Abstract
Abstract A basis of identities for an algebra is irredundant if each of its proper subsets fails to be a basis for the algebra. The first known examples of finite involution semigroups with infinite irredundant bases are exhibited. These involution semigroups satisfy several counterintuitive properties: their semigroup reducts do not have irredundant bases, they share reducts with some other finitely based involution semigroups, and they are direct products of finitely based involution semigroups.
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