Can anybody tell me if there's any general formulas for the following: If $i, j, k, l$ runs over a finite set, then
$(\wedge _j \vee _i x_{i,j})$ $\vee$ $(\wedge _k \vee _l y_{k,l}) = ?$
$(\wedge _j \vee _i x_{i,j})$ $+$ $(\wedge _k \vee _l y_{k,l}) = ?$
Also, I want to know how $A^{\vee \wedge} = A^{\wedge \vee}$ in the attached picture?
I'm using the fact that the underlying lattice in a Riesz space is distributive, i.e., it satisfies $$x \wedge (y \vee z) = (x \wedge y) \vee (x \wedge z),$$ and its (equivalent) dual $$x \vee (y \wedge z) = (x \vee y) \wedge (x \vee z).$$ Also, addition distributes over both lattice operations, i,e., the space satisfies $$(x \wedge y) + z = (x+z) \wedge (y+z),$$ and $$(x \vee y) + z = (x+z) \vee (y+z).$$
I'll use the fact that the lattice is distributive to answer or questions (1) and (3).
A completely distributive lattice satisfies $$\bigwedge_{i\in I}\bigvee_{j\in J}x_{i,j} = \bigvee_{f:I\to J}\bigwedge_{i\in I}x_{i,f(i)}.$$ Whether the underlying distributive lattice of a Riesz space is completely distributive or not, is not relevant here, since the joins and meets you ask about are finitary.
So I'll derive an expression for the general, finite case, base on that above.
For notational convenience, for each natural number $n \geq 1$, identify $n$ with the set $\{1,\ldots,n\}$.
I'll prove, by induction on $n$, that $$\bigwedge_{i=1}^n\bigvee_{j=1}^m a_{i,j} = \bigvee_{f:n\to m}\bigwedge_{i=1}^n a_{i,f(i)}.$$ The result is trivial if $n=1$; so suppose it's true for a certain $n$, and let's see it still holds for $n+1$. \begin{align} \bigvee_{f:n+1\to m}\bigwedge_{i=1}^{n+1}a_{i,f(i)} &=\bigvee_{f:n+1\to m}\left(\bigwedge_{i=1}^n a_{i,f(i)} \wedge a_{n+1,f(n+1)} \right)\\ &=\bigvee_{f:n+1\to m}\bigwedge_{i=1}^n a_{i,f(i)} \wedge \bigvee_{f:n+1\to m}a_{n+1,f(n+1)}, \end{align} where, in the last equality, it was used distributivity, as in the second equality in the expression $\bigvee_i\bigwedge_j(u_{ij}\wedge v_i) = \bigvee_i(\bigwedge_ju_{ij}\wedge v_i) = \bigvee_i\bigwedge_ju_{ij}\wedge\bigvee_iv_i$.
Now, $\bigvee_{f:n+1\to m} a_{n+1,f(n+1)} = \bigvee_{j=1}^m a_{n+1,j}$, because $f(n+1)$ takes all the values in that range. Also, $\bigvee_{f:n+1\to m}\bigwedge_{i=1}^n a_{i,f(i)}=\bigvee_{f:n\to m}\bigwedge_{i=1}^n a_{i,f(i)}$, because if $i \leq n$, then it doesn't matter the value $f$ takes at $n+1$, and so we can restrict $f$ to the domain $\{1,\ldots,n\}$. So \begin{align} \bigvee_{f:n+1\to m}\bigwedge_{i=1}^n a_{i,f(i)} \wedge \bigvee_{f:n+1\to m} a_{n+1,f(n+1)} &=\bigvee_{f:n\to m}\bigwedge_{i=1}^n a_{i,f(i)} \wedge \bigvee_{j=1}^m a_{n+1,j}\\ &=\bigwedge_{i=1}^n\bigvee_{j=1}^m a_{i,j} \wedge \bigvee_{j=1}^m a_{n+1,j}\\ &=\bigwedge_{i=1}^{n+1}\bigvee_{j=1}^m a_{i,j}, \end{align} where the second equality was the induction hypothesis.
This answers question (3).
Likewise we have the expression $$\bigvee_{i=1}^n\bigwedge_{j=1}^m a_{i,j} = \bigwedge_{f:n\to m}\bigvee_{i=1}^n a_{i,f(i)},$$ and since addition distributes over the lattice operations, $$\sum_{i=1}^n\bigwedge_{j=1}^m = \bigwedge_{f:n\to m}\sum_{i=1}^n a_{i,f(i)}$$ and $$\sum_{i=1}^n\bigvee_{j=1}^m = \bigvee_{f:n\to m}\sum_{i=1}^n a_{i,f(i)}.$$
It follows that $$\bigwedge_{f:n\to m}\bigvee_{i=1}^n\bigvee_{k=1}^p a_{i,f(i),k} = \bigvee_{i=1}^n\bigwedge_{j=1}^k\bigvee_{k=1}^p a_{i,j,k} = \bigvee_{i=1}^n\bigvee_{g:m\to p}\bigwedge_{j=1}^m a_{i,j,g(j)},$$ that is, given a join of meets of joins, you can transform it into a join of meets or a meet of joins, whichever suits you best. With $n=2$, this answers question (1).
The answer to question (2) should follow from \begin{align} \sum_{i=1}^n\bigwedge_{j=1}^m\bigvee_{k=1}^p a_{i,j,k} &=\bigwedge_{f:n\to m}\sum_{i=1}^n\bigvee_{k=1}^p a_{i,f(i),k}\\ &=\bigwedge_{f:n\to m}\bigvee_{g:n\to p}\sum_{i=1}^n a_{i,f(i),g(i)}, \end{align} but that depends in what shape you're interested in...
I mean, if $A$ is closed under $+$, then it follows from the above expression that so does $A^{\vee\wedge}=A^{\wedge\vee}$.
Update
The proof that $$\bigwedge_{i=1}^n\bigvee_{j=1}^m a_{i,j} = \bigvee_{f:n\to m}\bigwedge_{i=1}^n a_{i,f(i)}$$ above has what seems to me now a circular reasoning;
specifically, when I say that I'm using $$\bigvee_i(\bigwedge_ju_{ij}\wedge v_i) = \bigvee_i\bigwedge_ju_{ij}\wedge\bigvee_iv_i$$ this expression seems not to be easier to prove than the original one.
So I'm adding a new proof by induction, and this time I was more careful, so I trust it is correct.
\begin{align} \bigvee_{f:n+1\to m}\bigwedge_{i=1}^{n+1}a_{i,f(i)} &=\bigvee_{f:n+1\to m}\left(\bigwedge_{i=1}^{n}a_{i,f(i)} \wedge a_{n+1,f(n+1)}\right)\\ &=\bigvee_{j=1}^{m}\bigvee_{\substack{f:n+1\to m\\f(n+1)=j}}\left(\bigwedge_{i=1}^{n}a_{i,f(i)} \wedge a_{n+1,f(n+1)}\right)\\ &=\bigvee_{j=1}^{m}\bigvee_{\substack{f:n+1\to m\\f(n+1)=j}}\left(\bigwedge_{i=1}^{n}a_{i,f(i)} \wedge a_{n+1,j}\right)\\ &=\bigvee_{j=1}^{m}\left(\left( \bigvee_{\substack{f:n+1\to m\\f(n+1)=j}}\bigwedge_{i=1}^{n}a_{i,f(i)}\right) \wedge a_{n+1,j} \right) \tag{1}\\ &=\bigvee_{j=1}^{m}\left(\left( \bigvee_{f:n\to m}\bigwedge_{i=1}^{n}a_{i,f(i)}\right) \wedge a_{n+1,j} \right) \tag{2}\\ &=\left(\bigvee_{f:n\to m}\bigwedge_{i=1}^{n}a_{i,f(i)}\right)\wedge\bigvee_{j=1}^{m}a_{n+1,j} \tag{3}\\ &=\bigwedge_{i=1}^{n}\bigvee_{j=1}^{m}a_{i,j} \wedge \bigvee_{j=1}^{m}a_{n+1,j}\tag{4}\\ &=\bigwedge_{i=1}^{n+1}\bigvee_{j=1}^{m}a_{i,j}, \end{align} where
(1) is because $a_{n+1,j}$ depends neither on $j$ nor on $f$;
(2) is because $1 \leq i \leq n$ so we only take the restriction of $f:n+1\to m$ to the set $\{1,\ldots,n\}$, which is the same as considering functions $f:n\to m$;
(3) is because $\bigvee_{f:n\to m}\bigwedge_{i=1}^{n}a_{i,f(i)}$ doesn't depend on $j$;
(4) is by induction hypothesis.
In both (1) and (3) we're using the straightforward generalized version of distributivity which gives $\bigvee_{i=1}^m(a_i\wedge b)=(\bigvee_{i=1}^m a_i)\wedge b$.