Universal algebra: Revision history

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13 September 2014

11 February 2014

17 January 2014

21 June 2012

  • curprev 05:1405:14, 21 June 2012en>Magidin 20,003 bytes +20,003 not just arbitrary (e.g., we don't ask that a homomorphism of rings satisfy $f(a+b) = f(a)f(b)$, just because $+$ and $\cdot$ have the same arity).