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(b(x1))) → a(x1)
a(b(a(x1))) → b(b(a(x1)))
b(b(x1)) → a(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(b(x1))) → a(x1)
a(b(a(x1))) → b(b(a(x1)))
b(b(x1)) → a(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:
B(b(x1)) → A(a(a(x1)))
B(b(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(b(a(x1))) → B(b(a(x1)))
B(b(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(a(b(x1))) → a(x1)
a(b(a(x1))) → b(b(a(x1)))
b(b(x1)) → a(a(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:
B(b(x1)) → A(a(a(x1)))
B(b(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(b(a(x1))) → B(b(a(x1)))
B(b(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(a(b(x1))) → a(x1)
a(b(a(x1))) → b(b(a(x1)))
b(b(x1)) → a(a(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(x1)) → A(a(a(x1))) at position [0] we obtained the following new rules:
B(b(b(a(x0)))) → A(a(b(b(a(x0)))))
B(b(a(b(x0)))) → A(a(a(x0)))
B(b(b(x0))) → A(a(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(b(a(x1))) → B(b(a(x1)))
B(b(b(a(x0)))) → A(a(b(b(a(x0)))))
B(b(x1)) → A(a(x1))
B(b(a(b(x0)))) → A(a(a(x0)))
B(b(b(x0))) → A(a(x0))
The TRS R consists of the following rules:
a(a(b(x1))) → a(x1)
a(b(a(x1))) → b(b(a(x1)))
b(b(x1)) → a(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
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(x1))) → a(x1)
a(b(a(x1))) → b(b(a(x1)))
b(b(x1)) → a(a(a(x1)))
B(b(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(b(a(x1))) → B(b(a(x1)))
B(b(b(a(x0)))) → A(a(b(b(a(x0)))))
B(b(x1)) → A(a(x1))
B(b(a(b(x0)))) → A(a(a(x0)))
B(b(b(x0))) → A(a(x0))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(x1))) → a(x1)
a(b(a(x1))) → b(b(a(x1)))
b(b(x1)) → a(a(a(x1)))
B(b(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(b(a(x1))) → B(b(a(x1)))
B(b(b(a(x0)))) → A(a(b(b(a(x0)))))
B(b(x1)) → A(a(x1))
B(b(a(b(x0)))) → A(a(a(x0)))
B(b(b(x0))) → A(a(x0))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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(b(a(x))) → B1(b(x))
B1(a(b(B(x)))) → A1(a(A(x)))
A1(b(b(B(x)))) → B1(a(A(x)))
A1(b(A(x))) → A1(b(B(x)))
A1(b(b(B(x)))) → A1(A(x))
B1(B(x)) → A1(A(x))
A1(b(a(x))) → B1(x)
B1(b(B(x))) → A1(A(x))
A1(b(a(x))) → A1(b(b(x)))
B1(b(x)) → A1(a(a(x)))
A1(b(b(B(x)))) → B1(b(a(A(x))))
B1(b(x)) → A1(x)
A1(b(b(B(x)))) → A1(b(b(a(A(x)))))
B1(b(x)) → A1(a(x))
B1(a(b(B(x)))) → A1(A(x))
A1(b(A(x))) → B1(B(x))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ 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:
A1(b(a(x))) → B1(b(x))
B1(a(b(B(x)))) → A1(a(A(x)))
A1(b(b(B(x)))) → B1(a(A(x)))
A1(b(A(x))) → A1(b(B(x)))
A1(b(b(B(x)))) → A1(A(x))
B1(B(x)) → A1(A(x))
A1(b(a(x))) → B1(x)
B1(b(B(x))) → A1(A(x))
A1(b(a(x))) → A1(b(b(x)))
B1(b(x)) → A1(a(a(x)))
A1(b(b(B(x)))) → B1(b(a(A(x))))
B1(b(x)) → A1(x)
A1(b(b(B(x)))) → A1(b(b(a(A(x)))))
B1(b(x)) → A1(a(x))
B1(a(b(B(x)))) → A1(A(x))
A1(b(A(x))) → B1(B(x))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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 9 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ 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:
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
A1(b(a(x))) → A1(b(b(x)))
A1(b(b(B(x)))) → B1(b(a(A(x))))
B1(b(x)) → A1(x)
A1(b(b(B(x)))) → A1(b(b(a(A(x)))))
A1(b(A(x))) → A1(b(B(x)))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(A(x))) → A1(b(B(x))) at position [0] we obtained the following new rules:
A1(b(A(x0))) → A1(A(x0))
A1(b(A(x0))) → A1(a(A(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ 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:
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
A1(b(a(x))) → A1(b(b(x)))
A1(b(b(B(x)))) → B1(b(a(A(x))))
A1(b(A(x0))) → A1(a(A(x0)))
B1(b(x)) → A1(x)
A1(b(b(B(x)))) → A1(b(b(a(A(x)))))
A1(b(A(x0))) → A1(A(x0))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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
↳ 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:
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
A1(b(a(x))) → A1(b(b(x)))
A1(b(b(B(x)))) → B1(b(a(A(x))))
B1(b(x)) → A1(x)
A1(b(b(B(x)))) → A1(b(b(a(A(x)))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(b(B(x)))) → A1(b(b(a(A(x))))) at position [0] we obtained the following new rules:
A1(b(b(B(x0)))) → A1(b(A(x0)))
A1(b(b(B(y0)))) → A1(a(a(a(a(A(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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:
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
A1(b(a(x))) → A1(b(b(x)))
A1(b(b(B(x)))) → B1(b(a(A(x))))
B1(b(x)) → A1(x)
A1(b(b(B(x0)))) → A1(b(A(x0)))
A1(b(b(B(y0)))) → A1(a(a(a(a(A(y0))))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
A1(b(a(x))) → A1(b(b(x)))
A1(b(b(B(x)))) → B1(b(a(A(x))))
B1(b(x)) → A1(x)
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B1(b(x)) → A1(x) we obtained the following new rules:
B1(b(b(y_1))) → A1(b(y_1))
B1(b(b(b(B(y_0))))) → A1(b(b(B(y_0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
A1(b(a(x))) → A1(b(b(x)))
A1(b(b(B(x)))) → B1(b(a(A(x))))
B1(b(b(y_1))) → A1(b(y_1))
B1(b(b(b(B(y_0))))) → A1(b(b(B(y_0))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
A1(b(a(x))) → A1(b(b(x)))
B1(b(b(y_1))) → A1(b(y_1))
B1(b(b(b(B(y_0))))) → A1(b(b(B(y_0))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(a(x))) → A1(b(b(x))) at position [0] we obtained the following new rules:
A1(b(a(a(b(B(x0)))))) → A1(b(a(a(A(x0)))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(a(A(x0))))) → A1(b(A(x0)))
A1(b(a(b(B(x0))))) → A1(b(a(A(x0))))
A1(b(a(B(x0)))) → A1(a(A(x0)))
A1(b(a(B(x0)))) → A1(b(A(x0)))
A1(b(a(x0))) → A1(a(a(a(x0))))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ 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:
A1(b(a(x))) → B1(b(x))
A1(b(a(a(b(B(x0)))))) → A1(b(a(a(A(x0)))))
A1(b(a(b(B(x0))))) → A1(b(a(A(x0))))
A1(b(a(x0))) → A1(a(a(a(x0))))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(a(A(x0))))) → A1(b(A(x0)))
A1(b(a(x))) → B1(x)
A1(b(a(B(x0)))) → A1(a(A(x0)))
A1(b(a(B(x0)))) → A1(b(A(x0)))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
B1(b(b(b(B(y_0))))) → A1(b(b(B(y_0))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ 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:
A1(b(a(x))) → B1(b(x))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(a(b(B(x0)))))) → A1(b(a(a(A(x0)))))
A1(b(a(x))) → B1(x)
A1(b(a(b(B(x0))))) → A1(b(a(A(x0))))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
B1(b(b(b(B(y_0))))) → A1(b(b(B(y_0))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(B(y_0))))) → A1(b(b(B(y_0)))) at position [0] we obtained the following new rules:
B1(b(b(b(B(x0))))) → A1(b(A(x0)))
B1(b(b(b(B(y0))))) → A1(a(a(a(B(y0)))))
B1(b(b(b(B(x0))))) → A1(b(a(A(x0))))
B1(b(b(b(B(x0))))) → A1(a(A(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ 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:
A1(b(a(x))) → B1(b(x))
A1(b(a(a(b(B(x0)))))) → A1(b(a(a(A(x0)))))
A1(b(a(b(B(x0))))) → A1(b(a(A(x0))))
B1(b(b(b(B(x0))))) → A1(b(A(x0)))
B1(b(b(b(B(x0))))) → A1(b(a(A(x0))))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
B1(b(b(b(B(y0))))) → A1(a(a(a(B(y0)))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
B1(b(b(b(B(x0))))) → A1(a(A(x0)))
A1(b(a(x))) → B1(x)
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(a(x))) → B1(b(x))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(a(b(B(x0)))))) → A1(b(a(a(A(x0)))))
A1(b(a(x))) → B1(x)
A1(b(a(b(B(x0))))) → A1(b(a(A(x0))))
B1(b(b(b(B(x0))))) → A1(b(a(A(x0))))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A1(b(a(x))) → B1(b(x)) we obtained the following new rules:
A1(b(a(b(b(B(y_0)))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(y_0)))) → B1(b(b(y_0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(a(a(b(B(x0)))))) → A1(b(a(a(A(x0)))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(x))) → B1(x)
A1(b(a(b(b(B(y_0)))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(B(x0))))) → A1(b(a(A(x0))))
B1(b(b(b(B(x0))))) → A1(b(a(A(x0))))
A1(b(a(b(y_0)))) → B1(b(b(y_0)))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A1(b(a(x))) → B1(x) we obtained the following new rules:
A1(b(a(b(b(b(B(y_0))))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(b(y_0))))) → B1(b(b(y_0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(a(b(b(b(B(y_0))))))) → B1(b(b(b(B(y_0)))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(a(b(B(x0)))))) → A1(b(a(a(A(x0)))))
A1(b(a(b(b(B(y_0)))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(B(x0))))) → A1(b(a(A(x0))))
B1(b(b(b(B(x0))))) → A1(b(a(A(x0))))
A1(b(a(b(y_0)))) → B1(b(b(y_0)))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(b(y_0))))) → B1(b(b(y_0)))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(b(B(x0))))) → A1(b(a(A(x0)))) at position [0] we obtained the following new rules:
B1(b(b(b(B(x0))))) → A1(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(a(b(b(b(B(y_0))))))) → B1(b(b(b(B(y_0)))))
A1(b(a(a(b(B(x0)))))) → A1(b(a(a(A(x0)))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
B1(b(b(b(B(x0))))) → A1(A(x0))
A1(b(a(b(b(B(y_0)))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(B(x0))))) → A1(b(a(A(x0))))
A1(b(a(b(y_0)))) → B1(b(b(y_0)))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
A1(b(a(b(b(y_0))))) → B1(b(b(y_0)))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ 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:
A1(b(a(b(b(b(B(y_0))))))) → B1(b(b(b(B(y_0)))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(a(b(B(x0)))))) → A1(b(a(a(A(x0)))))
A1(b(a(b(b(B(y_0)))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(B(x0))))) → A1(b(a(A(x0))))
A1(b(a(b(y_0)))) → B1(b(b(y_0)))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(b(y_0))))) → B1(b(b(y_0)))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(a(a(b(B(x0)))))) → A1(b(a(a(A(x0))))) at position [0] we obtained the following new rules:
A1(b(a(a(b(B(y0)))))) → A1(a(A(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ 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:
A1(b(a(b(b(b(B(y_0))))))) → B1(b(b(b(B(y_0)))))
A1(b(a(a(b(B(y0)))))) → A1(a(A(y0)))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(b(b(B(y_0)))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(B(x0))))) → A1(b(a(A(x0))))
A1(b(a(b(y_0)))) → B1(b(b(y_0)))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
A1(b(a(b(b(y_0))))) → B1(b(b(y_0)))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ 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:
A1(b(a(b(b(b(B(y_0))))))) → B1(b(b(b(B(y_0)))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(b(b(B(y_0)))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(B(x0))))) → A1(b(a(A(x0))))
A1(b(a(b(y_0)))) → B1(b(b(y_0)))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(b(y_0))))) → B1(b(b(y_0)))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(a(b(B(x0))))) → A1(b(a(A(x0)))) at position [0] we obtained the following new rules:
A1(b(a(b(B(x0))))) → A1(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A1(b(a(b(B(x0))))) → A1(A(x0))
A1(b(a(b(b(b(B(y_0))))))) → B1(b(b(b(B(y_0)))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(b(b(B(y_0)))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(y_0)))) → B1(b(b(y_0)))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
A1(b(a(b(b(y_0))))) → B1(b(b(y_0)))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A1(b(a(b(b(b(B(y_0))))))) → B1(b(b(b(B(y_0)))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(b(b(B(y_0)))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(y_0)))) → B1(b(b(y_0)))
A1(b(a(B(x0)))) → A1(b(a(A(x0))))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(b(y_0))))) → B1(b(b(y_0)))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(a(B(x0)))) → A1(b(a(A(x0)))) at position [0] we obtained the following new rules:
A1(b(a(B(x0)))) → A1(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A1(b(a(b(b(b(B(y_0))))))) → B1(b(b(b(B(y_0)))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(b(b(B(y_0)))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(y_0)))) → B1(b(b(y_0)))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(B(x0)))) → A1(A(x0))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
A1(b(a(b(b(y_0))))) → B1(b(b(y_0)))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A1(b(a(b(b(b(B(y_0))))))) → B1(b(b(b(B(y_0)))))
A1(b(a(a(a(x0))))) → A1(b(a(x0)))
A1(b(a(b(b(B(y_0)))))) → B1(b(b(b(B(y_0)))))
A1(b(a(b(y_0)))) → B1(b(b(y_0)))
B1(b(b(y_1))) → A1(b(y_1))
A1(b(a(b(b(y_0))))) → B1(b(b(y_0)))
A1(b(a(b(x0)))) → A1(b(a(a(a(x0)))))
The TRS R consists of the following rules:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(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
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(x))) → a(x)
a(b(a(x))) → b(b(a(x)))
b(b(x)) → a(a(a(x)))
B(b(x)) → A(x)
A(a(b(x))) → A(x)
A(b(a(x))) → B(b(a(x)))
B(b(b(a(x)))) → A(a(b(b(a(x)))))
B(b(x)) → A(a(x))
B(b(a(b(x)))) → A(a(a(x)))
B(b(b(x))) → A(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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(b(x))) → a(x)
a(b(a(x))) → b(b(a(x)))
b(b(x)) → a(a(a(x)))
B(b(x)) → A(x)
A(a(b(x))) → A(x)
A(b(a(x))) → B(b(a(x)))
B(b(b(a(x)))) → A(a(b(b(a(x)))))
B(b(x)) → A(a(x))
B(b(a(b(x)))) → A(a(a(x)))
B(b(b(x))) → A(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
b(B(x)) → A(x)
b(a(A(x))) → A(x)
a(b(A(x))) → a(b(B(x)))
a(b(b(B(x)))) → a(b(b(a(A(x)))))
b(B(x)) → a(A(x))
b(a(b(B(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(x))) → a(x)
a(b(a(x))) → b(b(a(x)))
b(b(x)) → a(a(a(x)))
B(b(x)) → A(x)
A(a(b(x))) → A(x)
A(b(a(x))) → B(b(a(x)))
B(b(b(a(x)))) → A(a(b(b(a(x)))))
B(b(x)) → A(a(x))
B(b(a(b(x)))) → A(a(a(x)))
B(b(b(x))) → A(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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(b(x))) → a(x)
a(b(a(x))) → b(b(a(x)))
b(b(x)) → a(a(a(x)))
B(b(x)) → A(x)
A(a(b(x))) → A(x)
A(b(a(x))) → B(b(a(x)))
B(b(b(a(x)))) → A(a(b(b(a(x)))))
B(b(x)) → A(a(x))
B(b(a(b(x)))) → A(a(a(x)))
B(b(b(x))) → A(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(x1))) → a(x1)
a(b(a(x1))) → b(b(a(x1)))
b(b(x1)) → a(a(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(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(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(x1))) → a(x1)
a(b(a(x1))) → b(b(a(x1)))
b(b(x1)) → a(a(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(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(x))) → a(x)
a(b(a(x))) → a(b(b(x)))
b(b(x)) → a(a(a(x)))
Q is empty.