# De morgan's law in formal language

I found in some exercise in computation the following step:

I can't understand why is it equal terms, based of what I know about De morgan's law:

1. OR should be replaced by AND
2. where $$w=\varepsilon$$ come from

Let $$\Sigma$$ be the alphabet on which $$A_{TM}$$ is defined over, and assume w.l.o.g that:

1. $$A_{TM}\cup \overline{A_{TM}} = \Sigma^*$$, and
2. $$0, 1 \notin\Sigma.$$

The set $$J$$ can be written as:

$$J = \{0\}\cdot A_{TM} \ \cup \{1\}\cdot \overline{A_{TM}}$$ where $$J$$ is defined over the alphabet $$\Sigma \cup \{ 0, 1\}$$. Now, using De morgan's law, it holds that: $$\overline{J} = \overline{\{0\}\cdot A_{TM}} \ \ \cap \overline{\{1\}\cdot \overline{A_{TM}} } \\$$ $$=\left(\{w: \text{ w \neq \sigma\cdot u, \sigma\in\{0, 1\} and u\in \Sigma^*} \} \cup \{0\}\cdot \overline{A_{TM}} \ \cup \{1\}\cdot \Sigma^*\right) \ \cap \\ \left(\{w: \text{ w \neq \sigma\cdot u, \sigma\in\{0, 1\} and u\in \Sigma^*} \} \cup \{1\}\cdot A_{TM} \ \cup \{0\}\cdot \Sigma^*\right) \\$$ $$= \{w: \text{ w \neq \sigma\cdot u, \sigma\in\{0, 1\} and u\in \Sigma^*} \} \cup \left( \left[\{0\}\cdot \overline{A_{TM}} \ \cup \{1\}\cdot \Sigma^* \right] \cap \left[ \{1\}\cdot A_{TM} \ \cup \{0\}\cdot \Sigma^*\right] \right)$$

$$= \{w: \text{ w \neq \sigma\cdot u, \sigma\in\{0, 1\} and u\in \Sigma^*} \} \cup \left( \{ 0\}\cdot \overline{A_{TM}} \cup \{ 1\}\cdot A_{TM} \right)$$

So, we have that $$w \in \overline{J}$$ iff $$w$$ does not encode a bit followed by a word of the form $$\langle M, w \rangle$$, or $$w = 1x$$ for some $$x \in A_{TM}$$ or $$w = 0y$$ for some $$y \in \overline{A_{TM}}$$.

Finally, note that they're implicitly assuming w.l.o.g that every nonempty word in $$\{0, 1\}\cup \Sigma^*$$ is of the form "$$\text{bit} \cdot \langle M, w\rangle$$". This assumption is okay, as checking whether a word respects some encoding can be easily done by a TM.