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I found in some exercise in computation the following step: enter image description here enter image description here

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
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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.

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