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(a(a(a(x1)))) → b(b(x1))
b(b(a(a(x1)))) → a(a(b(b(x1))))
b(b(b(b(c(c(x1)))))) → c(c(a(a(x1))))
b(b(b(b(x1)))) → a(a(a(a(a(a(x1))))))
c(c(a(a(x1)))) → b(b(a(a(c(c(x1))))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(a(x1)))) → b(b(x1))
b(b(a(a(x1)))) → a(a(b(b(x1))))
b(b(b(b(c(c(x1)))))) → c(c(a(a(x1))))
b(b(b(b(x1)))) → a(a(a(a(a(a(x1))))))
c(c(a(a(x1)))) → b(b(a(a(c(c(x1))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(a(a(x1)))) → b(b(x1))
b(b(a(a(x1)))) → a(a(b(b(x1))))
b(b(b(b(c(c(x1)))))) → c(c(a(a(x1))))
b(b(b(b(x1)))) → a(a(a(a(a(a(x1))))))
c(c(a(a(x1)))) → b(b(a(a(c(c(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(a(a(x)))) → b(b(x))
a(a(b(b(x)))) → b(b(a(a(x))))
c(c(b(b(b(b(x)))))) → a(a(c(c(x))))
b(b(b(b(x)))) → a(a(a(a(a(a(x))))))
a(a(c(c(x)))) → c(c(a(a(b(b(x))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(a(x)))) → b(b(x))
a(a(b(b(x)))) → b(b(a(a(x))))
c(c(b(b(b(b(x)))))) → a(a(c(c(x))))
b(b(b(b(x)))) → a(a(a(a(a(a(x))))))
a(a(c(c(x)))) → c(c(a(a(b(b(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(a(a(x1)))) → b(b(x1))
b(b(a(a(x1)))) → a(a(b(b(x1))))
b(b(b(b(c(c(x1)))))) → c(c(a(a(x1))))
b(b(b(b(x1)))) → a(a(a(a(a(a(x1))))))
c(c(a(a(x1)))) → b(b(a(a(c(c(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(a(a(x)))) → b(b(x))
a(a(b(b(x)))) → b(b(a(a(x))))
c(c(b(b(b(b(x)))))) → a(a(c(c(x))))
b(b(b(b(x)))) → a(a(a(a(a(a(x))))))
a(a(c(c(x)))) → c(c(a(a(b(b(x))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(a(x)))) → b(b(x))
a(a(b(b(x)))) → b(b(a(a(x))))
c(c(b(b(b(b(x)))))) → a(a(c(c(x))))
b(b(b(b(x)))) → a(a(a(a(a(a(x))))))
a(a(c(c(x)))) → c(c(a(a(b(b(x))))))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(a(x1)))) → b(b(x1))
b(b(a(a(x1)))) → a(a(b(b(x1))))
b(b(b(b(c(c(x1)))))) → c(c(a(a(x1))))
b(b(b(b(x1)))) → a(a(a(a(a(a(x1))))))
c(c(a(a(x1)))) → b(b(a(a(c(c(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:
b(b(b(b(x1)))) → a(a(a(a(a(a(x1))))))
c(c(a(a(x1)))) → b(b(a(a(c(c(x1))))))
Used ordering:
Polynomial interpretation [25]:
POL(a(x1)) = 1 + x1
POL(b(x1)) = 2 + x1
POL(c(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(a(x1)))) → b(b(x1))
b(b(a(a(x1)))) → a(a(b(b(x1))))
b(b(b(b(c(c(x1)))))) → c(c(a(a(x1))))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(a(x1)))) → b(b(x1))
b(b(a(a(x1)))) → a(a(b(b(x1))))
b(b(b(b(c(c(x1)))))) → c(c(a(a(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:
a(a(a(a(x1)))) → b(b(x1))
b(b(b(b(c(c(x1)))))) → c(c(a(a(x1))))
Used ordering:
Polynomial interpretation [25]:
POL(a(x1)) = 2 + 2·x1
POL(b(x1)) = 2 + 2·x1
POL(c(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(a(x1)))) → a(a(b(b(x1))))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(a(x1)))) → a(a(b(b(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:
b(b(a(a(x1)))) → a(a(b(b(x1))))
Used ordering:
Polynomial interpretation [25]:
POL(a(x1)) = 2 + 2·x1
POL(b(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RisEmptyProof
Q restricted rewrite system:
R is empty.
Q is empty.
The TRS R is empty. Hence, termination is trivially proven.