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Let $L$ be a lattice and let $a,b,c,d \in L$. Show that:

  1. $\theta(a,b) \subseteq \theta(c,d)$ iff $\langle a,b\rangle \in \theta(c,d)$

  2. $\theta(a,b)=\theta(a \wedge b, a \vee b)$

Where $\theta$ is a congruence. Definition: given a lattice $L$, a congruence $\theta$ on $L$ is an equivalence relation, compatible with the lattice operations, i.e. if $x_1\theta x_2$ and $y_1 \theta y_2$, then $x_1 \wedge x_2 \theta y_1 \wedge y_2$ and $x_1 \vee x_2 \theta y_1 \vee y_2$.

I specially have a problem with the first part, how to prove that it is a subset. The part of $\Leftarrow$ seems "evident", I have 2 elements, $a,b$, I assume they belong to the congruence, so they have to be inside, but I don't get the more formal way to explain this. For the $\Rightarrow$ part, I assume that $\theta (a,b) \subseteq \theta (c,d)$, so I think I have to argue that all the elements of a subset have a certain characteristic, so in particular $a,b$ have it. Again, if this is right, I am having trouble translating it into a proper proof.

About the second one, I can only think of proving it in a way similar to let $a= a\wedge b$ so $b = a\vee b$ (I have a lemma that $a= a\wedge b$ iff $b = a\vee b$). But it seems like there is something missing. I would very much appreciate your help to sort this out, and your input on how I can write this properly.

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  • $\begingroup$ What does $\theta(a,b)$ mean? $\endgroup$ – bof Jan 29 '15 at 12:49
  • $\begingroup$ The notation is a bit weird, honestly, it's part of what got me stuck, it doesn't fit with that on most sources. For what I understand, $\langle a,b \rangle$ means exactly the elements a and b, while $\theta (a,b)$ would mean the elements that are delimited by a and b that are part of the congruence. Does that sound plausible? Thank you for your answer. $\endgroup$ – Sara Jan 29 '15 at 12:56
  • $\begingroup$ I can't say it's plausible or not, because I still don't know what it means. "The elements that are delimited by $a$ and $b$ that are part of the congruence" is Greek to me. Sorry. $\endgroup$ – bof Jan 29 '15 at 14:01
  • $\begingroup$ $\theta(a, b)$ is the usual notation for the congruence generated by the pair $\langle a, b \rangle$, but I'm not quite sure. $\endgroup$ – Random Jack Jan 29 '15 at 15:40
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First of all, lattice $L$ in your question is not supposed to be distributive (but the title uses this word). If $\theta(a, b)$ means the congruence generated by $\langle a, b \rangle$ then:

  1. If $\theta(a, b) \subseteq \theta(c, d)$ then (since $\theta(a, b)$ is generated by $\langle a, b \rangle$) we have $$\langle a, b \rangle \in \theta(a, b) \subseteq \theta(c, d) \Rightarrow \langle a, b \rangle \in \theta(c, d).$$ Conversely, if $\langle a, b \rangle \in \theta(c, d)$ then since $\theta(a, b)$ is the smallest congruence containing pair $\langle a, b \rangle$ and $\theta(c, d)$ is some congruence containing this pair we have $\theta(a, b) \subseteq \theta(c, d)$.

  2. We want to show two inclusions, which are equivalent (by the first part) to showing that generating pair belongs to congruence. To show that $\langle a, b \rangle \in \theta(a \wedge b, a \vee b)$ observe that: $$a = a \vee (a \wedge b) \theta(a \wedge b, a \vee b) a \vee (a \vee b) = a \vee b,$$ $$b = b \vee (a \wedge b) \theta(a \wedge b, a \vee b) b \vee (a \vee b) = a \vee b,$$ which means that $a \theta(a \wedge b, a \vee b) b$ and hence $\langle a, b \rangle \in \theta(a \wedge b, a \vee b)$. To show that $\langle a \wedge b, a \vee b \rangle \in \theta(a, b)$ observe that: $$ a \wedge b \theta(a, b) a \wedge a = a = a \vee a \theta(a, b) a \vee b,$$ which means that $a \wedge b\theta(a, b)a \vee b$, and hence $\langle a \wedge b, a \vee b \rangle \in \theta(a, b)$. By the first part we have both inclusions and hence $\theta(a, b) = \theta(a \wedge b, a \vee b).$

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  • $\begingroup$ Hi thank you very much for the notation explanation and your answer, everything is much clearer now! $\endgroup$ – Sara Jan 30 '15 at 17:08

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