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(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(c(x1))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(c(x1))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(c(x1))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(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(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(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(x1)))) → A(a(x1))
A(c(x1)) → A(x1)
A(a(b(b(x1)))) → A(x1)
A(a(b(b(x1)))) → A(a(a(x1)))
C(b(x1)) → C(x1)
A(c(x1)) → C(a(x1))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(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:
A(a(b(b(x1)))) → A(a(x1))
A(c(x1)) → A(x1)
A(a(b(b(x1)))) → A(x1)
A(a(b(b(x1)))) → A(a(a(x1)))
C(b(x1)) → C(x1)
A(c(x1)) → C(a(x1))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(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 2 SCCs with 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(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
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → C(x1)
R is empty.
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
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → C(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:
C(b(x1)) → C(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(C(x1)) = 2·x1
POL(b(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ 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.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(b(x1)))) → A(a(x1))
A(c(x1)) → A(x1)
A(a(b(b(x1)))) → A(x1)
A(a(b(b(x1)))) → A(a(a(x1)))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(c(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:
A(c(x1)) → A(x1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = x1
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(x1)
A(a(b(b(x1)))) → A(a(a(x1)))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(c(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(b(b(x1)))) → A(a(a(x1))) at position [0] we obtained the following new rules:
A(a(b(b(c(x0))))) → A(a(c(a(x0))))
A(a(b(b(a(b(b(x0))))))) → A(a(b(b(b(a(a(a(x0))))))))
A(a(b(b(b(b(x0)))))) → A(b(b(b(a(a(a(x0)))))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(a(b(b(x0))))))) → A(a(b(b(b(a(a(a(x0))))))))
A(a(b(b(c(x0))))) → A(a(c(a(x0))))
A(a(b(b(x1)))) → A(x1)
A(a(b(b(b(b(x0)))))) → A(b(b(b(a(a(a(x0)))))))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(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 1 SCC with 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(a(b(b(x0))))))) → A(a(b(b(b(a(a(a(x0))))))))
A(a(b(b(c(x0))))) → A(a(c(a(x0))))
A(a(b(b(x1)))) → A(x1)
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(c(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A(a(b(b(x1)))) → A(x1) we obtained the following new rules:
A(a(b(b(a(y_2))))) → A(a(y_2))
A(a(b(b(a(b(b(a(y_2)))))))) → A(a(b(b(a(y_2)))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(c(x0))))) → A(a(c(a(x0))))
A(a(b(b(a(b(b(x0))))))) → A(a(b(b(b(a(a(a(x0))))))))
A(a(b(b(a(y_2))))) → A(a(y_2))
A(a(b(b(a(b(b(a(y_2)))))))) → A(a(b(b(a(y_2)))))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(c(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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(c(x1))
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(c(x0))))) → A(a(c(a(x0))))
A(a(b(b(a(b(b(x0))))))) → A(a(b(b(b(a(a(a(x0))))))))
A(a(b(b(a(y_2))))) → A(a(y_2))
A(a(b(b(a(b(b(a(y_2)))))))) → A(a(b(b(a(y_2)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(c(x1))
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(c(x0))))) → A(a(c(a(x0))))
A(a(b(b(a(b(b(x0))))))) → A(a(b(b(b(a(a(a(x0))))))))
A(a(b(b(a(y_2))))) → A(a(y_2))
A(a(b(b(a(b(b(a(y_2)))))))) → A(a(b(b(a(y_2)))))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(b(a(A(x)))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(b(a(A(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(b(a(A(x)))))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(b(x)))) → b(b(b(a(a(a(x))))))
a(c(x)) → c(a(x))
c(b(x)) → b(c(x))
A(a(b(b(x)))) → A(a(x))
A(a(b(b(c(x))))) → A(a(c(a(x))))
A(a(b(b(a(b(b(x))))))) → A(a(b(b(b(a(a(a(x))))))))
A(a(b(b(a(x))))) → A(a(x))
A(a(b(b(a(b(b(a(x)))))))) → A(a(b(b(a(x)))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(x)))) → b(b(b(a(a(a(x))))))
a(c(x)) → c(a(x))
c(b(x)) → b(c(x))
A(a(b(b(x)))) → A(a(x))
A(a(b(b(c(x))))) → A(a(c(a(x))))
A(a(b(b(a(b(b(x))))))) → A(a(b(b(b(a(a(a(x))))))))
A(a(b(b(a(x))))) → A(a(x))
A(a(b(b(a(b(b(a(x)))))))) → A(a(b(b(a(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(b(a(A(x)))))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(b(x)))) → b(b(b(a(a(a(x))))))
a(c(x)) → c(a(x))
c(b(x)) → b(c(x))
A(a(b(b(x)))) → A(a(x))
A(a(b(b(c(x))))) → A(a(c(a(x))))
A(a(b(b(a(b(b(x))))))) → A(a(b(b(b(a(a(a(x))))))))
A(a(b(b(a(x))))) → A(a(x))
A(a(b(b(a(b(b(a(x)))))))) → A(a(b(b(a(x)))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(x)))) → b(b(b(a(a(a(x))))))
a(c(x)) → c(a(x))
c(b(x)) → b(c(x))
A(a(b(b(x)))) → A(a(x))
A(a(b(b(c(x))))) → A(a(c(a(x))))
A(a(b(b(a(b(b(x))))))) → A(a(b(b(b(a(a(a(x))))))))
A(a(b(b(a(x))))) → A(a(x))
A(a(b(b(a(b(b(a(x)))))))) → A(a(b(b(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(b(a(b(b(a(A(x))))))) → A1(b(b(b(a(A(x))))))
B(b(a(a(x)))) → A1(b(b(b(x))))
C(a(x)) → A1(c(x))
B(b(a(b(b(a(A(x))))))) → A1(a(a(b(b(b(a(A(x))))))))
C(a(x)) → C(x)
B(b(a(a(x)))) → A1(a(a(b(b(b(x))))))
C(b(b(a(A(x))))) → A1(c(a(A(x))))
B(b(a(a(x)))) → B(b(x))
B(b(a(a(x)))) → B(x)
B(b(a(a(x)))) → B(b(b(x)))
B(b(a(b(b(a(A(x))))))) → A1(a(b(b(b(a(A(x)))))))
B(b(a(a(x)))) → A1(a(b(b(b(x)))))
B(c(x)) → C(b(x))
C(b(b(a(A(x))))) → C(a(A(x)))
B(b(a(b(b(a(A(x))))))) → B(b(b(a(A(x)))))
B(c(x)) → B(x)
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(b(a(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(b(b(a(A(x))))))) → A1(b(b(b(a(A(x))))))
B(b(a(a(x)))) → A1(b(b(b(x))))
C(a(x)) → A1(c(x))
B(b(a(b(b(a(A(x))))))) → A1(a(a(b(b(b(a(A(x))))))))
C(a(x)) → C(x)
B(b(a(a(x)))) → A1(a(a(b(b(b(x))))))
C(b(b(a(A(x))))) → A1(c(a(A(x))))
B(b(a(a(x)))) → B(b(x))
B(b(a(a(x)))) → B(x)
B(b(a(a(x)))) → B(b(b(x)))
B(b(a(b(b(a(A(x))))))) → A1(a(b(b(b(a(A(x)))))))
B(b(a(a(x)))) → A1(a(b(b(b(x)))))
B(c(x)) → C(b(x))
C(b(b(a(A(x))))) → C(a(A(x)))
B(b(a(b(b(a(A(x))))))) → B(b(b(a(A(x)))))
B(c(x)) → B(x)
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(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 2 SCCs with 9 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x)) → C(x)
C(b(b(a(A(x))))) → C(a(A(x)))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(b(a(A(x)))))
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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x)) → C(x)
C(b(b(a(A(x))))) → C(a(A(x)))
R is empty.
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
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x)) → C(x)
C(b(b(a(A(x))))) → C(a(A(x)))
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:
C(a(x)) → C(x)
C(b(b(a(A(x))))) → C(a(A(x)))
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2 + 2·x1
POL(C(x1)) = 2·x1
POL(a(x1)) = 2 + 2·x1
POL(b(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ 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.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(x)))) → B(x)
B(b(a(a(x)))) → B(b(b(x)))
B(b(a(a(x)))) → B(b(x))
B(b(a(b(b(a(A(x))))))) → B(b(b(a(A(x)))))
B(c(x)) → B(x)
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(b(a(A(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:
B(c(x)) → B(x)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(B(x1)) = x1
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(x)))) → B(x)
B(b(a(a(x)))) → B(b(b(x)))
B(b(a(a(x)))) → B(b(x))
B(b(a(b(b(a(A(x))))))) → B(b(b(a(A(x)))))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(b(a(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(a(b(b(a(A(x))))))) → B(b(b(a(A(x))))) at position [0] we obtained the following new rules:
B(b(a(b(b(a(A(x0))))))) → B(a(A(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(x)))) → B(x)
B(b(a(a(x)))) → B(b(b(x)))
B(b(a(a(x)))) → B(b(x))
B(b(a(b(b(a(A(x0))))))) → B(a(A(x0)))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(x)))) → B(x)
B(b(a(a(x)))) → B(b(b(x)))
B(b(a(a(x)))) → B(b(x))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(b(a(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(a(a(x)))) → B(b(b(x))) at position [0] we obtained the following new rules:
B(b(a(a(a(A(x0)))))) → B(a(A(x0)))
B(b(a(a(a(b(b(a(A(x0))))))))) → B(a(a(a(b(b(b(a(A(x0)))))))))
B(b(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(x0)))))))
B(b(a(a(b(a(a(x0))))))) → B(b(a(a(a(b(b(b(x0))))))))
B(b(a(a(b(a(b(b(a(A(x0)))))))))) → B(b(a(a(a(b(b(b(a(A(x0))))))))))
B(b(a(a(c(x0))))) → B(b(c(b(x0))))
B(b(a(a(b(a(A(x0))))))) → B(b(a(A(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(a(b(b(a(A(x0))))))))) → B(a(a(a(b(b(b(a(A(x0)))))))))
B(b(a(a(a(A(x0)))))) → B(a(A(x0)))
B(b(a(a(x)))) → B(x)
B(b(a(a(a(a(x0)))))) → B(a(a(a(b(b(b(x0)))))))
B(b(a(a(b(a(a(x0))))))) → B(b(a(a(a(b(b(b(x0))))))))
B(b(a(a(b(a(b(b(a(A(x0)))))))))) → B(b(a(a(a(b(b(b(a(A(x0))))))))))
B(b(a(a(x)))) → B(b(x))
B(b(a(a(c(x0))))) → B(b(c(b(x0))))
B(b(a(a(b(a(A(x0))))))) → B(b(a(A(x0))))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(x)))) → B(x)
B(b(a(a(b(a(a(x0))))))) → B(b(a(a(a(b(b(b(x0))))))))
B(b(a(a(b(a(b(b(a(A(x0)))))))))) → B(b(a(a(a(b(b(b(a(A(x0))))))))))
B(b(a(a(x)))) → B(b(x))
B(b(a(a(c(x0))))) → B(b(c(b(x0))))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(b(a(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B(b(a(a(x)))) → B(x) we obtained the following new rules:
B(b(a(a(b(y_2))))) → B(b(y_2))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(b(y_2))))) → B(b(y_2))
B(b(a(a(b(a(a(x0))))))) → B(b(a(a(a(b(b(b(x0))))))))
B(b(a(a(b(a(b(b(a(A(x0)))))))))) → B(b(a(a(a(b(b(b(a(A(x0))))))))))
B(b(a(a(x)))) → B(b(x))
B(b(a(a(c(x0))))) → B(b(c(b(x0))))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
b(b(a(A(x)))) → a(A(x))
c(b(b(a(A(x))))) → a(c(a(A(x))))
b(b(a(b(b(a(A(x))))))) → a(a(a(b(b(b(a(A(x))))))))
a(b(b(a(A(x))))) → a(A(x))
a(b(b(a(b(b(a(A(x)))))))) → a(b(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
a(a(b(b(x1)))) → b(b(b(a(a(a(x1))))))
a(c(x1)) → c(a(x1))
c(b(x1)) → b(c(x1))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(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(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
b(c(x)) → c(b(x))
Q is empty.