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