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Nov 22 at 20:38 vote accept user107952
S Nov 19 at 17:12 history bounty ended user107952
S Nov 19 at 17:12 history notice removed user107952
Nov 19 at 6:08 answer added Keith Kearnes timeline score: 9
Nov 13 at 1:17 comment added C7X If this question has a negative answer, it may be possible to resolve it by looking through the implication graph produced by Terence Tao's recent equational theories project, searching for pairs of nodes without least upper bounds or greatest lower bounds. The table of implications is available through this page.
Nov 12 at 23:30 comment added Peter Taylor In general, $p(\vec{x})=q(\vec{x})$ implies $r(p(\vec{x}), q(\vec{x}), \vec{y}) = r(q(\vec{x}), p(\vec{x}), \vec{y})$. By using $r_i$ of arity four or more we can generate arbitrarily many lower bounds on a pair of equations, and in general there's no reason to suppose that these lower bounds are mutually comparable.
Nov 12 at 11:47 comment added Peter Taylor You have least and greatest elements $x=x$ and $x=y$, so the question is whether two elements can have multiple incomparable common maximal lower / minimal upper bounds.
S Nov 11 at 18:25 history bounty started user107952
S Nov 11 at 18:25 history notice added user107952 Draw attention
Oct 27 at 19:07 comment added user107952 @bof I don't know, but if it has neither, that would answer my question.
Oct 27 at 0:46 comment added bof Do the associative law and the commutative law have a least upper bound and/or a greatest lower bound in your partial order?
Oct 26 at 17:21 history edited LSpice CC BY-SA 4.0
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Oct 26 at 16:44 history asked user107952 CC BY-SA 4.0