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