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