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(a(x1)) → x1
a(b(b(b(x1)))) → b(b(b(a(b(a(x1))))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(b(b(x1)))) → b(b(b(a(b(a(x1))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → x1
a(b(b(b(x1)))) → b(b(b(a(b(a(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
b(b(b(a(x)))) → a(b(a(b(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:
a(a(x)) → x
b(b(b(a(x)))) → a(b(a(b(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:
A(b(b(b(x1)))) → A(x1)
A(b(b(b(x1)))) → A(b(a(x1)))
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(b(b(x1)))) → b(b(b(a(b(a(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(b(x1)))) → A(x1)
A(b(b(b(x1)))) → A(b(a(x1)))
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(b(b(x1)))) → b(b(b(a(b(a(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(b(x1)))) → A(b(a(x1))) at position [0,0] we obtained the following new rules:
A(b(b(b(a(x0))))) → A(b(x0))
A(b(b(b(b(b(b(x0))))))) → A(b(b(b(b(a(b(a(x0))))))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(b(x1)))) → A(x1)
A(b(b(b(b(b(b(x0))))))) → A(b(b(b(b(a(b(a(x0))))))))
A(b(b(b(a(x0))))) → A(b(x0))
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(b(b(x1)))) → b(b(b(a(b(a(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
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(b(b(x1)))) → b(b(b(a(b(a(x1))))))
A(b(b(b(x1)))) → A(x1)
A(b(b(b(b(b(b(x0))))))) → A(b(b(b(b(a(b(a(x0))))))))
A(b(b(b(a(x0))))) → A(b(x0))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → x1
a(b(b(b(x1)))) → b(b(b(a(b(a(x1))))))
A(b(b(b(x1)))) → A(x1)
A(b(b(b(b(b(b(x0))))))) → A(b(b(b(b(a(b(a(x0))))))))
A(b(b(b(a(x0))))) → A(b(x0))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
b(b(b(a(x)))) → a(b(a(b(b(b(x))))))
b(b(b(A(x)))) → A(x)
b(b(b(b(b(b(A(x))))))) → a(b(a(b(b(b(b(A(x))))))))
a(b(b(b(A(x))))) → b(A(x))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → x
b(b(b(a(x)))) → a(b(a(b(b(b(x))))))
b(b(b(A(x)))) → A(x)
b(b(b(b(b(b(A(x))))))) → a(b(a(b(b(b(b(A(x))))))))
a(b(b(b(A(x))))) → b(A(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(b(a(x)))) → B(b(b(x)))
B(b(b(b(b(b(A(x))))))) → A1(b(a(b(b(b(b(A(x))))))))
B(b(b(b(b(b(A(x))))))) → A1(b(b(b(b(A(x))))))
B(b(b(a(x)))) → B(x)
B(b(b(a(x)))) → A1(b(a(b(b(b(x))))))
B(b(b(a(x)))) → A1(b(b(b(x))))
B(b(b(b(b(b(A(x))))))) → B(a(b(b(b(b(A(x)))))))
B(b(b(a(x)))) → B(b(x))
B(b(b(a(x)))) → B(a(b(b(b(x)))))
The TRS R consists of the following rules:
a(a(x)) → x
b(b(b(a(x)))) → a(b(a(b(b(b(x))))))
b(b(b(A(x)))) → A(x)
b(b(b(b(b(b(A(x))))))) → a(b(a(b(b(b(b(A(x))))))))
a(b(b(b(A(x))))) → b(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(a(x)))) → B(b(b(x)))
B(b(b(b(b(b(A(x))))))) → A1(b(a(b(b(b(b(A(x))))))))
B(b(b(b(b(b(A(x))))))) → A1(b(b(b(b(A(x))))))
B(b(b(a(x)))) → B(x)
B(b(b(a(x)))) → A1(b(a(b(b(b(x))))))
B(b(b(a(x)))) → A1(b(b(b(x))))
B(b(b(b(b(b(A(x))))))) → B(a(b(b(b(b(A(x)))))))
B(b(b(a(x)))) → B(b(x))
B(b(b(a(x)))) → B(a(b(b(b(x)))))
The TRS R consists of the following rules:
a(a(x)) → x
b(b(b(a(x)))) → a(b(a(b(b(b(x))))))
b(b(b(A(x)))) → A(x)
b(b(b(b(b(b(A(x))))))) → a(b(a(b(b(b(b(A(x))))))))
a(b(b(b(A(x))))) → b(A(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 4 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(a(x)))) → B(b(b(x)))
B(b(b(a(x)))) → B(x)
B(b(b(b(b(b(A(x))))))) → B(a(b(b(b(b(A(x)))))))
B(b(b(a(x)))) → B(b(x))
B(b(b(a(x)))) → B(a(b(b(b(x)))))
The TRS R consists of the following rules:
a(a(x)) → x
b(b(b(a(x)))) → a(b(a(b(b(b(x))))))
b(b(b(A(x)))) → A(x)
b(b(b(b(b(b(A(x))))))) → a(b(a(b(b(b(b(A(x))))))))
a(b(b(b(A(x))))) → b(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(b(b(b(b(A(x))))))) → B(a(b(b(b(b(A(x))))))) at position [0] we obtained the following new rules:
B(b(b(b(b(b(A(x0))))))) → B(a(b(A(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(a(x)))) → B(b(b(x)))
B(b(b(a(x)))) → B(x)
B(b(b(a(x)))) → B(b(x))
B(b(b(a(x)))) → B(a(b(b(b(x)))))
B(b(b(b(b(b(A(x0))))))) → B(a(b(A(x0))))
The TRS R consists of the following rules:
a(a(x)) → x
b(b(b(a(x)))) → a(b(a(b(b(b(x))))))
b(b(b(A(x)))) → A(x)
b(b(b(b(b(b(A(x))))))) → a(b(a(b(b(b(b(A(x))))))))
a(b(b(b(A(x))))) → b(A(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
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(a(x)))) → B(b(b(x)))
B(b(b(a(x)))) → B(x)
B(b(b(a(x)))) → B(b(x))
B(b(b(a(x)))) → B(a(b(b(b(x)))))
The TRS R consists of the following rules:
a(a(x)) → x
b(b(b(a(x)))) → a(b(a(b(b(b(x))))))
b(b(b(A(x)))) → A(x)
b(b(b(b(b(b(A(x))))))) → a(b(a(b(b(b(b(A(x))))))))
a(b(b(b(A(x))))) → b(A(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
a(a(x1)) → x1
a(b(b(b(x1)))) → b(b(b(a(b(a(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
b(b(b(a(x)))) → a(b(a(b(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:
a(a(x)) → x
b(b(b(a(x)))) → a(b(a(b(b(b(x))))))
Q is empty.