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(a(b(x1)))) → b(a(b(b(a(a(x1))))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(a(b(x1)))) → b(a(b(b(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:
A(b(a(b(x1)))) → A(a(x1))
A(b(a(b(x1)))) → A(b(b(a(a(x1)))))
A(b(a(b(x1)))) → A(x1)
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(a(b(x1)))) → b(a(b(b(a(a(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(a(b(x1)))) → A(a(x1))
A(b(a(b(x1)))) → A(b(b(a(a(x1)))))
A(b(a(b(x1)))) → A(x1)
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(a(b(x1)))) → b(a(b(b(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
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(a(b(x1)))) → A(a(x1))
A(b(a(b(x1)))) → A(x1)
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(a(b(x1)))) → b(a(b(b(a(a(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(a(b(x1)))) → A(a(x1)) at position [0] we obtained the following new rules:
A(b(a(b(b(a(b(x0))))))) → A(b(a(b(b(a(a(x0)))))))
A(b(a(b(a(x0))))) → A(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(a(b(b(a(b(x0))))))) → A(b(a(b(b(a(a(x0)))))))
A(b(a(b(a(x0))))) → A(x0)
A(b(a(b(x1)))) → A(x1)
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(a(b(x1)))) → b(a(b(b(a(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(a(b(x1)))) → b(a(b(b(a(a(x1))))))
A(b(a(b(b(a(b(x0))))))) → A(b(a(b(b(a(a(x0)))))))
A(b(a(b(a(x0))))) → A(x0)
A(b(a(b(x1)))) → A(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → x1
a(b(a(b(x1)))) → b(a(b(b(a(a(x1))))))
A(b(a(b(b(a(b(x0))))))) → A(b(a(b(b(a(a(x0)))))))
A(b(a(b(a(x0))))) → A(x0)
A(b(a(b(x1)))) → A(x1)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
b(a(b(b(a(b(A(x))))))) → a(a(b(b(a(b(A(x)))))))
a(b(a(b(A(x))))) → A(x)
b(a(b(A(x)))) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
b(a(b(b(a(b(A(x))))))) → a(a(b(b(a(b(A(x)))))))
a(b(a(b(A(x))))) → A(x)
b(a(b(A(x)))) → 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(a(b(b(a(b(A(x))))))) → A1(a(b(b(a(b(A(x)))))))
B(a(b(a(x)))) → A1(b(b(a(b(x)))))
B(a(b(a(x)))) → B(a(b(x)))
B(a(b(a(x)))) → A1(a(b(b(a(b(x))))))
B(a(b(a(x)))) → B(x)
B(a(b(a(x)))) → B(b(a(b(x))))
B(a(b(a(x)))) → A1(b(x))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
b(a(b(b(a(b(A(x))))))) → a(a(b(b(a(b(A(x)))))))
a(b(a(b(A(x))))) → A(x)
b(a(b(A(x)))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(b(a(b(A(x))))))) → A1(a(b(b(a(b(A(x)))))))
B(a(b(a(x)))) → A1(b(b(a(b(x)))))
B(a(b(a(x)))) → B(a(b(x)))
B(a(b(a(x)))) → A1(a(b(b(a(b(x))))))
B(a(b(a(x)))) → B(x)
B(a(b(a(x)))) → B(b(a(b(x))))
B(a(b(a(x)))) → A1(b(x))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
b(a(b(b(a(b(A(x))))))) → a(a(b(b(a(b(A(x)))))))
a(b(a(b(A(x))))) → A(x)
b(a(b(A(x)))) → 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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(a(x)))) → B(a(b(x)))
B(a(b(a(x)))) → B(x)
B(a(b(a(x)))) → B(b(a(b(x))))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
b(a(b(b(a(b(A(x))))))) → a(a(b(b(a(b(A(x)))))))
a(b(a(b(A(x))))) → A(x)
b(a(b(A(x)))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(b(a(x)))) → B(b(a(b(x)))) at position [0] we obtained the following new rules:
B(a(b(a(a(b(A(x0))))))) → B(b(A(x0)))
B(a(b(a(a(x0))))) → B(a(a(b(b(a(b(x0)))))))
B(a(b(a(a(b(A(x0))))))) → B(b(a(A(x0))))
B(a(b(a(A(x0))))) → B(A(x0))
B(a(b(a(a(b(a(x0))))))) → B(b(a(a(a(b(b(a(b(x0)))))))))
B(a(b(a(a(b(b(a(b(A(x0)))))))))) → B(b(a(a(a(b(b(a(b(A(x0))))))))))
B(a(b(a(b(a(b(A(x0)))))))) → B(a(a(b(b(a(b(A(x0))))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(a(a(b(A(x0))))))) → B(b(A(x0)))
B(a(b(a(a(x0))))) → B(a(a(b(b(a(b(x0)))))))
B(a(b(a(a(b(A(x0))))))) → B(b(a(A(x0))))
B(a(b(a(x)))) → B(a(b(x)))
B(a(b(a(A(x0))))) → B(A(x0))
B(a(b(a(x)))) → B(x)
B(a(b(a(a(b(a(x0))))))) → B(b(a(a(a(b(b(a(b(x0)))))))))
B(a(b(a(a(b(b(a(b(A(x0)))))))))) → B(b(a(a(a(b(b(a(b(A(x0))))))))))
B(a(b(a(b(a(b(A(x0)))))))) → B(a(a(b(b(a(b(A(x0))))))))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
b(a(b(b(a(b(A(x))))))) → a(a(b(b(a(b(A(x)))))))
a(b(a(b(A(x))))) → A(x)
b(a(b(A(x)))) → 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 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(a(a(x0))))) → B(a(a(b(b(a(b(x0)))))))
B(a(b(a(x)))) → B(a(b(x)))
B(a(b(a(x)))) → B(x)
B(a(b(a(a(b(a(x0))))))) → B(b(a(a(a(b(b(a(b(x0)))))))))
B(a(b(a(a(b(b(a(b(A(x0)))))))))) → B(b(a(a(a(b(b(a(b(A(x0))))))))))
B(a(b(a(b(a(b(A(x0)))))))) → B(a(a(b(b(a(b(A(x0))))))))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
b(a(b(b(a(b(A(x))))))) → a(a(b(b(a(b(A(x)))))))
a(b(a(b(A(x))))) → A(x)
b(a(b(A(x)))) → 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(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
b(a(b(b(a(b(A(x))))))) → a(a(b(b(a(b(A(x)))))))
a(b(a(b(A(x))))) → A(x)
b(a(b(A(x)))) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
a(b(a(b(x)))) → b(a(b(b(a(a(x))))))
A(b(a(b(b(a(b(x))))))) → A(b(a(b(b(a(a(x)))))))
A(b(a(b(a(x))))) → A(x)
A(b(a(b(x)))) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → x
a(b(a(b(x)))) → b(a(b(b(a(a(x))))))
A(b(a(b(b(a(b(x))))))) → A(b(a(b(b(a(a(x)))))))
A(b(a(b(a(x))))) → A(x)
A(b(a(b(x)))) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
b(a(b(b(a(b(A(x))))))) → a(a(b(b(a(b(A(x)))))))
a(b(a(b(A(x))))) → A(x)
b(a(b(A(x)))) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
a(b(a(b(x)))) → b(a(b(b(a(a(x))))))
A(b(a(b(b(a(b(x))))))) → A(b(a(b(b(a(a(x)))))))
A(b(a(b(a(x))))) → A(x)
A(b(a(b(x)))) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → x
a(b(a(b(x)))) → b(a(b(b(a(a(x))))))
A(b(a(b(b(a(b(x))))))) → A(b(a(b(b(a(a(x)))))))
A(b(a(b(a(x))))) → A(x)
A(b(a(b(x)))) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → x1
a(b(a(b(x1)))) → b(a(b(b(a(a(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → x1
a(b(a(b(x1)))) → b(a(b(b(a(a(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
Q is empty.