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:
C(C(x1)) → c(c(x1))
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → x1
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
C(C(x1)) → c(c(x1))
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → x1
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
C(C(x1)) → c(c(x1))
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → x1
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
C(C(x1)) → c(c(x1))
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → x1
Used ordering:
Polynomial interpretation [25]:
POL(B(x1)) = x1
POL(C(x1)) = 2 + x1
POL(b(x1)) = x1
POL(c(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
c(c(c(c(x1)))) → x1
Used ordering:
Polynomial interpretation [25]:
POL(B(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = 2 + 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
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:
B1(b(b(b(x1)))) → B2(B(x1))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(b(b(c(c(B(B(c(c(b(b(x1)))))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(b(c(c(B(B(c(c(b(b(x1))))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(x1)
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(b(b(x1))))
B2(B(B(B(x1)))) → B1(x1)
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(B(c(c(b(b(x1))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(B(B(c(c(b(b(x1)))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(B(B(c(c(b(b(x1))))))))
B2(B(B(B(x1)))) → B1(b(x1))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(c(c(B(B(c(c(b(b(x1)))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(b(x1))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(b(b(x1)))
B1(b(b(b(x1)))) → B2(x1)
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(c(c(b(b(x1)))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(c(c(b(b(c(c(B(B(c(c(b(b(x1)))))))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))
The TRS R consists of the following rules:
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(b(x1)))) → B2(B(x1))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(b(b(c(c(B(B(c(c(b(b(x1)))))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(b(c(c(B(B(c(c(b(b(x1))))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(x1)
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(b(b(x1))))
B2(B(B(B(x1)))) → B1(x1)
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(B(c(c(b(b(x1))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(B(B(c(c(b(b(x1)))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(B(B(c(c(b(b(x1))))))))
B2(B(B(B(x1)))) → B1(b(x1))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(c(c(B(B(c(c(b(b(x1)))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(b(x1))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(b(b(x1)))
B1(b(b(b(x1)))) → B2(x1)
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(c(c(b(b(x1)))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(c(c(b(b(c(c(B(B(c(c(b(b(x1)))))))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))
The TRS R consists of the following rules:
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → 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 8 less nodes.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B2(B(B(B(x1)))) → B1(b(x1))
B1(b(b(b(x1)))) → B2(B(x1))
B1(b(b(b(x1)))) → B2(x1)
B2(B(B(B(x1)))) → B1(x1)
The TRS R consists of the following rules:
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → 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
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B2(B(B(B(x1)))) → B1(b(x1))
B1(b(b(b(x1)))) → B2(B(x1))
B1(b(b(b(x1)))) → B2(x1)
B2(B(B(B(x1)))) → B1(x1)
The TRS R consists of the following rules:
B(B(B(B(x1)))) → b(b(x1))
b(b(b(b(x1)))) → B(B(x1))
B(B(b(b(x1)))) → x1
b(b(B(B(x1)))) → 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
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B2(B(B(B(x1)))) → B1(b(x1))
B1(b(b(b(x1)))) → B2(B(x1))
B1(b(b(b(x1)))) → B2(x1)
B2(B(B(B(x1)))) → B1(x1)
The TRS R consists of the following rules:
B(B(B(B(x1)))) → b(b(x1))
b(b(b(b(x1)))) → B(B(x1))
B(B(b(b(x1)))) → x1
b(b(B(B(x1)))) → 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:
B2(B(B(B(x1)))) → B1(b(x1))
B1(b(b(b(x1)))) → B2(B(x1))
B1(b(b(b(x1)))) → B2(x1)
B2(B(B(B(x1)))) → B1(x1)
Strictly oriented rules of the TRS R:
B(B(B(B(x1)))) → b(b(x1))
b(b(b(b(x1)))) → B(B(x1))
B(B(b(b(x1)))) → x1
b(b(B(B(x1)))) → x1
Used ordering: POLO with Polynomial interpretation [25]:
POL(B(x1)) = 2 + 2·x1
POL(B1(x1)) = 2 + 2·x1
POL(B2(x1)) = 2·x1
POL(b(x1)) = 1 + 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QDP
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
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(b(b(c(c(B(B(c(c(b(b(x1)))))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(b(b(x1)))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(b(b(x1))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(B(B(c(c(b(b(x1)))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(B(B(c(c(b(b(x1))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))
The TRS R consists of the following rules:
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → 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:
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(b(b(c(c(B(B(c(c(b(b(x1)))))))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(b(b(x1)))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(b(b(x1))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(B(B(c(c(b(b(x1)))))))
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(B(B(c(c(b(b(x1))))))))
Used ordering: POLO with Polynomial interpretation [25]:
POL(B(x1)) = x1
POL(C(x1)) = 2·x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C(c(B(B(c(c(b(b(c(c(x1)))))))))) → C(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))
The TRS R consists of the following rules:
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.