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(x1)) → x1
b(b(a(x1))) → 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(a(x1)) → x1
a(b(x1)) → x1
b(b(a(x1))) → 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:
B(b(a(x1))) → B(b(x1))
B(b(a(x1))) → B(x1)
B(b(a(x1))) → A(b(b(x1)))
B(b(a(x1))) → B(a(b(b(x1))))
B(b(a(x1))) → A(b(a(b(b(x1)))))
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(x1)) → x1
b(b(a(x1))) → a(b(a(b(b(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:
B(b(a(x1))) → B(b(x1))
B(b(a(x1))) → B(x1)
B(b(a(x1))) → A(b(b(x1)))
B(b(a(x1))) → B(a(b(b(x1))))
B(b(a(x1))) → A(b(a(b(b(x1)))))
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(x1)) → x1
b(b(a(x1))) → 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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(x1))) → B(b(x1))
B(b(a(x1))) → B(x1)
B(b(a(x1))) → B(a(b(b(x1))))
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(x1)) → x1
b(b(a(x1))) → a(b(a(b(b(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(a(x1))) → B(a(b(b(x1)))) at position [0] we obtained the following new rules:
B(b(a(y0))) → B(b(y0))
B(b(a(b(a(x0))))) → B(a(b(a(b(a(b(b(x0))))))))
B(b(a(a(x0)))) → B(a(a(b(a(b(b(x0)))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(x1))) → B(b(x1))
B(b(a(x1))) → B(x1)
B(b(a(a(x0)))) → B(a(a(b(a(b(b(x0)))))))
B(b(a(b(a(x0))))) → B(a(b(a(b(a(b(b(x0))))))))
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(x1)) → x1
b(b(a(x1))) → 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
↳ 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(x1)) → x1
b(b(a(x1))) → a(b(a(b(b(x1)))))
B(b(a(x1))) → B(b(x1))
B(b(a(x1))) → B(x1)
B(b(a(a(x0)))) → B(a(a(b(a(b(b(x0)))))))
B(b(a(b(a(x0))))) → B(a(b(a(b(a(b(b(x0))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → x1
a(b(x1)) → x1
b(b(a(x1))) → a(b(a(b(b(x1)))))
B(b(a(x1))) → B(b(x1))
B(b(a(x1))) → B(x1)
B(b(a(a(x0)))) → B(a(a(b(a(b(b(x0)))))))
B(b(a(b(a(x0))))) → B(a(b(a(b(a(b(b(x0))))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(b(a(B(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(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(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:
A(b(a(b(B(x))))) → A(b(a(b(a(B(x))))))
A(a(b(B(x)))) → B1(b(a(b(a(a(B(x)))))))
A(b(b(x))) → B1(a(x))
A(b(a(b(B(x))))) → A(B(x))
A(b(b(x))) → B1(a(b(a(x))))
A(b(b(x))) → A(b(a(x)))
A(a(b(B(x)))) → B1(a(b(a(a(B(x))))))
A(b(a(b(B(x))))) → B1(b(a(b(a(b(a(B(x))))))))
A(b(b(x))) → B1(b(a(b(a(x)))))
A(b(a(b(B(x))))) → B1(a(b(a(B(x)))))
A(b(b(x))) → A(x)
A(a(b(B(x)))) → B1(a(a(B(x))))
A(b(a(b(B(x))))) → B1(a(b(a(b(a(B(x)))))))
A(a(b(B(x)))) → A(b(a(a(B(x)))))
A(a(b(B(x)))) → A(B(x))
A(a(b(B(x)))) → A(a(B(x)))
A(b(a(b(B(x))))) → B1(a(B(x)))
A(b(a(b(B(x))))) → A(b(a(B(x))))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(b(a(B(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:
A(b(a(b(B(x))))) → A(b(a(b(a(B(x))))))
A(a(b(B(x)))) → B1(b(a(b(a(a(B(x)))))))
A(b(b(x))) → B1(a(x))
A(b(a(b(B(x))))) → A(B(x))
A(b(b(x))) → B1(a(b(a(x))))
A(b(b(x))) → A(b(a(x)))
A(a(b(B(x)))) → B1(a(b(a(a(B(x))))))
A(b(a(b(B(x))))) → B1(b(a(b(a(b(a(B(x))))))))
A(b(b(x))) → B1(b(a(b(a(x)))))
A(b(a(b(B(x))))) → B1(a(b(a(B(x)))))
A(b(b(x))) → A(x)
A(a(b(B(x)))) → B1(a(a(B(x))))
A(b(a(b(B(x))))) → B1(a(b(a(b(a(B(x)))))))
A(a(b(B(x)))) → A(b(a(a(B(x)))))
A(a(b(B(x)))) → A(B(x))
A(a(b(B(x)))) → A(a(B(x)))
A(b(a(b(B(x))))) → B1(a(B(x)))
A(b(a(b(B(x))))) → A(b(a(B(x))))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(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 13 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:
A(b(b(x))) → A(x)
A(b(a(b(B(x))))) → A(b(a(b(a(B(x))))))
A(a(b(B(x)))) → A(b(a(a(B(x)))))
A(b(b(x))) → A(b(a(x)))
A(b(a(b(B(x))))) → A(b(a(B(x))))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(b(a(B(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(b(B(x)))) → A(b(a(a(B(x))))) at position [0] we obtained the following new rules:
A(a(b(B(y0)))) → A(a(B(y0)))
A(a(b(B(y0)))) → A(b(B(y0)))
↳ 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:
A(b(b(x))) → A(x)
A(b(a(b(B(x))))) → A(b(a(b(a(B(x))))))
A(a(b(B(y0)))) → A(a(B(y0)))
A(b(b(x))) → A(b(a(x)))
A(a(b(B(y0)))) → A(b(B(y0)))
A(b(a(b(B(x))))) → A(b(a(B(x))))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(x))) → A(x)
A(b(a(b(B(x))))) → A(b(a(b(a(B(x))))))
A(b(b(x))) → A(b(a(x)))
A(b(a(b(B(x))))) → A(b(a(B(x))))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(b(a(B(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(a(b(B(x))))) → A(b(a(b(a(B(x)))))) at position [0] we obtained the following new rules:
A(b(a(b(B(y0))))) → A(b(a(B(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(x))) → A(x)
A(b(b(x))) → A(b(a(x)))
A(b(a(b(B(x))))) → A(b(a(B(x))))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(b(a(B(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(a(b(B(x))))) → A(b(a(B(x)))) at position [0] we obtained the following new rules:
A(b(a(b(B(y0))))) → A(B(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(x))) → A(x)
A(b(a(b(B(y0))))) → A(B(y0))
A(b(b(x))) → A(b(a(x)))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(x))) → A(x)
A(b(b(x))) → A(b(a(x)))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(b(a(B(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(x))) → A(b(a(x))) at position [0] we obtained the following new rules:
A(b(b(x0))) → A(x0)
A(b(b(b(B(x0))))) → A(b(b(B(x0))))
A(b(b(b(B(x0))))) → A(b(B(x0)))
A(b(b(a(b(B(x0)))))) → A(b(b(b(a(b(a(a(B(x0)))))))))
A(b(b(a(x0)))) → A(b(x0))
A(b(b(b(b(x0))))) → A(b(b(b(a(b(a(x0)))))))
A(b(b(b(a(b(B(x0))))))) → A(b(b(b(a(b(a(b(a(B(x0))))))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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:
A(b(b(b(B(x0))))) → A(b(b(B(x0))))
A(b(b(x))) → A(x)
A(b(b(b(B(x0))))) → A(b(B(x0)))
A(b(b(a(b(B(x0)))))) → A(b(b(b(a(b(a(a(B(x0)))))))))
A(b(b(a(x0)))) → A(b(x0))
A(b(b(b(a(b(B(x0))))))) → A(b(b(b(a(b(a(b(a(B(x0))))))))))
A(b(b(b(b(x0))))) → A(b(b(b(a(b(a(x0)))))))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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:
A(b(b(b(B(x0))))) → A(b(b(B(x0))))
A(b(b(x))) → A(x)
A(b(b(a(b(B(x0)))))) → A(b(b(b(a(b(a(a(B(x0)))))))))
A(b(b(a(x0)))) → A(b(x0))
A(b(b(b(a(b(B(x0))))))) → A(b(b(b(a(b(a(b(a(B(x0))))))))))
A(b(b(b(b(x0))))) → A(b(b(b(a(b(a(x0)))))))
The TRS R consists of the following rules:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(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
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(b(a(B(x))))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
a(b(x)) → x
b(b(a(x))) → a(b(a(b(b(x)))))
B(b(a(x))) → B(b(x))
B(b(a(x))) → B(x)
B(b(a(a(x)))) → B(a(a(b(a(b(b(x)))))))
B(b(a(b(a(x))))) → B(a(b(a(b(a(b(b(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(x)) → x
b(b(a(x))) → a(b(a(b(b(x)))))
B(b(a(x))) → B(b(x))
B(b(a(x))) → B(x)
B(b(a(a(x)))) → B(a(a(b(a(b(b(x)))))))
B(b(a(b(a(x))))) → B(a(b(a(b(a(b(b(x))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
a(b(B(x))) → b(B(x))
a(b(B(x))) → B(x)
a(a(b(B(x)))) → b(b(a(b(a(a(B(x)))))))
a(b(a(b(B(x))))) → b(b(a(b(a(b(a(B(x))))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
a(b(x)) → x
b(b(a(x))) → a(b(a(b(b(x)))))
B(b(a(x))) → B(b(x))
B(b(a(x))) → B(x)
B(b(a(a(x)))) → B(a(a(b(a(b(b(x)))))))
B(b(a(b(a(x))))) → B(a(b(a(b(a(b(b(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(x)) → x
b(b(a(x))) → a(b(a(b(b(x)))))
B(b(a(x))) → B(b(x))
B(b(a(x))) → B(x)
B(b(a(a(x)))) → B(a(a(b(a(b(b(x)))))))
B(b(a(b(a(x))))) → B(a(b(a(b(a(b(b(x))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → x1
a(b(x1)) → x1
b(b(a(x1))) → a(b(a(b(b(x1)))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(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(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → x1
a(b(x1)) → x1
b(b(a(x1))) → a(b(a(b(b(x1)))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(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(a(x)) → x
b(a(x)) → x
a(b(b(x))) → b(b(a(b(a(x)))))
Q is empty.