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(s(X), X) → f(X, a(X))
f(X, c(X)) → f(s(X), X)
f(X, X) → c(X)
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(X), X) → f(X, a(X))
f(X, c(X)) → f(s(X), X)
f(X, X) → c(X)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(X), X) → f(X, a(X))
f(X, c(X)) → f(s(X), X)
f(X, X) → c(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(X, c(X)) → f(s(X), X)
f(X, X) → c(X)
Used ordering:
Polynomial interpretation [25]:
POL(a(x1)) = x1
POL(c(x1)) = 1 + 2·x1
POL(f(x1, x2)) = 2 + 2·x1 + x2
POL(s(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(X), X) → f(X, a(X))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(X), X) → f(X, 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(s(X), X) → f(X, a(X))
Used ordering:
Polynomial interpretation [25]:
POL(a(x1)) = 1 + 2·x1
POL(f(x1, x2)) = 2·x1 + x2
POL(s(x1)) = 2 + 2·x1
↳ 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.