Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → 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(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → 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(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
The set Q consists of the following terms:
A(b(x0))
B(a(x0))
A(a(x0))
B(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:
B1(a(x1)) → B1(x1)
B1(a(x1)) → A1(B(x1))
A1(b(x1)) → B1(A(x1))
A1(b(x1)) → A1(x1)
The TRS R consists of the following rules:
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
The set Q consists of the following terms:
A(b(x0))
B(a(x0))
A(a(x0))
B(b(x0))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x1)) → B1(x1)
B1(a(x1)) → A1(B(x1))
A1(b(x1)) → B1(A(x1))
A1(b(x1)) → A1(x1)
The TRS R consists of the following rules:
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
The set Q consists of the following terms:
A(b(x0))
B(a(x0))
A(a(x0))
B(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x1)) → B1(x1)
B1(a(x1)) → A1(B(x1))
A1(b(x1)) → B1(A(x1))
A1(b(x1)) → A1(x1)
The TRS R consists of the following rules:
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
Q is empty.
We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule A1(b(x1)) → B1(A(x1)) at position [0] we obtained the following new rules:
A1(b(a(x0))) → B1(x0)
A1(b(b(x0))) → B1(b(a(B(A(x0)))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x1)) → B1(x1)
A1(b(a(x0))) → B1(x0)
B1(a(x1)) → A1(B(x1))
A1(b(b(x0))) → B1(b(a(B(A(x0)))))
A1(b(x1)) → A1(x1)
The TRS R consists of the following rules:
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
The set Q consists of the following terms:
A(b(x0))
B(a(x0))
A(a(x0))
B(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x1)) → B1(x1)
A1(b(a(x0))) → B1(x0)
B1(a(x1)) → A1(B(x1))
A1(b(x1)) → A1(x1)
The TRS R consists of the following rules:
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
The set Q consists of the following terms:
A(b(x0))
B(a(x0))
A(a(x0))
B(b(x0))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(x1)) → A1(B(x1)) at position [0] we obtained the following new rules:
B1(a(a(x0))) → A1(a(b(A(B(x0)))))
B1(a(b(x0))) → A1(x0)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x1)) → B1(x1)
A1(b(a(x0))) → B1(x0)
B1(a(a(x0))) → A1(a(b(A(B(x0)))))
B1(a(b(x0))) → A1(x0)
A1(b(x1)) → A1(x1)
The TRS R consists of the following rules:
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
The set Q consists of the following terms:
A(b(x0))
B(a(x0))
A(a(x0))
B(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x1)) → B1(x1)
A1(b(a(x0))) → B1(x0)
B1(a(b(x0))) → A1(x0)
A1(b(x1)) → A1(x1)
The TRS R consists of the following rules:
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
The set Q consists of the following terms:
A(b(x0))
B(a(x0))
A(a(x0))
B(b(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x1)) → B1(x1)
A1(b(a(x0))) → B1(x0)
B1(a(b(x0))) → A1(x0)
A1(b(x1)) → A1(x1)
R is empty.
The set Q consists of the following terms:
A(b(x0))
B(a(x0))
A(a(x0))
B(b(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
A(b(x0))
B(a(x0))
A(a(x0))
B(b(x0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x1)) → B1(x1)
A1(b(a(x0))) → B1(x0)
B1(a(b(x0))) → A1(x0)
A1(b(x1)) → A1(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B1(a(x1)) → B1(x1) we obtained the following new rules:
B1(a(a(b(y_0)))) → B1(a(b(y_0)))
B1(a(a(y_0))) → B1(a(y_0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(a(x0))) → B1(x0)
B1(a(a(b(y_0)))) → B1(a(b(y_0)))
B1(a(a(y_0))) → B1(a(y_0))
B1(a(b(x0))) → A1(x0)
A1(b(x1)) → A1(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A1(b(a(x0))) → B1(x0) we obtained the following new rules:
A1(b(a(a(a(b(y_0)))))) → B1(a(a(b(y_0))))
A1(b(a(a(a(y_0))))) → B1(a(a(y_0)))
A1(b(a(a(b(y_0))))) → B1(a(b(y_0)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(a(a(a(b(y_0)))))) → B1(a(a(b(y_0))))
A1(b(a(a(a(y_0))))) → B1(a(a(y_0)))
A1(b(a(a(b(y_0))))) → B1(a(b(y_0)))
B1(a(a(b(y_0)))) → B1(a(b(y_0)))
B1(a(a(y_0))) → B1(a(y_0))
B1(a(b(x0))) → A1(x0)
A1(b(x1)) → A1(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B1(a(b(x0))) → A1(x0) we obtained the following new rules:
B1(a(b(b(a(a(a(y_0))))))) → A1(b(a(a(a(y_0)))))
B1(a(b(b(a(a(a(b(y_0)))))))) → A1(b(a(a(a(b(y_0))))))
B1(a(b(b(a(a(b(y_0))))))) → A1(b(a(a(b(y_0)))))
B1(a(b(b(y_0)))) → A1(b(y_0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(b(a(a(a(y_0))))))) → A1(b(a(a(a(y_0)))))
A1(b(a(a(a(b(y_0)))))) → B1(a(a(b(y_0))))
A1(b(a(a(a(y_0))))) → B1(a(a(y_0)))
B1(a(b(b(a(a(a(b(y_0)))))))) → A1(b(a(a(a(b(y_0))))))
B1(a(a(b(y_0)))) → B1(a(b(y_0)))
A1(b(a(a(b(y_0))))) → B1(a(b(y_0)))
B1(a(b(b(a(a(b(y_0))))))) → A1(b(a(a(b(y_0)))))
B1(a(a(y_0))) → B1(a(y_0))
B1(a(b(b(y_0)))) → A1(b(y_0))
A1(b(x1)) → A1(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A1(b(x1)) → A1(x1) we obtained the following new rules:
A1(b(b(a(a(a(y_0)))))) → A1(b(a(a(a(y_0)))))
A1(b(b(a(a(a(b(y_0))))))) → A1(b(a(a(a(b(y_0))))))
A1(b(b(a(a(b(y_0)))))) → A1(b(a(a(b(y_0)))))
A1(b(b(y_0))) → A1(b(y_0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(b(a(a(a(y_0))))))) → A1(b(a(a(a(y_0)))))
B1(a(b(b(a(a(a(b(y_0)))))))) → A1(b(a(a(a(b(y_0))))))
B1(a(a(b(y_0)))) → B1(a(b(y_0)))
B1(a(a(y_0))) → B1(a(y_0))
A1(b(b(a(a(a(y_0)))))) → A1(b(a(a(a(y_0)))))
B1(a(b(b(y_0)))) → A1(b(y_0))
A1(b(b(a(a(b(y_0)))))) → A1(b(a(a(b(y_0)))))
A1(b(b(y_0))) → A1(b(y_0))
A1(b(a(a(a(b(y_0)))))) → B1(a(a(b(y_0))))
A1(b(a(a(a(y_0))))) → B1(a(a(y_0)))
A1(b(a(a(b(y_0))))) → B1(a(b(y_0)))
B1(a(b(b(a(a(b(y_0))))))) → A1(b(a(a(b(y_0)))))
A1(b(b(a(a(a(b(y_0))))))) → A1(b(a(a(a(b(y_0))))))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
B1(a(b(b(a(a(a(y_0))))))) → A1(b(a(a(a(y_0)))))
B1(a(b(b(a(a(a(b(y_0)))))))) → A1(b(a(a(a(b(y_0))))))
B1(a(a(b(y_0)))) → B1(a(b(y_0)))
A1(b(b(a(a(a(y_0)))))) → A1(b(a(a(a(y_0)))))
B1(a(b(b(y_0)))) → A1(b(y_0))
A1(b(b(a(a(b(y_0)))))) → A1(b(a(a(b(y_0)))))
A1(b(b(y_0))) → A1(b(y_0))
A1(b(a(a(a(b(y_0)))))) → B1(a(a(b(y_0))))
A1(b(a(a(a(y_0))))) → B1(a(a(y_0)))
A1(b(a(a(b(y_0))))) → B1(a(b(y_0)))
B1(a(b(b(a(a(b(y_0))))))) → A1(b(a(a(b(y_0)))))
A1(b(b(a(a(a(b(y_0))))))) → A1(b(a(a(a(b(y_0))))))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A1(x1)) = 1 + x1
POL(B1(x1)) = 1 + 2·x1
POL(a(x1)) = 2·x1
POL(b(x1)) = 1 + x1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(y_0))) → B1(a(y_0))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
B1(a(a(y_0))) → B1(a(y_0))
Used ordering: POLO with Polynomial interpretation [25]:
POL(B1(x1)) = 2·x1
POL(a(x1)) = 1 + 2·x1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We have reversed the following QTRS:
The set of rules R is
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
b(A(x)) → A(B(a(b(x))))
a(B(x)) → B(A(b(a(x))))
a(A(x)) → x
b(B(x)) → 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:
b(A(x)) → A(B(a(b(x))))
a(B(x)) → B(A(b(a(x))))
a(A(x)) → x
b(B(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
A(b(x1)) → b(a(B(A(x1))))
B(a(x1)) → a(b(A(B(x1))))
A(a(x1)) → x1
B(b(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
b(A(x)) → A(B(a(b(x))))
a(B(x)) → B(A(b(a(x))))
a(A(x)) → x
b(B(x)) → 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:
b(A(x)) → A(B(a(b(x))))
a(B(x)) → B(A(b(a(x))))
a(A(x)) → x
b(B(x)) → x
Q is empty.