α -r e 2 s at , and bt ≥ α 2i-r e 2 α -r e 2 s at . Since α r e 1 a ≥ e 1 at, there exists v ∈ S such that at = α r e 1 av. Then e 1 sat . We can prove similarly the other inequations. Thus it follows from the equations above and Lemma 7(ii) that α 2i-r α -r e 2 s at = α 2i-r f 2 α -r f 2 s at . T
On multiplicative bases in commutative semigroups
✍ Scribed by Vladimi´r Pusˇ
- Publisher
- Elsevier Science
- Year
- 1992
- Tongue
- English
- Weight
- 851 KB
- Volume
- 13
- Category
- Article
- ISSN
- 0195-6698
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