True or false? $\bigwedge_{i \in I} \bigvee_{j \in J}a_{i,j} \equiv \bigvee_{j \in J} \bigwedge_{i \in I}a_{i,j}$ Let $I = \left\{i_1,...,i_m\right\}$ and $J = \left\{j_1,...,j_n\right\}$ be finite sets, and for every $i \in I$ and $j \in J$, there is a formula $a_{i,j}$ given. Do you always have that 
$$\bigwedge_{i \in I} \bigvee_{j \in J}a_{i,j} \equiv \bigvee_{j \in J} \bigwedge_{i \in I}a_{i,j}$$

In my other question I ask how can write as example $$\bigwedge_{i=1}^{2} \bigvee_{j=1}^{2}a_{ij}$$
You write it like this:
$$ (a_{11} \lor a_{12} \lor a_{13} \lor a_{14}) \land (a_{21} \lor a_{22} \lor a_{23} \lor a_{24})$$
Now when we swap the symbol (look at right side of equivalence symbol) we have:
$$ (a_{11} \land a_{12} \land a_{13} \land a_{14}) \lor (a_{21} \land a_{22} \land a_{23} \land a_{24})$$
From this example we see that not same because sign is all opposite, never the same.. And also the index $i,j$ can be different, then also no same.
That why this is false?
 A: 
From this example we see that not same because sign is all opposite, never the same.

You have indeed described a counterexample, but I don't think you have an argument that it is a counterexample: just because two propositional sentences look different, doesn't mean that they are. E.g. $$(a\vee b)\wedge (c\vee d)$$ and $$(a\wedge c)\vee (a\wedge d)\vee (b\wedge c)\vee (b\wedge d)$$ are equivalent.
To show that the two sentences are not equivalent, you need to find a truth assignment making one true and the other false. HINT: note that as long as (say) $a_{11}$ and $a_{22}$ are true, the first sentence is true; is that enough to make the second sentence true as well, or not?
A: Suppose $I=J=\{1,2\}.$ Let $a_{1,1}=a,\;a_{1,2}=b,\;a_{2,1}=c,\;a_{2,2}=d.$ Then $$\land_{i\in I}\lor_{j\in J}a_{i,j}\iff [\;(a_{1,1}\lor a_{1,2})\land (a_{2,1}\lor a_{2,2})\;]\iff [\;(a\lor b)\land (c\lor d)\;]$$ which says that at least one of $a,b$ is true and at least one of $c,d$ is true. And $$\lor_{j\in J}\land_{i\in I}a_{i,j}\iff [\;(a_{1,1}\land a_{1,2})\lor (a_{2,1}\land a_{2,2}\;]\iff [\;(a\land b)\lor (c\land d)\;]$$ which says that  $a, b$ are both true or $c,d$ are both true . 
In particular if $b\iff \neg a$ and $d\iff \neg c$ then the LHS in your Q is true and the RHS is false. 
Another counter-example is to let $I=J=\Bbb N$ and let $a_{i,j}$ be "$i=j$". 
A: Those other answers are too sensible. Let's start at the base case: $I=\emptyset$ and $J=\emptyset$. Then $\bigwedge_{i\in I}\bigvee_{j\in J}a_{i,j}$ is $\top$ and $\bigvee_{j\in J}\bigwedge_{i\in I}a_{i,j}$ is $\bot$.
