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On page 30 of Birkhoff's Lattice Theory, Lemma 3 states that in distributive lattices \begin{gather*} \bigvee_{\alpha\in A}\left\{\bigwedge_{S_\alpha}x_i\right\}=\bigwedge_{\delta\in D}\left\{\bigvee_{T_\delta}x_j\right\} \end{gather*} but I'm totally confused by his proof, starting with "On the other hand". I can see the two expansions but I'm not sure how they imply the result. If someone could enlighten me that'd be great.

I also tried to do it differently and tried showing that the reverse to \begin{gather*} \bigvee_{\alpha\in A}\left\{\bigwedge_{S_\alpha}x_i\right\}\preccurlyeq\bigwedge_{S_\alpha}\left\{\bigvee_{\alpha\in A}x_i\right\} \end{gather*} holds for distributive lattices but that too to no avail.

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  • $\begingroup$ I also found the proof rather confusing. Meanwhile, you understand that the first sentence of the proof (until the point you refer to) is just to say that the result applies trivially to polynomials which are just projections ($p(x_1,\ldots,x_n)=x_i$) and the equality of joins of meets with meets of joins is to say that whatever form the polynomial comes with, it'll reduce to one of those. $\endgroup$
    – amrsa
    Commented Jul 8 at 21:10
  • $\begingroup$ Now that's the part which I found confusing and I suppose you'd better use an expression such as in this wikipedia article and prove it by induction on the cardinality of $J$. $\endgroup$
    – amrsa
    Commented Jul 8 at 21:11
  • $\begingroup$ @amrsa Thanks for the comment; and you're right, it's probably easier that way. $\endgroup$
    – user9871234
    Commented Jul 9 at 21:53

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