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(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
Q is empty.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(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:
B(b(b(c(x1)))) → A(x1)
B(b(b(c(x1)))) → A(b(a(a(x1))))
A(x1) → B(x1)
B(b(b(c(x1)))) → B(a(a(x1)))
B(b(b(c(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(x1)))) → A(x1)
B(b(b(c(x1)))) → A(b(a(a(x1))))
A(x1) → B(x1)
B(b(b(c(x1)))) → B(a(a(x1)))
B(b(b(c(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ Rewriting
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(x1)))) → A(x1)
B(b(b(c(x1)))) → A(b(a(a(x1))))
A(x1) → B(x1)
B(b(b(c(x1)))) → B(a(a(x1)))
B(b(b(c(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
Q is empty.
We have to consider all (P,Q,R)-chains.
By rewriting [15] the rule B(b(b(c(x1)))) → A(b(a(a(x1)))) at position [0,0] we obtained the following new rules:
B(b(b(c(x1)))) → A(b(b(a(x1))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(x1)))) → A(x1)
B(b(b(c(x1)))) → A(b(b(a(x1))))
A(x1) → B(x1)
B(b(b(c(x1)))) → B(a(a(x1)))
B(b(b(c(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule B(b(b(c(x1)))) → B(a(a(x1))) at position [0] we obtained the following new rules:
B(b(b(c(x1)))) → B(b(a(x1)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(x1)))) → A(x1)
B(b(b(c(x1)))) → B(b(a(x1)))
A(x1) → B(x1)
B(b(b(c(x1)))) → A(b(b(a(x1))))
B(b(b(c(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule B(b(b(c(x1)))) → A(a(x1)) at position [0] we obtained the following new rules:
B(b(b(c(x1)))) → A(b(x1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(x1)))) → A(x1)
B(b(b(c(x1)))) → B(b(a(x1)))
B(b(b(c(x1)))) → A(b(b(a(x1))))
A(x1) → B(x1)
B(b(b(c(x1)))) → A(b(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule B(b(b(c(x1)))) → A(b(b(a(x1)))) at position [0,0,0] we obtained the following new rules:
B(b(b(c(x1)))) → A(b(b(b(x1))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(x1)))) → A(x1)
B(b(b(c(x1)))) → B(b(a(x1)))
A(x1) → B(x1)
B(b(b(c(x1)))) → A(b(b(b(x1))))
B(b(b(c(x1)))) → A(b(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule B(b(b(c(x1)))) → B(b(a(x1))) at position [0,0] we obtained the following new rules:
B(b(b(c(x1)))) → B(b(b(x1)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(x1)))) → A(x1)
A(x1) → B(x1)
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → A(b(b(b(x1))))
B(b(b(c(x1)))) → B(b(b(x1)))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A(x1) → B(x1) we obtained the following new rules:
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(x1)))) → A(x1)
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(b(b(b(x1))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(b(c(x1)))) → A(b(b(b(x1)))) at position [0] we obtained the following new rules:
B(b(b(c(b(c(x0)))))) → A(b(c(c(a(b(a(a(x0))))))))
B(b(b(c(c(x0))))) → A(c(c(a(b(a(a(x0)))))))
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(a(b(a(a(x0)))))))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(x1)))) → A(x1)
B(b(b(c(b(c(x0)))))) → A(b(c(c(a(b(a(a(x0))))))))
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(c(x0))))) → A(c(c(a(b(a(a(x0)))))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(a(b(a(a(x0)))))))))
B(b(b(c(x1)))) → B(b(b(x1)))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(x1)))) → A(x1)
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(a(b(a(a(x0)))))))))
B(b(b(c(x1)))) → B(b(b(x1)))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(a(b(a(a(x0))))))))) at position [0,0,0,0,0] we obtained the following new rules:
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(b(b(a(a(x0)))))))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(x1)))) → A(x1)
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(b(b(a(a(x0)))))))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(b(b(a(a(x0))))))))) at position [0,0,0,0,0,0,0] we obtained the following new rules:
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(b(b(b(a(x0)))))))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(b(b(b(a(x0)))))))))
B(b(b(c(x1)))) → A(x1)
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(b(b(b(a(x0))))))))) at position [0,0,0,0,0,0,0,0] we obtained the following new rules:
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(b(b(b(b(x0)))))))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(b(b(b(b(x0)))))))))
B(b(b(c(x1)))) → A(x1)
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B(b(b(c(x1)))) → A(x1) we obtained the following new rules:
B(b(b(c(b(b(c(y_0))))))) → A(b(b(c(y_0))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(b(b(b(b(x0)))))))))
B(b(b(c(b(b(c(y_0))))))) → A(b(b(c(y_0))))
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q consists of the following terms:
a(x0)
b(b(b(c(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(b(b(b(b(x0)))))))))
B(b(b(c(b(b(c(y_0))))))) → A(b(b(c(y_0))))
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))
The set Q consists of the following terms:
a(x0)
b(b(b(c(x0))))
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
B(b(b(c(b(b(c(x0))))))) → A(b(b(c(c(b(b(b(b(x0)))))))))
B(b(b(c(b(b(c(y_0))))))) → A(b(b(c(y_0))))
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))
The set Q is {a(x0), b(b(b(c(x0))))}.
We have obtained the following QTRS:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
The set Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(b(b(c(x)))) → c(c(a(b(a(a(x))))))
B(b(b(c(b(b(c(x))))))) → A(b(b(c(c(b(b(b(b(x)))))))))
B(b(b(c(b(b(c(x))))))) → A(b(b(c(x))))
A(b(b(c(x)))) → B(b(b(c(x))))
B(b(b(c(x)))) → A(b(x))
B(b(b(c(x)))) → B(b(b(x)))
The set Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
b(b(b(c(x)))) → c(c(a(b(a(a(x))))))
B(b(b(c(b(b(c(x))))))) → A(b(b(c(c(b(b(b(b(x)))))))))
B(b(b(c(b(b(c(x))))))) → A(b(b(c(x))))
A(b(b(c(x)))) → B(b(b(c(x))))
B(b(b(c(x)))) → A(b(x))
B(b(b(c(x)))) → B(b(b(x)))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
C(b(b(c(b(b(B(x))))))) → C(b(b(A(x))))
C(b(b(A(x)))) → C(b(b(B(x))))
C(b(b(b(x)))) → C(x)
C(b(b(b(x)))) → A1(c(c(x)))
C(b(b(c(b(b(B(x))))))) → C(c(b(b(A(x)))))
C(b(b(b(x)))) → A1(b(a(c(c(x)))))
C(b(b(b(x)))) → A1(a(b(a(c(c(x))))))
C(b(b(b(x)))) → C(c(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(c(b(b(B(x))))))) → C(b(b(A(x))))
C(b(b(A(x)))) → C(b(b(B(x))))
C(b(b(b(x)))) → C(x)
C(b(b(b(x)))) → A1(c(c(x)))
C(b(b(c(b(b(B(x))))))) → C(c(b(b(A(x)))))
C(b(b(b(x)))) → A1(b(a(c(c(x)))))
C(b(b(b(x)))) → A1(a(b(a(c(c(x))))))
C(b(b(b(x)))) → C(c(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 5 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(b(x)))) → C(x)
C(b(b(c(b(b(B(x))))))) → C(c(b(b(A(x)))))
C(b(b(b(x)))) → C(c(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(b(b(x)))) → C(c(x)) at position [0] we obtained the following new rules:
C(b(b(b(b(b(B(x0))))))) → C(b(b(B(x0))))
C(b(b(b(b(b(A(x0))))))) → C(c(b(b(B(x0)))))
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(c(b(b(A(x0)))))
C(b(b(b(b(b(B(x0))))))) → C(b(A(x0)))
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(b(b(b(b(c(c(b(b(A(x0))))))))))
C(b(b(b(b(b(b(x0))))))) → C(a(a(b(a(c(c(x0)))))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(b(x)))) → C(x)
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(b(b(b(b(c(c(b(b(A(x0))))))))))
C(b(b(b(b(b(B(x0))))))) → C(b(A(x0)))
C(b(b(c(b(b(B(x))))))) → C(c(b(b(A(x)))))
C(b(b(b(b(b(A(x0))))))) → C(c(b(b(B(x0)))))
C(b(b(b(b(b(B(x0))))))) → C(b(b(B(x0))))
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(c(b(b(A(x0)))))
C(b(b(b(b(b(b(x0))))))) → C(a(a(b(a(c(c(x0)))))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(b(x)))) → C(x)
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(b(b(b(b(c(c(b(b(A(x0))))))))))
C(b(b(c(b(b(B(x))))))) → C(c(b(b(A(x)))))
C(b(b(b(b(b(A(x0))))))) → C(c(b(b(B(x0)))))
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(c(b(b(A(x0)))))
C(b(b(b(b(b(b(x0))))))) → C(a(a(b(a(c(c(x0)))))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(b(c(b(b(B(x))))))) → C(c(b(b(A(x))))) at position [0] we obtained the following new rules:
C(b(b(c(b(b(B(x0))))))) → C(c(b(b(B(x0)))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(b(x)))) → C(x)
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(b(b(b(b(c(c(b(b(A(x0))))))))))
C(b(b(b(b(b(A(x0))))))) → C(c(b(b(B(x0)))))
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(c(b(b(A(x0)))))
C(b(b(c(b(b(B(x0))))))) → C(c(b(b(B(x0)))))
C(b(b(b(b(b(b(x0))))))) → C(a(a(b(a(c(c(x0)))))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(b(b(b(b(A(x0))))))) → C(c(b(b(B(x0))))) at position [0] we obtained the following new rules:
C(b(b(b(b(b(A(x0))))))) → C(b(b(B(x0))))
C(b(b(b(b(b(A(x0))))))) → C(b(A(x0)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(b(x)))) → C(x)
C(b(b(b(b(b(A(x0))))))) → C(b(b(B(x0))))
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(b(b(b(b(c(c(b(b(A(x0))))))))))
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(c(b(b(A(x0)))))
C(b(b(c(b(b(B(x0))))))) → C(c(b(b(B(x0)))))
C(b(b(b(b(b(A(x0))))))) → C(b(A(x0)))
C(b(b(b(b(b(b(x0))))))) → C(a(a(b(a(c(c(x0)))))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(b(x)))) → C(x)
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(b(b(b(b(c(c(b(b(A(x0))))))))))
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(c(b(b(A(x0)))))
C(b(b(c(b(b(B(x0))))))) → C(c(b(b(B(x0)))))
C(b(b(b(b(b(b(x0))))))) → C(a(a(b(a(c(c(x0)))))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(b(c(b(b(B(x0))))))) → C(c(b(b(B(x0))))) at position [0] we obtained the following new rules:
C(b(b(c(b(b(B(x0))))))) → C(b(A(x0)))
C(b(b(c(b(b(B(x0))))))) → C(b(b(B(x0))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(c(b(b(B(x0))))))) → C(b(A(x0)))
C(b(b(b(x)))) → C(x)
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(b(b(b(b(c(c(b(b(A(x0))))))))))
C(b(b(c(b(b(B(x0))))))) → C(b(b(B(x0))))
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(c(b(b(A(x0)))))
C(b(b(b(b(b(b(x0))))))) → C(a(a(b(a(c(c(x0)))))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(b(x)))) → C(x)
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(b(b(b(b(c(c(b(b(A(x0))))))))))
C(b(b(b(b(b(c(b(b(B(x0)))))))))) → C(c(b(b(A(x0)))))
C(b(b(b(b(b(b(x0))))))) → C(a(a(b(a(c(c(x0)))))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
c(b(b(c(b(b(B(x))))))) → b(b(b(b(c(c(b(b(A(x)))))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(x)))) → b(b(B(x)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(b(b(c(x)))) → c(c(a(b(a(a(x))))))
B(b(b(c(b(b(c(x))))))) → A(b(b(c(c(b(b(b(b(x)))))))))
B(b(b(c(b(b(c(x))))))) → A(b(b(c(x))))
A(b(b(c(x)))) → B(b(b(c(x))))
B(b(b(c(x)))) → A(b(x))
B(b(b(c(x)))) → B(b(b(x)))
The set Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
b(b(b(c(x)))) → c(c(a(b(a(a(x))))))
B(b(b(c(b(b(c(x))))))) → A(b(b(c(c(b(b(b(b(x)))))))))
B(b(b(c(b(b(c(x))))))) → A(b(b(c(x))))
A(b(b(c(x)))) → B(b(b(c(x))))
B(b(b(c(x)))) → A(b(x))
B(b(b(c(x)))) → B(b(b(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
The set Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
b(b(b(c(x1)))) → c(c(a(b(a(a(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
The set Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(b(x)))) → a(a(b(a(c(c(x))))))
Q is empty.