# Generalized associative law of union

$$\Large\bigcup\limits_{k\in\bigcup\limits_{i\in I}J_i}A_k=\bigcup\limits_{i\in I}\bigg(\bigcup\limits_{k\in J_i}A_k\bigg)$$

My attempt:

$\large x\in\bigcup\limits_{k\in\bigcup\limits_{i\in I}J_i}A_k\iff (\exists k\in\bigcup\limits_{i\in I}J_i)(x\in A_k)\iff [(\exists i\in I) (k\in J_i)](x\in A_k)$

$\large x\in \bigcup\limits_{i\in I}\bigg(\bigcup\limits_{k\in J_i}A_k\bigg) \iff (\exists i\in I)(x\in \bigcup\limits_{k\in J_i}A_k\bigg) \iff (\exists i\in I)[(\exists k\in J_i)(x\in A_k)]$

I think the equality holds if and only if we show that $$[(\exists i\in I) (k\in J_i)](x\in A_k) \iff (\exists i\in I)[(\exists k\in J_i)(x\in A_k)]$$

My questions:

1. Are my above transformations fine?

2. How do I proceed to prove the last statement?

I don't understand what $$(\exists i\in I) (k\in J_i)](x\in A_k)$$ is supposed to mean; what is $$k$$ in this formula, exactly?

The statement that we want to prove appears to be something like $$[(\exists i\in I) (\exists k\in J_i)](x\in A_k) \iff (\exists i\in I)[(\exists k\in J_i)(x\in A_k)],$$ although it probably isn't exactly this because the second quantifier in the right side doesn't look right.

Nitpicking at this level seems fairly pointless to me. If you really are trying to be extremely formal about it because that's what you've chosen to do with your life, then I'm already concerned with what $$(\exists i \in I)P(i)$$ actually means. The formalism I am used to would write this $$(\exists i)((i\in I) \land P(i))$$. Under this formalism the first line of $$\iff$$'s should be $$(\exists k)[[(\exists i)((i\in I) \land (k\in J_i))]\land (x\in A_k) ]$$ and the second line should be $$(\exists i)[i\in I \land [(\exists k)((k\in J_i)\land (x \in A_k))]].$$ The equivalence of these two formulas follows by commuting quantifiers and conjunctions; whatever formalism you're using should allow a rule like $$(\exists z)P(z) \land Q(y) \iff (\exists z)(P(z)\land Q(y)).$$

• Hi @Eric! Since I'm just exposed to set theory. I present the explicit transformations below. Please have a check on it! Thank you so much! – Le Anh Dung Oct 11 '18 at 2:01
• $$(\exists k)[[(\exists i)((i\in I) \land (k\in J_i))]\land (x\in A_k) ]$$ $$\iff (\exists k)[[(\exists i)(i\in I) \land (\exists i)(k\in J_i)]\land (x\in A_k) ]$$ $$\iff (\exists k)[(\exists i)(i\in I) \land (\exists i)(k\in J_i)\land (x\in A_k) ]$$ $$\iff (\exists k)[(\exists i)(i\in I) \land (\exists i)(k\in J_i)\land (x\in A_k) ]$$ $$\iff (\exists k)(\exists i)[(i\in I) \land (k\in J_i)\land (x\in A_k) ]$$ […] – Le Anh Dung Oct 11 '18 at 2:01
• […] $$\iff (\exists i)(\exists k)[(i\in I) \land (k\in J_i)\land (x\in A_k) ]$$ $$\iff (\exists i)[(\exists k)(i\in I) \land (\exists k)(k\in J_i)\land (\exists k)(x\in A_k) ]$$ $$\iff (\exists i)[(i\in I) \land (\exists k)(k\in J_i)\land (\exists k)(x\in A_k) ]$$ $$\iff (\exists i)[(i\in I) \land (\exists k)[(k\in J_i)\land (x\in A_k)]]$$ – Le Anh Dung Oct 11 '18 at 2:01

On the basis of Eric Stucky's answer, I present some detail here.

LHS $$\iff(\exists k)[[(\exists i)((i\in I) \land (k\in J_i))]\land (x\in A_k) ]$$

$$\iff(\exists k)[[(\exists i)(i\in I) \land (\exists i)(k\in J_i)]\land (x\in A_k) ]$$

$$\iff(\exists k)[(\exists i)(i\in I) \land (\exists i)(k\in J_i)\land (x\in A_k) ]$$

$$\iff(\exists k)[(\exists i)(i\in I) \land (\exists i)(k\in J_i)\land (x\in A_k) ]$$

$$\iff(\exists k)(\exists i)[(i\in I) \land (k\in J_i)\land (x\in A_k) ]$$

$$\iff(\exists i)(\exists k)[(i\in I) \land (k\in J_i)\land (x\in A_k) ]$$

$$\iff(\exists i)[(\exists k)(i\in I) \land (\exists k)(k\in J_i)\land (\exists k)(x\in A_k) ]$$

$$\iff(\exists i)[(i\in I) \land (\exists k)(k\in J_i)\land (\exists k)(x\in A_k) ]$$

$$\iff(\exists i)[(i\in I) \land (\exists k)[(k\in J_i)\land (x\in A_k)]]\iff$$ RHS