) → B1(a(B(x)))
B1(a(b(B(x)))) → B1(b(a(B(x))))
B1(a(b(B(x)))) → B1(b(b(a(B(x)))))
B1(a(b(b(x)))) → B1(b(b(a(b(x)))))
B1(a(b(B(x)))) → B1(a(B(x)))
B1(a(b(b(x)))) → B1(b(b(b(a(b(x))))))
B1(b(a(B(x)))) → B1(b(B(x)))
B1(b(a(b(x)))) → B1(b(x))
B1(b(b(x))) → B1(a(b(a(a(b(b(x)))))))
B1(b(a(B(x)))) → B1(B(x))
B1(a(b(B(x)))) → B1(b(b(B(x))))
B1(a(b(b(x)))) → B1(a(b(x)))
B1(a(b(b(x)))) → B1(b(a(b(x))))
B1(b(a(b(x)))) → B1(b(b(x)))
B1(a(b(B(x)))) → B1(b(B(x)))
B1(a(b(B(x)))) → B1(b(b(b(a(B(x))))))
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 14 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(b(x)))) → B1(b(b(x)))
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(a(b(x)))) → B1(b(b(x))) at position [0] we obtained the following new rules:
B1(b(a(b(b(b(x0)))))) → B1(b(b(a(b(a(a(b(b(x0)))))))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(B(x0))))
B1(b(a(b(a(b(x0)))))) → B1(b(b(b(x0))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(b(x0)))))
B1(b(a(b(b(B(x0)))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(b(a(B(x0))))))))
B1(b(a(b(B(x0))))) → B1(b(a(B(x0))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(a(b(b(x0))))))) → B1(b(b(b(b(b(a(b(x0))))))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(a(B(x0)))))) → B1(b(b(B(x0))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(B(x0))))))
B1(b(a(b(b(x0))))) → B1(b(a(b(a(a(b(b(x0))))))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(b(b(b(x0)))))) → B1(b(b(a(b(a(a(b(b(x0)))))))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(B(x0))))
B1(b(a(b(a(b(x0)))))) → B1(b(b(b(x0))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(b(x0)))))
B1(b(a(b(b(B(x0)))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(b(a(B(x0))))))))
B1(b(a(b(B(x0))))) → B1(b(a(B(x0))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(a(b(b(x0))))))) → B1(b(b(b(b(b(a(b(x0))))))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(a(B(x0)))))) → B1(b(b(B(x0))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(B(x0))))))
B1(b(a(b(b(x0))))) → B1(b(a(b(a(a(b(b(x0))))))))
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(b(b(b(x0)))))) → B1(b(b(a(b(a(a(b(b(x0)))))))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(B(x0))))
B1(b(a(b(a(b(x0)))))) → B1(b(b(b(x0))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(b(x0)))))
B1(b(a(b(b(B(x0)))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(b(a(B(x0))))))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(a(b(b(x0))))))) → B1(b(b(b(b(b(a(b(x0))))))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(a(B(x0)))))) → B1(b(b(B(x0))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(B(x0))))))
B1(b(a(b(b(x0))))) → B1(b(a(b(a(a(b(b(x0))))))))
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(a(b(a(b(B(x0))))))) → B1(b(b(B(x0)))) at position [0] we obtained the following new rules:
B1(b(a(b(a(b(B(x0))))))) → B1(b(a(B(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(b(b(b(x0)))))) → B1(b(b(a(b(a(a(b(b(x0)))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(b(b(x0))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(b(x0)))))
B1(b(a(b(b(B(x0)))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(b(a(B(x0))))))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(b(a(b(B(x0))))))) → B1(b(a(B(x0))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(a(b(b(x0))))))) → B1(b(b(b(b(b(a(b(x0))))))))
B1(b(a(b(a(B(x0)))))) → B1(b(b(B(x0))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(B(x0))))))
B1(b(a(b(b(x0))))) → B1(b(a(b(a(a(b(b(x0))))))))
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(b(b(b(x0)))))) → B1(b(b(a(b(a(a(b(b(x0)))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(b(b(x0))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(b(x0)))))
B1(b(a(b(b(B(x0)))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(a(b(b(x0))))))) → B1(b(b(b(b(b(a(b(x0))))))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(b(a(B(x0))))))))
B1(b(a(b(a(B(x0)))))) → B1(b(b(B(x0))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(B(x0))))))
B1(b(a(b(b(x0))))) → B1(b(a(b(a(a(b(b(x0))))))))
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(a(b(a(B(x0)))))) → B1(b(b(B(x0)))) at position [0] we obtained the following new rules:
B1(b(a(b(a(B(x0)))))) → B1(b(a(B(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(b(b(b(x0)))))) → B1(b(b(a(b(a(a(b(b(x0)))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(b(b(x0))))
B1(b(a(b(a(B(x0)))))) → B1(b(a(B(x0))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(b(x0)))))
B1(b(a(b(b(B(x0)))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(b(a(B(x0))))))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(a(b(b(x0))))))) → B1(b(b(b(b(b(a(b(x0))))))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(B(x0))))))
B1(b(a(b(b(x0))))) → B1(b(a(b(a(a(b(b(x0))))))))
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(b(b(b(x0)))))) → B1(b(b(a(b(a(a(b(b(x0)))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(b(b(x0))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(b(x0)))))
B1(b(a(b(b(B(x0)))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(b(B(x0)))))
B1(b(a(b(a(b(b(x0))))))) → B1(b(b(b(b(b(a(b(x0))))))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(b(a(B(x0))))))))
B1(b(a(b(a(b(B(x0))))))) → B1(b(b(b(b(B(x0))))))
B1(b(a(b(b(x0))))) → B1(b(a(b(a(a(b(b(x0))))))))
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(b(x)))) → B1(a(b(x)))
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(b(x)))) → b(a(b(b(b(b(x))))))
b(a(b(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(b(a(a(b(a(b(x)))))))
B(b(a(b(x)))) → B(b(b(b(x))))
B(b(a(b(x)))) → B(a(b(b(b(b(x))))))
B(a(b(b(x)))) → B(b(b(x)))
B(b(a(b(x)))) → B(b(x))
B(b(a(b(x)))) → B(b(b(x)))
B(b(b(x))) → B(a(b(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(b(b(b(x))))))
b(a(b(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(b(a(a(b(a(b(x)))))))
B(b(a(b(x)))) → B(b(b(b(x))))
B(b(a(b(x)))) → B(a(b(b(b(b(x))))))
B(a(b(b(x)))) → B(b(b(x)))
B(b(a(b(x)))) → B(b(x))
B(b(a(b(x)))) → B(b(b(x)))
B(b(b(x))) → B(a(b(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
b(a(b(B(x)))) → b(b(b(B(x))))
b(a(b(B(x)))) → b(b(b(b(a(B(x))))))
b(b(a(B(x)))) → b(b(B(x)))
b(a(b(B(x)))) → b(B(x))
b(a(b(B(x)))) → b(b(B(x)))
b(b(B(x))) → b(a(B(x)))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(b(x)))) → b(a(b(b(b(b(x))))))
b(a(b(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(b(a(a(b(a(b(x)))))))
B(b(a(b(x)))) → B(b(b(b(x))))
B(b(a(b(x)))) → B(a(b(b(b(b(x))))))
B(a(b(b(x)))) → B(b(b(x)))
B(b(a(b(x)))) → B(b(x))
B(b(a(b(x)))) → B(b(b(x)))
B(b(b(x))) → B(a(b(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(b(b(b(x))))))
b(a(b(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(b(a(a(b(a(b(x)))))))
B(b(a(b(x)))) → B(b(b(b(x))))
B(b(a(b(x)))) → B(a(b(b(b(b(x))))))
B(a(b(b(x)))) → B(b(b(x)))
B(b(a(b(x)))) → B(b(x))
B(b(a(b(x)))) → B(b(b(x)))
B(b(b(x))) → B(a(b(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(a(b(x1)))) → b(a(b(b(b(b(x1))))))
b(a(b(b(x1)))) → b(b(b(x1)))
b(b(b(x1))) → b(b(a(a(b(a(b(x1)))))))
The set Q is empty.
We have obtained the following QTRS:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(b(a(b(x))))))
b(b(a(b(x)))) → b(b(b(x)))
b(b(b(x))) → b(a(b(a(a(b(b(x)))))))
Q is empty.