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6 hours ago history edited Florian CC BY-SA 4.0
correct a grammar mistake
yesterday comment added Florian @Emil Jeřábek Thanks for your helpful and important suggestions!
yesterday history edited Florian CC BY-SA 4.0
simplified some conditions, and added some conditions
yesterday comment added Emil Jeřábek ... with limit $\sqrt2$, and, say, $x=2$, $y=3$, $z=4$, then $\bigwedge\{x,y\lor z,x_1,x_2,x_3,\dots\}$ does not exist. So in what sense is this “nonexistent element” less than or equal $2=(x\land y)\lor(x\land z)$?
yesterday comment added Emil Jeřábek ... is not a term over lattices, as there is no $\to$ operation in the language of lattices. So I’m not sure what to make of that. Second, despite having been prompted for it already several times, you still haven’t explained what is supposed to happen WHEN $\bigwedge\Gamma$ DOES NOT EXIST in the lattice in question. This is absolutely crucial. For example, you claim that $\bigwedge\{x,y\lor z,x_1,x_2,x_3,\dots\}\le(x\land y)\lor(x\land z)$ holds in distributive lattices. But $(\mathbb Q,\le)$ is a distributive lattice, and when I take for $x_1,x_2,\dots$ a descending rational sequence ...
yesterday comment added Emil Jeřábek Hmm, so the question actually asks something completely different from what it looked originally. However, it is still fairly unclear. First, “formulas” are things like $\forall x\,\exists y\,x<y$ that are either true or false; they do not evaluate in any meaningful way to elements of the lattice. It seems that when you write “formulas”, you actually mean “terms over the lattice signature”, which do indeed take values in the lattice. Certainly $y\lor z$ and $(x\land y)\lor(x\land z)$ are terms. (I’m not going to use the square symbols as they are difficult to type.) However, “$x\to y$” ...
yesterday history edited Florian CC BY-SA 4.0
added 4 characters in body
yesterday history edited Florian CC BY-SA 4.0
further clarification
yesterday comment added Emil Jeřábek You are making the question more and more unclear. I assume “kind of lattice” just means a class $C$ of lattices, in which case the only way I can read “instance” of “this kind” is that is refers to lattices $L\in C$. But then saying “$\bigwedge\Gamma\le y$ holds for all $L\in C$” makes no sense whatsoever, as $\Gamma$, $\bigwedge\Gamma$, and $y$ only belong to one particular lattice. And you still did not clarify what happens when the meet of $\Gamma$ does not exist.
yesterday history edited Florian CC BY-SA 4.0
to make some expressions more precise
yesterday history became hot network question
yesterday comment added Emil Jeřábek Simultaneously cross-posted at math.stackexchange.com/questions/5115256/… . Please, do not do that.
yesterday answer added Emil Jeřábek timeline score: 5
yesterday comment added Emil Jeřábek If the lattice is not complete, how is the definition of logic compactness involving $\bigsqcap\Gamma$ intended to be interpreted in the first place?
S 2 days ago review First questions
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S 2 days ago history asked Florian CC BY-SA 4.0