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)))) → a(b(b(b(b(a(a(a(x1))))))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
Q is empty.
We were given the following TRS:
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
By stripping symbols from the only rule of the system, we obtained the following TRS:
a(a(b(x1))) → b(b(b(b(a(a(a(x1)))))))
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS
↳ QTRS Reverse
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(x1))) → b(b(b(b(a(a(a(x1)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ QTRS
↳ Strip Symbols Proof
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
Q is empty.
We were given the following TRS:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
By stripping symbols from the only rule of the system, we obtained the following TRS:
b(a(a(x))) → a(a(a(b(b(b(b(x)))))))
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ QTRS
↳ Strip Symbols Proof
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(x))) → a(a(a(b(b(b(b(x)))))))
Q is empty.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
The set Q consists of the following terms:
a(a(a(b(x0))))
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)))) → A(x1)
A(a(a(b(x1)))) → A(a(a(x1)))
A(a(a(b(x1)))) → A(b(b(b(b(a(a(a(x1))))))))
A(a(a(b(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
The set Q consists of the following terms:
a(a(a(b(x0))))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(x1)))) → A(x1)
A(a(a(b(x1)))) → A(a(a(x1)))
A(a(a(b(x1)))) → A(b(b(b(b(a(a(a(x1))))))))
A(a(a(b(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
The set Q consists of the following terms:
a(a(a(b(x0))))
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
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(x1)))) → A(x1)
A(a(a(b(x1)))) → A(a(a(x1)))
A(a(a(b(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
The set Q consists of the following terms:
a(a(a(b(x0))))
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(x1)))) → A(a(a(x1)))
A(a(a(b(x1)))) → A(x1)
A(a(a(b(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
Q is empty.
We have to consider all (P,Q,R)-chains.
By forward instantiating [14] the rule A(a(a(b(x1)))) → A(x1) we obtained the following new rules:
A(a(a(b(a(a(b(y_0))))))) → A(a(a(b(y_0))))
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(x1)))) → A(a(a(x1)))
A(a(a(b(a(a(b(y_0))))))) → A(a(a(b(y_0))))
A(a(a(b(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
The set Q consists of the following terms:
a(a(a(b(x0))))
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
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
A(a(a(b(x1)))) → A(a(a(x1)))
A(a(a(b(a(a(b(y_0))))))) → A(a(a(b(y_0))))
A(a(a(b(x1)))) → A(a(x1))
The set Q consists of the following terms:
a(a(a(b(x0))))
We have reversed the following QTRS:
The set of rules R is
a(a(a(b(x1)))) → a(b(b(b(b(a(a(a(x1))))))))
A(a(a(b(x1)))) → A(a(a(x1)))
A(a(a(b(a(a(b(y_0))))))) → A(a(a(b(y_0))))
A(a(a(b(x1)))) → A(a(x1))
The set Q is {a(a(a(b(x0))))}.
We have obtained the following QTRS:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(x)))) → a(A(x))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(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:
B(a(a(a(x)))) → B(b(a(x)))
B(a(a(a(x)))) → B(b(b(b(a(x)))))
B(a(a(a(x)))) → B(b(b(a(x))))
B(a(a(a(x)))) → B(a(x))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(x)))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(x)))) → B(b(a(x)))
B(a(a(a(x)))) → B(b(b(b(a(x)))))
B(a(a(a(x)))) → B(b(b(a(x))))
B(a(a(a(x)))) → B(a(x))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(x)))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(a(a(x)))) → B(b(b(b(a(x))))) at position [0] we obtained the following new rules:
B(a(a(a(a(A(x0)))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(A(x0)))))) → B(b(b(a(A(x0)))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(b(a(a(A(x0)))))))
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(A(x0)))))) → B(b(b(a(A(x0)))))
B(a(a(a(x)))) → B(b(a(x)))
B(a(a(a(x)))) → B(b(b(a(x))))
B(a(a(a(a(A(x0)))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(b(a(a(A(x0)))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(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
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(x)))) → B(b(a(x)))
B(a(a(a(x)))) → B(b(b(a(x))))
B(a(a(a(a(A(x0)))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(b(a(a(A(x0)))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(x)))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(a(a(x)))) → B(b(a(x))) at position [0] we obtained the following new rules:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(A(x0)))))) → B(a(a(A(x0))))
B(a(a(a(a(A(x0)))))) → B(a(A(x0)))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(A(x0)))))) → B(a(a(A(x0))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(x)))) → B(b(b(a(x))))
B(a(a(a(a(A(x0)))))) → B(a(A(x0)))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(A(x0)))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(b(a(a(A(x0)))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(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
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ 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:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(x)))) → B(b(b(a(x))))
B(a(a(a(a(A(x0)))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(b(a(a(A(x0)))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(x)))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(a(a(x)))) → B(b(b(a(x)))) at position [0] we obtained the following new rules:
B(a(a(a(a(A(x0)))))) → B(b(a(A(x0))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(A(x0)))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(a(x0)))))) → B(b(a(a(a(b(b(b(b(a(x0))))))))))
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ 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:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(A(x0)))))) → B(b(a(A(x0))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(A(x0)))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(a(x0)))))) → B(b(a(a(a(b(b(b(b(a(x0))))))))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(A(x0)))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(b(a(a(A(x0)))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(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
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ 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:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(A(x0)))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(a(x0)))))) → B(b(a(a(a(b(b(b(b(a(x0))))))))))
B(a(a(a(a(A(x0)))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(b(a(a(A(x0)))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(x)))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(a(a(a(A(x0)))))) → B(b(b(a(a(A(x0)))))) at position [0] we obtained the following new rules:
B(a(a(a(a(A(x0)))))) → B(b(a(A(x0))))
B(a(a(a(a(A(x0)))))) → B(b(a(a(A(x0)))))
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ 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:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(A(x0)))))) → B(b(a(A(x0))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(A(x0)))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(a(x0)))))) → B(b(a(a(a(b(b(b(b(a(x0))))))))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(b(a(a(A(x0)))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(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
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(A(x0)))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(a(x0)))))) → B(b(a(a(a(b(b(b(b(a(x0))))))))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(b(a(a(A(x0)))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(x)))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(b(a(a(A(x0))))))) at position [0] we obtained the following new rules:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(A(x0)))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(A(x0)))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(a(x0)))))) → B(b(a(a(a(b(b(b(b(a(x0))))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(A(x0)))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(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
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(A(x0)))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(a(x0)))))) → B(b(a(a(a(b(b(b(b(a(x0))))))))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(x)))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(a(a(A(x0))))) at position [0] we obtained the following new rules:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(a(a(A(x0))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(a(A(x0)))
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(a(A(x0)))
B(a(a(a(a(A(x0)))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(a(x0)))))) → B(b(a(a(a(b(b(b(b(a(x0))))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(a(a(A(x0))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(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
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(A(x0)))))) → B(b(a(a(A(x0)))))
B(a(a(a(a(a(x0)))))) → B(b(a(a(a(b(b(b(b(a(x0))))))))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(x)))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(a(a(a(A(x0)))))) → B(b(a(a(A(x0))))) at position [0] we obtained the following new rules:
B(a(a(a(a(A(x0)))))) → B(a(a(A(x0))))
B(a(a(a(a(A(x0)))))) → B(a(A(x0)))
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
B(a(a(a(a(A(x0)))))) → B(a(a(A(x0))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(a(x0)))))) → B(b(a(a(a(b(b(b(b(a(x0))))))))))
B(a(a(a(a(A(x0)))))) → B(a(A(x0)))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(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
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(a(b(a(a(A(x0))))))))) → B(b(b(a(a(A(x0))))))
B(a(a(a(a(a(x0)))))) → B(b(b(a(a(a(b(b(b(b(a(x0)))))))))))
B(a(a(a(a(a(x0)))))) → B(b(a(a(a(b(b(b(b(a(x0))))))))))
B(a(a(a(x)))) → B(a(x))
B(a(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(b(a(x0)))))))))
The TRS R consists of the following rules:
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(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(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(x)))) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(a(a(b(x)))) → a(b(b(b(b(a(a(a(x))))))))
A(a(a(b(x)))) → A(a(a(x)))
A(a(a(b(a(a(b(x))))))) → A(a(a(b(x))))
A(a(a(b(x)))) → A(a(x))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(b(x)))) → a(b(b(b(b(a(a(a(x))))))))
A(a(a(b(x)))) → A(a(a(x)))
A(a(a(b(a(a(b(x))))))) → A(a(a(b(x))))
A(a(a(b(x)))) → A(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(a(a(x)))) → a(a(a(b(b(b(b(a(x))))))))
b(a(a(A(x)))) → a(a(A(x)))
b(a(a(b(a(a(A(x))))))) → b(a(a(A(x))))
b(a(a(A(x)))) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(a(a(b(x)))) → a(b(b(b(b(a(a(a(x))))))))
A(a(a(b(x)))) → A(a(a(x)))
A(a(a(b(a(a(b(x))))))) → A(a(a(b(x))))
A(a(a(b(x)))) → A(a(x))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ Strip Symbols Proof
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(b(x)))) → a(b(b(b(b(a(a(a(x))))))))
A(a(a(b(x)))) → A(a(a(x)))
A(a(a(b(a(a(b(x))))))) → A(a(a(b(x))))
A(a(a(b(x)))) → A(a(x))
Q is empty.