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(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(b(x1))))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(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:
A(b(b(a(x1)))) → A(b(a(b(b(x1)))))
A(b(b(a(x1)))) → A(b(b(x1)))
A(b(b(a(x1)))) → A(a(b(a(b(b(x1))))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(a(x1)))) → A(b(a(b(b(x1)))))
A(b(b(a(x1)))) → A(b(b(x1)))
A(b(b(a(x1)))) → A(a(b(a(b(b(x1))))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(a(x1)))) → A(a(b(a(b(b(x1)))))) at position [0] we obtained the following new rules:
A(b(b(a(a(x0))))) → A(a(b(a(a(b(a(b(b(x0)))))))))
A(b(b(a(y0)))) → A(b(a(b(b(y0)))))
A(b(b(a(y0)))) → A(a(b(b(b(y0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(a(a(x0))))) → A(a(b(a(a(b(a(b(b(x0)))))))))
A(b(b(a(x1)))) → A(b(a(b(b(x1)))))
A(b(b(a(y0)))) → A(a(b(b(b(y0)))))
A(b(b(a(x1)))) → A(b(b(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(a(x1)))) → A(b(a(b(b(x1))))) at position [0,0] we obtained the following new rules:
A(b(b(a(a(x0))))) → A(b(a(a(b(a(b(b(x0))))))))
A(b(b(a(y0)))) → A(b(b(b(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(a(a(x0))))) → A(a(b(a(a(b(a(b(b(x0)))))))))
A(b(b(a(y0)))) → A(a(b(b(b(y0)))))
A(b(b(a(x1)))) → A(b(b(x1)))
A(b(b(a(y0)))) → A(b(b(b(y0))))
A(b(b(a(a(x0))))) → A(b(a(a(b(a(b(b(x0))))))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(a(a(x0))))) → A(a(b(a(a(b(a(b(b(x0)))))))))
A(b(b(a(y0)))) → A(a(b(b(b(y0)))))
A(b(b(a(x1)))) → A(b(b(x1)))
A(b(b(a(a(x0))))) → A(b(a(a(b(a(b(b(x0))))))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(a(y0)))) → A(a(b(b(b(y0))))) at position [0] we obtained the following new rules:
A(b(b(a(y0)))) → A(b(b(b(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(a(a(x0))))) → A(a(b(a(a(b(a(b(b(x0)))))))))
A(b(b(a(x1)))) → A(b(b(x1)))
A(b(b(a(y0)))) → A(b(b(b(y0))))
A(b(b(a(a(x0))))) → A(b(a(a(b(a(b(b(x0))))))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(a(a(x0))))) → A(a(b(a(a(b(a(b(b(x0)))))))))
A(b(b(a(x1)))) → A(b(b(x1)))
A(b(b(a(a(x0))))) → A(b(a(a(b(a(b(b(x0))))))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(b(x1))))))
A(b(b(a(a(x0))))) → A(a(b(a(a(b(a(b(b(x0)))))))))
A(b(b(a(x1)))) → A(b(b(x1)))
A(b(b(a(a(x0))))) → A(b(a(a(b(a(b(b(x0))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(b(x1))))))
A(b(b(a(a(x0))))) → A(a(b(a(a(b(a(b(b(x0)))))))))
A(b(b(a(x1)))) → A(b(b(x1)))
A(b(b(a(a(x0))))) → A(b(a(a(b(a(b(b(x0))))))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(a(A(x)))))))))
a(b(b(A(x)))) → b(b(A(x)))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(A(x))))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(a(A(x)))))))))
a(b(b(A(x)))) → b(b(A(x)))
a(a(b(b(A(x))))) → b(b(a(b(a(a(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:
A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(x)))) → A1(b(a(a(x))))
A1(a(b(b(A(x))))) → A1(a(b(a(A(x)))))
A1(b(b(a(x)))) → A1(a(x))
A1(a(b(b(A(x))))) → A1(b(A(x)))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))
A1(a(b(b(A(x))))) → A1(a(b(A(x))))
A1(a(b(b(A(x))))) → A1(b(a(A(x))))
A1(a(b(b(A(x))))) → A1(A(x))
The TRS R consists of the following rules:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(a(A(x)))))))))
a(b(b(A(x)))) → b(b(A(x)))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(A(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(x)))) → A1(b(a(a(x))))
A1(a(b(b(A(x))))) → A1(a(b(a(A(x)))))
A1(b(b(a(x)))) → A1(a(x))
A1(a(b(b(A(x))))) → A1(b(A(x)))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))
A1(a(b(b(A(x))))) → A1(a(b(A(x))))
A1(a(b(b(A(x))))) → A1(b(a(A(x))))
A1(a(b(b(A(x))))) → A1(A(x))
The TRS R consists of the following rules:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(a(A(x)))))))))
a(b(b(A(x)))) → b(b(A(x)))
a(a(b(b(A(x))))) → b(b(a(b(a(a(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(x)))) → A1(b(a(a(x))))
A1(a(b(b(A(x))))) → A1(a(b(a(A(x)))))
A1(b(b(a(x)))) → A1(a(x))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))
A1(a(b(b(A(x))))) → A1(a(b(A(x))))
A1(a(b(b(A(x))))) → A1(b(a(A(x))))
The TRS R consists of the following rules:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(a(A(x)))))))))
a(b(b(A(x)))) → b(b(A(x)))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(A(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(b(b(A(x))))) → A1(a(b(a(A(x))))) at position [0] we obtained the following new rules:
A1(a(b(b(A(y0))))) → A1(a(b(A(y0))))
A1(a(b(b(A(y0))))) → A1(b(a(A(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(x)))) → A1(b(a(a(x))))
A1(b(b(a(x)))) → A1(a(x))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))
A1(a(b(b(A(x))))) → A1(b(a(A(x))))
A1(a(b(b(A(x))))) → A1(a(b(A(x))))
The TRS R consists of the following rules:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(a(A(x)))))))))
a(b(b(A(x)))) → b(b(A(x)))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(A(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(b(b(A(x))))) → A1(b(a(A(x)))) at position [0,0] we obtained the following new rules:
A1(a(b(b(A(y0))))) → A1(b(A(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(x)))) → A1(b(a(a(x))))
A1(b(b(a(x)))) → A1(a(x))
A1(a(b(b(A(y0))))) → A1(b(A(y0)))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))
A1(a(b(b(A(x))))) → A1(a(b(A(x))))
The TRS R consists of the following rules:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(a(A(x)))))))))
a(b(b(A(x)))) → b(b(A(x)))
a(a(b(b(A(x))))) → b(b(a(b(a(a(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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(x)))) → A1(b(a(a(x))))
A1(b(b(a(x)))) → A1(a(x))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))
A1(a(b(b(A(x))))) → A1(a(b(A(x))))
The TRS R consists of the following rules:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(a(A(x)))))))))
a(b(b(A(x)))) → b(b(A(x)))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(A(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(b(b(A(x))))) → A1(a(b(A(x)))) at position [0] we obtained the following new rules:
A1(a(b(b(A(y0))))) → A1(b(A(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(x)))) → A1(b(a(a(x))))
A1(b(b(a(x)))) → A1(a(x))
A1(a(b(b(A(y0))))) → A1(b(A(y0)))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))
The TRS R consists of the following rules:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(a(A(x)))))))))
a(b(b(A(x)))) → b(b(A(x)))
a(a(b(b(A(x))))) → b(b(a(b(a(a(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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(x)))) → A1(b(a(a(x))))
A1(b(b(a(x)))) → A1(a(x))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))
The TRS R consists of the following rules:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
a(a(b(b(A(x))))) → b(b(a(b(a(a(b(a(A(x)))))))))
a(b(b(A(x)))) → b(b(A(x)))
a(a(b(b(A(x))))) → b(b(a(b(a(a(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(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(b(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(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(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(b(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(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(x) → x
a(b(b(a(x)))) → b(b(a(b(a(a(x))))))
Q is empty.