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