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:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
c(a(x1)) → a(b(c(x1)))
c(b(x1)) → b(a(c(x1)))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
c(a(x1)) → a(b(c(x1)))
c(b(x1)) → b(a(c(x1)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
c(a(x1)) → a(b(c(x1)))
c(b(x1)) → b(a(c(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(c(x)) → c(b(a(x)))
b(c(x)) → c(a(b(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(c(x)) → c(b(a(x)))
b(c(x)) → c(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(b(a(x1))) → B(b(x1))
C(b(x1)) → B(a(c(x1)))
B(b(a(x1))) → B(x1)
C(a(x1)) → B(c(x1))
C(a(x1)) → C(x1)
B(b(a(x1))) → B(b(b(x1)))
B(a(b(x1))) → B(a(x1))
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
c(a(x1)) → a(b(c(x1)))
c(b(x1)) → b(a(c(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(x1))) → B(b(x1))
C(b(x1)) → B(a(c(x1)))
B(b(a(x1))) → B(x1)
C(a(x1)) → B(c(x1))
C(a(x1)) → C(x1)
B(b(a(x1))) → B(b(b(x1)))
B(a(b(x1))) → B(a(x1))
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
c(a(x1)) → a(b(c(x1)))
c(b(x1)) → b(a(c(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(x1))) → B(a(x1))
The TRS R consists of the following rules:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
c(a(x1)) → a(b(c(x1)))
c(b(x1)) → b(a(c(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(x1))) → B(a(x1))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
B(a(b(x1))) → B(a(x1))
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(B(x1)) = 2·x1
POL(a(x1)) = 2·x1
POL(b(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(x1))) → B(a(x1))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(x1))) → B(b(x1))
B(b(a(x1))) → B(x1)
B(b(a(x1))) → B(b(b(x1)))
The TRS R consists of the following rules:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
c(a(x1)) → a(b(c(x1)))
c(b(x1)) → b(a(c(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(x1))) → B(b(x1))
B(b(a(x1))) → B(x1)
B(b(a(x1))) → B(b(b(x1)))
The TRS R consists of the following rules:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
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:
B(b(a(x1))) → B(b(x1))
B(b(a(x1))) → B(x1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(B(x1)) = x1
POL(a(x1)) = 2 + 2·x1
POL(b(x1)) = 2 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(x1))) → B(b(b(x1)))
The TRS R consists of the following rules:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(a(x1))) → B(b(b(x1))) at position [0] we obtained the following new rules:
B(b(a(b(a(x0))))) → B(b(b(b(b(x0)))))
B(b(a(a(x0)))) → B(b(b(b(x0))))
B(b(a(a(b(x0))))) → B(b(a(b(a(x0)))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(b(a(x0))))) → B(b(b(b(b(x0)))))
B(b(a(a(b(x0))))) → B(b(a(b(a(x0)))))
B(b(a(a(x0)))) → B(b(b(b(x0))))
The TRS R consists of the following rules:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
Q is empty.
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
B(b(a(b(a(x0))))) → B(b(b(b(b(x0)))))
B(b(a(a(b(x0))))) → B(b(a(b(a(x0)))))
B(b(a(a(x0)))) → B(b(b(b(x0))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
B(b(a(b(a(x0))))) → B(b(b(b(b(x0)))))
B(b(a(a(b(x0))))) → B(b(a(b(a(x0)))))
B(b(a(a(x0)))) → B(b(b(b(x0))))
The set Q is empty.
We have obtained the following QTRS:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(b(b(B(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(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:
A(a(b(B(x)))) → B1(b(b(B(x))))
B1(a(b(x))) → A(x)
B1(a(b(x))) → A(b(a(x)))
A(b(a(b(B(x))))) → B1(b(b(b(B(x)))))
A(a(b(B(x)))) → B1(b(B(x)))
A(b(a(b(B(x))))) → B1(b(B(x)))
B1(a(a(b(B(x))))) → A(b(a(b(B(x)))))
B1(a(b(x))) → B1(a(x))
B1(a(a(b(B(x))))) → B1(a(b(B(x))))
A(b(a(b(B(x))))) → B1(b(b(B(x))))
A(b(b(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(b(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(B(x)))) → B1(b(b(B(x))))
B1(a(b(x))) → A(x)
B1(a(b(x))) → A(b(a(x)))
A(b(a(b(B(x))))) → B1(b(b(b(B(x)))))
A(a(b(B(x)))) → B1(b(B(x)))
A(b(a(b(B(x))))) → B1(b(B(x)))
B1(a(a(b(B(x))))) → A(b(a(b(B(x)))))
B1(a(b(x))) → B1(a(x))
B1(a(a(b(B(x))))) → B1(a(b(B(x))))
A(b(a(b(B(x))))) → B1(b(b(B(x))))
A(b(b(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(x))) → A(x)
B1(a(b(x))) → A(b(a(x)))
B1(a(a(b(B(x))))) → A(b(a(b(B(x)))))
B1(a(b(x))) → B1(a(x))
B1(a(a(b(B(x))))) → B1(a(b(B(x))))
A(b(b(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(b(b(B(x))))
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(b(B(x))))) → B1(a(b(B(x))))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(B(x1)) = 2 + 2·x1
POL(B1(x1)) = 2·x1
POL(a(x1)) = 2·x1
POL(b(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(x))) → A(b(a(x)))
B1(a(b(x))) → A(x)
B1(a(b(x))) → B1(a(x))
B1(a(a(b(B(x))))) → A(b(a(b(B(x)))))
A(b(b(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(b(b(B(x))))
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(x))) → A(x)
B1(a(b(x))) → B1(a(x))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = x1
POL(B(x1)) = x1
POL(B1(x1)) = x1
POL(a(x1)) = 1 + 2·x1
POL(b(x1)) = 1 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(x))) → A(b(a(x)))
B1(a(a(b(B(x))))) → A(b(a(b(B(x)))))
A(b(b(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(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 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(x))) → A(b(a(x)))
A(b(b(x))) → B1(b(b(x)))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(b(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(x))) → B1(b(b(x))) at position [0] we obtained the following new rules:
A(b(b(a(b(x0))))) → B1(b(a(b(a(x0)))))
A(b(b(a(a(b(B(x0))))))) → B1(b(a(b(a(b(B(x0)))))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(x))) → A(b(a(x)))
A(b(b(a(b(x0))))) → B1(b(a(b(a(x0)))))
A(b(b(a(a(b(B(x0))))))) → B1(b(a(b(a(b(B(x0)))))))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(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 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(x))) → A(b(a(x)))
A(b(b(a(b(x0))))) → B1(b(a(b(a(x0)))))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(b(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(b(x))) → A(b(a(x))) at position [0] we obtained the following new rules:
B1(a(b(a(b(B(x0)))))) → A(a(b(a(b(B(x0))))))
B1(a(b(b(x0)))) → A(a(b(a(x0))))
B1(a(b(a(b(B(x0)))))) → A(b(b(b(b(B(x0))))))
B1(a(b(b(b(x0))))) → A(b(b(b(b(x0)))))
B1(a(b(b(a(b(B(x0))))))) → A(b(b(b(b(b(B(x0)))))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(b(x0)))) → A(a(b(a(x0))))
B1(a(b(a(b(B(x0)))))) → A(a(b(a(b(B(x0))))))
A(b(b(a(b(x0))))) → B1(b(a(b(a(x0)))))
B1(a(b(a(b(B(x0)))))) → A(b(b(b(b(B(x0))))))
B1(a(b(b(a(b(B(x0))))))) → A(b(b(b(b(b(B(x0)))))))
B1(a(b(b(b(x0))))) → A(b(b(b(b(x0)))))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(b(x0)))) → A(a(b(a(x0))))
B1(a(b(a(b(B(x0)))))) → A(a(b(a(b(B(x0))))))
A(b(b(a(b(x0))))) → B1(b(a(b(a(x0)))))
B1(a(b(b(b(x0))))) → A(b(b(b(b(x0)))))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(b(b(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(b(a(b(B(x0)))))) → A(a(b(a(b(B(x0)))))) at position [0] we obtained the following new rules:
B1(a(b(a(b(B(y0)))))) → A(a(a(b(a(B(y0))))))
B1(a(b(a(b(B(x0)))))) → A(b(b(b(b(B(x0))))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(b(x0)))) → A(a(b(a(x0))))
A(b(b(a(b(x0))))) → B1(b(a(b(a(x0)))))
B1(a(b(a(b(B(x0)))))) → A(b(b(b(b(B(x0))))))
B1(a(b(a(b(B(y0)))))) → A(a(a(b(a(B(y0))))))
B1(a(b(b(b(x0))))) → A(b(b(b(b(x0)))))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(b(x0)))) → A(a(b(a(x0))))
A(b(b(a(b(x0))))) → B1(b(a(b(a(x0)))))
B1(a(b(b(b(x0))))) → A(b(b(b(b(x0)))))
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(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
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(b(b(B(x))))
The set Q is empty.
We have obtained the following QTRS:
b(a(b(x))) → a(b(a(x)))
b(b(a(x))) → b(b(b(x)))
B(b(a(b(a(x))))) → B(b(b(b(b(x)))))
B(b(a(a(b(x))))) → B(b(a(b(a(x)))))
B(b(a(a(x)))) → B(b(b(b(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
b(b(a(x))) → b(b(b(x)))
B(b(a(b(a(x))))) → B(b(b(b(b(x)))))
B(b(a(a(b(x))))) → B(b(a(b(a(x)))))
B(b(a(a(x)))) → B(b(b(b(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(b(a(b(B(x))))) → b(b(b(b(B(x)))))
b(a(a(b(B(x))))) → a(b(a(b(B(x)))))
a(a(b(B(x)))) → b(b(b(B(x))))
The set Q is empty.
We have obtained the following QTRS:
b(a(b(x))) → a(b(a(x)))
b(b(a(x))) → b(b(b(x)))
B(b(a(b(a(x))))) → B(b(b(b(b(x)))))
B(b(a(a(b(x))))) → B(b(a(b(a(x)))))
B(b(a(a(x)))) → B(b(b(b(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
b(b(a(x))) → b(b(b(x)))
B(b(a(b(a(x))))) → B(b(b(b(b(x)))))
B(b(a(a(b(x))))) → B(b(a(b(a(x)))))
B(b(a(a(x)))) → B(b(b(b(x))))
Q is empty.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(x1))) → B(b(x1))
B(b(a(x1))) → B(x1)
B(b(a(x1))) → B(b(b(x1)))
The TRS R consists of the following rules:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x1)) → C(x1)
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
b(a(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
c(a(x1)) → a(b(c(x1)))
c(b(x1)) → b(a(c(x1)))
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(b(x1))) → a(b(a(x1)))
b(b(a(x1))) → b(b(b(x1)))
c(a(x1)) → a(b(c(x1)))
c(b(x1)) → b(a(c(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(c(x)) → c(b(a(x)))
b(c(x)) → c(a(b(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(x))) → a(b(a(x)))
a(b(b(x))) → b(b(b(x)))
a(c(x)) → c(b(a(x)))
b(c(x)) → c(a(b(x)))
Q is empty.