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(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(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(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(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(a(a(b(x1)))) → B(a(b(a(x1))))
A(a(a(b(x1)))) → A(x1)
B(b(a(x1))) → A(b(a(x1)))
A(a(a(b(x1)))) → A(b(a(x1)))
B(b(a(x1))) → A(a(b(a(x1))))
A(a(a(b(x1)))) → B(a(x1))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(a(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(a(a(b(x1)))) → B(a(b(a(x1))))
A(a(a(b(x1)))) → A(x1)
B(b(a(x1))) → A(b(a(x1)))
A(a(a(b(x1)))) → A(b(a(x1)))
B(b(a(x1))) → A(a(b(a(x1))))
A(a(a(b(x1)))) → B(a(x1))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(a(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(a(b(x1)))) → B(a(b(a(x1)))) at position [0] we obtained the following new rules:
A(a(a(b(a(a(b(x0))))))) → B(a(b(b(a(b(a(x0)))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(x1)))) → A(x1)
A(a(a(b(a(a(b(x0))))))) → B(a(b(b(a(b(a(x0)))))))
B(b(a(x1))) → A(b(a(x1)))
A(a(a(b(x1)))) → A(b(a(x1)))
A(a(a(b(x1)))) → B(a(x1))
B(b(a(x1))) → A(a(b(a(x1))))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(a(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(a(b(x1)))) → A(b(a(x1))) at position [0] we obtained the following new rules:
A(a(a(b(a(a(b(x0))))))) → A(b(b(a(b(a(x0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(a(a(b(x0))))))) → A(b(b(a(b(a(x0))))))
A(a(a(b(x1)))) → A(x1)
B(b(a(x1))) → A(b(a(x1)))
A(a(a(b(a(a(b(x0))))))) → B(a(b(b(a(b(a(x0)))))))
B(b(a(x1))) → A(a(b(a(x1))))
A(a(a(b(x1)))) → B(a(x1))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(a(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(a(b(x1)))) → B(a(x1)) at position [0] we obtained the following new rules:
A(a(a(b(a(a(b(x0))))))) → B(b(a(b(a(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(a(a(b(x0))))))) → A(b(b(a(b(a(x0))))))
A(a(a(b(a(a(b(x0))))))) → B(b(a(b(a(x0)))))
A(a(a(b(x1)))) → A(x1)
A(a(a(b(a(a(b(x0))))))) → B(a(b(b(a(b(a(x0)))))))
B(b(a(x1))) → A(b(a(x1)))
B(b(a(x1))) → A(a(b(a(x1))))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(a(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(a(x1))) → A(a(b(a(x1)))) at position [0] we obtained the following new rules:
B(b(a(a(a(b(x0)))))) → A(a(b(b(a(b(a(x0)))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(a(a(b(x0))))))) → A(b(b(a(b(a(x0))))))
B(b(a(a(a(b(x0)))))) → A(a(b(b(a(b(a(x0)))))))
A(a(a(b(x1)))) → A(x1)
A(a(a(b(a(a(b(x0))))))) → B(b(a(b(a(x0)))))
B(b(a(x1))) → A(b(a(x1)))
A(a(a(b(a(a(b(x0))))))) → B(a(b(b(a(b(a(x0)))))))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(a(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(a(x1))) → A(b(a(x1))) at position [0] we obtained the following new rules:
B(b(a(a(a(b(x0)))))) → A(b(b(a(b(a(x0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(a(a(b(x0))))))) → A(b(b(a(b(a(x0))))))
A(a(a(b(a(a(b(x0))))))) → B(b(a(b(a(x0)))))
A(a(a(b(x1)))) → A(x1)
B(b(a(a(a(b(x0)))))) → A(a(b(b(a(b(a(x0)))))))
A(a(a(b(a(a(b(x0))))))) → B(a(b(b(a(b(a(x0)))))))
B(b(a(a(a(b(x0)))))) → A(b(b(a(b(a(x0))))))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(a(x1))))
A(a(a(b(a(a(b(x0))))))) → A(b(b(a(b(a(x0))))))
A(a(a(b(a(a(b(x0))))))) → B(b(a(b(a(x0)))))
A(a(a(b(x1)))) → A(x1)
B(b(a(a(a(b(x0)))))) → A(a(b(b(a(b(a(x0)))))))
A(a(a(b(a(a(b(x0))))))) → B(a(b(b(a(b(a(x0)))))))
B(b(a(a(a(b(x0)))))) → A(b(b(a(b(a(x0))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(a(x1))))
A(a(a(b(a(a(b(x0))))))) → A(b(b(a(b(a(x0))))))
A(a(a(b(a(a(b(x0))))))) → B(b(a(b(a(x0)))))
A(a(a(b(x1)))) → A(x1)
B(b(a(a(a(b(x0)))))) → A(a(b(b(a(b(a(x0)))))))
A(a(a(b(a(a(b(x0))))))) → B(a(b(b(a(b(a(x0)))))))
B(b(a(a(a(b(x0)))))) → A(b(b(a(b(a(x0))))))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(a(b(x)))) → b(a(b(a(x))))
b(b(a(x))) → a(a(b(a(x))))
A(a(a(b(a(a(b(x))))))) → A(b(b(a(b(a(x))))))
A(a(a(b(a(a(b(x))))))) → B(b(a(b(a(x)))))
A(a(a(b(x)))) → A(x)
B(b(a(a(a(b(x)))))) → A(a(b(b(a(b(a(x)))))))
A(a(a(b(a(a(b(x))))))) → B(a(b(b(a(b(a(x)))))))
B(b(a(a(a(b(x)))))) → A(b(b(a(b(a(x))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(b(x)))) → b(a(b(a(x))))
b(b(a(x))) → a(a(b(a(x))))
A(a(a(b(a(a(b(x))))))) → A(b(b(a(b(a(x))))))
A(a(a(b(a(a(b(x))))))) → B(b(a(b(a(x)))))
A(a(a(b(x)))) → A(x)
B(b(a(a(a(b(x)))))) → A(a(b(b(a(b(a(x)))))))
A(a(a(b(a(a(b(x))))))) → B(a(b(b(a(b(a(x)))))))
B(b(a(a(a(b(x)))))) → A(b(b(a(b(a(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(a(b(x)))) → b(a(b(a(x))))
b(b(a(x))) → a(a(b(a(x))))
A(a(a(b(a(a(b(x))))))) → A(b(b(a(b(a(x))))))
A(a(a(b(a(a(b(x))))))) → B(b(a(b(a(x)))))
A(a(a(b(x)))) → A(x)
B(b(a(a(a(b(x)))))) → A(a(b(b(a(b(a(x)))))))
A(a(a(b(a(a(b(x))))))) → B(a(b(b(a(b(a(x)))))))
B(b(a(a(a(b(x)))))) → A(b(b(a(b(a(x))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(b(x)))) → b(a(b(a(x))))
b(b(a(x))) → a(a(b(a(x))))
A(a(a(b(a(a(b(x))))))) → A(b(b(a(b(a(x))))))
A(a(a(b(a(a(b(x))))))) → B(b(a(b(a(x)))))
A(a(a(b(x)))) → A(x)
B(b(a(a(a(b(x)))))) → A(a(b(b(a(b(a(x)))))))
A(a(a(b(a(a(b(x))))))) → B(a(b(b(a(b(a(x)))))))
B(b(a(a(a(b(x)))))) → A(b(b(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:
B1(a(a(a(x)))) → B1(a(b(x)))
B1(a(a(b(a(a(A(x))))))) → B1(B(x))
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(a(b(B(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(a(b(b(A(x))))))
B1(a(a(b(a(a(A(x))))))) → A1(b(a(b(b(a(B(x)))))))
B1(a(a(a(b(B(x)))))) → B1(a(A(x)))
B1(a(a(a(b(B(x)))))) → B1(b(A(x)))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(a(A(x))))))
B1(a(a(b(a(a(A(x))))))) → B1(a(B(x)))
B1(a(a(a(x)))) → A1(b(a(b(x))))
A1(b(b(x))) → A1(a(x))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(A(x)))))
B1(a(a(a(b(B(x)))))) → A1(b(a(b(b(A(x))))))
A1(b(b(x))) → A1(b(a(a(x))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(a(b(B(x)))))) → A1(b(a(b(b(a(A(x)))))))
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(a(b(B(x)))))) → B1(b(a(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(b(B(x)))))) → B1(A(x))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(A(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(B(x)))
B1(a(a(b(a(a(A(x))))))) → B1(b(A(x)))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(B(x))))
B1(a(a(a(b(B(x)))))) → A1(A(x))
B1(a(a(b(a(a(A(x))))))) → B1(b(a(B(x))))
B1(a(a(b(a(a(A(x))))))) → A1(B(x))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(a(B(x))))))
B1(a(a(b(a(a(A(x))))))) → B1(A(x))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(x)))) → B1(a(b(x)))
B1(a(a(b(a(a(A(x))))))) → B1(B(x))
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(a(b(B(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(a(b(b(A(x))))))
B1(a(a(b(a(a(A(x))))))) → A1(b(a(b(b(a(B(x)))))))
B1(a(a(a(b(B(x)))))) → B1(a(A(x)))
B1(a(a(a(b(B(x)))))) → B1(b(A(x)))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(a(A(x))))))
B1(a(a(b(a(a(A(x))))))) → B1(a(B(x)))
B1(a(a(a(x)))) → A1(b(a(b(x))))
A1(b(b(x))) → A1(a(x))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(A(x)))))
B1(a(a(a(b(B(x)))))) → A1(b(a(b(b(A(x))))))
A1(b(b(x))) → A1(b(a(a(x))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(a(b(B(x)))))) → A1(b(a(b(b(a(A(x)))))))
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(a(b(B(x)))))) → B1(b(a(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(b(B(x)))))) → B1(A(x))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(A(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(B(x)))
B1(a(a(b(a(a(A(x))))))) → B1(b(A(x)))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(B(x))))
B1(a(a(a(b(B(x)))))) → A1(A(x))
B1(a(a(b(a(a(A(x))))))) → B1(b(a(B(x))))
B1(a(a(b(a(a(A(x))))))) → A1(B(x))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(a(B(x))))))
B1(a(a(b(a(a(A(x))))))) → B1(A(x))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(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 21 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(x)))) → B1(a(b(x)))
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(A(x)))))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(a(A(x))))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(A(x)))))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(a(B(x))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(a(x)))) → B1(a(b(x))) at position [0] we obtained the following new rules:
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(B(x0)))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(A(x0))))))) → B1(a(A(x0)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(B(x0)))))))
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(a(a(A(x0))))))) → B1(a(A(x0)))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(A(x)))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(a(A(x))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(A(x)))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(a(B(x))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(A(x)))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(a(A(x))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(A(x)))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(a(B(x))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(A(x))))) at position [0] we obtained the following new rules:
B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(A(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(a(A(x))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(A(y0))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(A(x)))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(a(B(x))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(b(a(a(A(x))))))) → B1(a(b(b(a(B(x)))))) at position [0] we obtained the following new rules:
B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(a(B(y0)))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(a(A(x))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(a(B(y0)))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(A(y0))))))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(A(x)))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(a(b(B(x)))))) → B1(a(b(b(a(A(x)))))) at position [0] we obtained the following new rules:
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(a(A(y0)))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(a(A(y0)))))))
B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(a(B(y0)))))))
B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(A(y0))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(b(B(x)))))) → B1(a(b(b(A(x)))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(a(b(B(x)))))) → B1(a(b(b(A(x))))) at position [0] we obtained the following new rules:
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(A(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(a(B(y0)))))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(a(A(y0)))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(A(y0))))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(A(y0))))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(A(y0)))))) at position [0] we obtained the following new rules:
B1(a(a(b(a(a(A(x0))))))) → B1(a(A(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(b(a(a(A(x0))))))) → B1(a(A(x0)))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(a(A(y0)))))))
B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(a(B(y0)))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(A(y0))))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(a(A(y0)))))))
B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(a(B(y0)))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(A(y0))))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(a(a(B(y0))))))) at position [0] we obtained the following new rules:
B1(a(a(b(a(a(A(y0))))))) → B1(a(a(b(a(b(B(y0)))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(a(A(y0)))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(A(y0))))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
B1(a(a(b(a(a(A(y0))))))) → B1(a(a(b(a(b(B(y0)))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(a(A(y0)))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(A(y0))))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(a(A(y0))))))) at position [0] we obtained the following new rules:
B1(a(a(a(b(B(y0)))))) → B1(a(a(b(a(b(A(y0)))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
B1(a(a(a(b(B(y0)))))) → B1(a(a(b(a(b(A(y0)))))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(A(y0))))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(A(y0))))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(a(b(B(y0)))))) → B1(a(b(a(a(A(y0)))))) at position [0] we obtained the following new rules:
B1(a(a(a(b(B(x0)))))) → B1(a(A(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(b(B(x0)))))) → B1(a(A(x0)))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(x)))) → A1(b(x))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B1(a(a(a(x)))) → A1(b(x)) we obtained the following new rules:
B1(a(a(a(b(y_0))))) → A1(b(b(y_0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(y_0))))) → A1(b(b(y_0)))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
A1(b(b(x))) → A1(x)
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(b(A(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A1(b(b(x))) → A1(x) we obtained the following new rules:
A1(b(b(b(b(y_0))))) → A1(b(b(y_0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(a(b(B(x)))))) → A1(b(b(a(A(x)))))
B1(a(a(a(x)))) → B1(x)
B1(a(a(b(a(a(A(x))))))) → A1(b(b(a(B(x)))))
B1(a(a(a(b(B(x)))))) → A1(b(b(A(x))))
B1(a(a(b(a(a(A(x))))))) → A1(b(b(A(x))))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(A(x0))))))))
B1(a(a(a(a(a(a(x0))))))) → B1(a(a(b(a(b(x0))))))
B1(a(a(a(b(y_0))))) → A1(b(b(y_0)))
B1(a(a(a(b(x0))))) → B1(a(b(a(a(x0)))))
A1(b(b(b(b(y_0))))) → A1(b(b(y_0)))
A1(b(b(x))) → B1(a(a(x)))
B1(a(a(a(a(a(a(b(B(x0))))))))) → B1(a(a(b(a(b(b(a(A(x0)))))))))
B1(a(a(a(a(a(b(a(a(A(x0)))))))))) → B1(a(a(b(a(b(b(a(B(x0)))))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(A(x))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(B(x)))))
b(a(a(A(x)))) → A(x)
b(a(a(a(b(B(x)))))) → a(b(a(b(b(a(A(x)))))))
b(a(a(b(a(a(A(x))))))) → a(b(a(b(b(a(B(x)))))))
b(a(a(a(b(B(x)))))) → a(b(a(b(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(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(a(x1))))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → 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:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(a(b(x1)))) → b(a(b(a(x1))))
b(b(a(x1))) → a(a(b(a(x1))))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → 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:
b(a(a(a(x)))) → a(b(a(b(x))))
a(b(b(x))) → a(b(a(a(x))))
Q is empty.