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:
f(a, x) → g(a, x)
g(a, x) → f(b, x)
f(a, x) → f(b, x)
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, x) → g(a, x)
g(a, x) → f(b, x)
f(a, x) → f(b, x)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, x) → g(a, x)
g(a, x) → f(b, x)
f(a, x) → f(b, x)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
g(a, x) → f(b, x)
f(a, x) → f(b, x)
Used ordering:
Polynomial interpretation [25]:
POL(a) = 1
POL(b) = 0
POL(f(x1, x2)) = 1 + x1 + x2
POL(g(x1, x2)) = 2·x1 + x2
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, x) → g(a, x)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, x) → g(a, x)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
f(a, x) → g(a, x)
Used ordering:
Polynomial interpretation [25]:
POL(a) = 1
POL(f(x1, x2)) = 2 + 2·x1 + 2·x2
POL(g(x1, x2)) = 1 + 2·x1 + 2·x2
↳ 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.