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Let A be an algebra and let $\alpha,\beta,\gamma$ congruence in A. How can I prove that the commutator is meet semidistributive in the second variable ( i.e. $[\beta,\alpha]=[\beta,\gamma]$ implies $[\beta,\alpha]=[\beta,\gamma\vee\alpha]$ ) ?

Def. For any $\alpha,\beta \in$ Con(A) there is a smallest congreunce $\delta$ with C$(\alpha,\beta;\delta)$; such congruence is called the commutator of $\alpha$ and $\beta$ and is denoted by $[\alpha,\beta]$.

C(x,y;z) is the centralizer

Seurat
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    Please provide more context. What are $\alpha$, $\beta$, and $\gamma$ here? The word "commutator" can mean different things in different contexts. – Alex Kruckman Oct 08 '22 at 15:44
  • Edited @AlexKruckman – Seurat Oct 08 '22 at 16:11
  • If you are only interested in the right semidistributivity of the commutator for congruence modular varieties, see the 1987 commutator theory book of Freese and McKenzie. If you are interested in the right semidistributivity of the commutator for varieties with a Taylor term, see https://arxiv.org/abs/2202.10009 – Keith Kearnes Oct 08 '22 at 17:21
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    You didn't define the centralizer. I suppose many people won't know what it is, even among those minimally acquainted with Universal Algebra. Anyway, I suppose the comment of Keith Kearnes will sort you out. – amrsa Oct 08 '22 at 18:17
  • Now it's clear. Thank you @KeithKearnes – Seurat Oct 08 '22 at 18:36

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