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There is a finite fragment $T_{\mathbb{L}}$ of
$\zf$[!] that proves that all of the operations
occuring in the definition of $\mathbb{L}$, i.e.
$\Fml, X^{< \omega}, \models, D, \mathcal{D}$ are
well-defined and \gls{abs}.

Thus, if $M$ is a transitive model of
$T_{\mathbb{L}}$, then
\[
  \forall \alpha \in \Ord \cap M,
  \mathbb{L}_\alpha \in M
\]
and thus $\mathbb{L}_{\alpha} \subseteq M$.

So $\bigcup_{\alpha \in \Ord \cap M}
\mathbb{L}_\alpha \subseteq M$.

\subsubsection*{Axioms of ZF}

\textbf{Structural axioms:}
\begin{itemize}
  \item Extensionality:
    \[
      \forall x, \forall y,
      (\forall w (w \in x \leftrightarrow w \in y)
      \to x = y)
    .\]
  \item Foundation:
    \[
      \forall x(\exists v, (v \in
      x) \to \exists m, (m \in x \wedge \forall w,
      \neg (w \in m \wedge w \in x)))
    .\]
  \item Infinity:
    \[
      \exists i, \exists e, (\forall
      v, \neg v \in e \wedge e \in i) \wedge \forall
      x, (x \in i \to \exists s, (s \in i \wedge
      \forall w, (w \in s \leftrightarrow w \in x
      \vee w = x)))
    .\]
\end{itemize}
\textbf{Functional axioms:}
\begin{itemize}
  \item Pairing:
    \[
      \forall x, \forall y, \exists p,
      \forall w, (w \in p \leftrightarrow w = x \vee
      w = y)
    .\]
  \item Union:
    \[
      \forall x, \exists v, \forall w,
      (w \in v \leftrightarrow \exists z, (z \in x
      \wedge w \in z))
    .\]
  \item Powerset:
    \[
      \forall x, \exists p, \forall
      w, (w \in p \leftrightarrow \forall v, (v \in
      w \to v \in x))
    .\]
  \item Separation $\varphi$:
    \[
      \forall \ol{p},
      \forall x, \exists s, \forall w, (w \in s
      \leftrightarrow w \in x \wedge \varphi(w,
      \ol{p}))
    .\]
  \item Replacement $\varphi$:
    \[
      \forall \ol{p},
      [\forall x, \forall y, \forall z (\varphi(x,
      y, \ol{p}) \wedge \varphi(x, z, \ol{p}) \to y
      = z)] \to \forall x, \exists r, \forall w, (w
      \in r \leftrightarrow \exists y, (y \in x
      \wedge \varphi(y, w, \ol{p})))
    .\]
\end{itemize}

Now we check that these hold in $\mathbb{L}$.

In Lecture 2, we proved Extensionality and
Foundation in all transitive structures, so also
in $\mathbb{L}$.

Note that $\omega$ satisfies the condition of the
axiom of infinity, so any $M$ transitive with
$\omega \in M$ will satisfy the axiom of infinity.
TODO

Now do pairing and union.

Since the definitions of pairs and unions
\begin{align*}
  z
  &= \{x, y\} \\
  z
  &= \bigcup x
\end{align*}
are \gls{abs} for transitive models, it's enough
to show that
\begin{align*}
  \forall x, y \in \mathbb{L},
  &\{x,y\} \in \mathbb{L} \\
  \forall x \in \mathbb{L}
  &\bigcup x \in \mathbb{L}
\end{align*}
If $x, y \in \mathbb{L}_\alpha$, $\varphi(w,
x, y) \defeq w = x \vee w = y$,
\begin{align*}
  D(\varphi, (x, y), L_\alpha)
  &= \{w \in L_\alpha : L_\alpha \models
  \varphi(w, x, y)\} \\
  &= \{w \in L_\alpha : L_\alpha \models w = x
  \wedge w \}
  TODO
\end{align*}

Powerset axiom.

\[
  \forall x, \exists p,
  \ub{\forall w, (w \in p
  \leftrightarrow w \subseteq x)}_{*}
\]
The problem here is that $*$ is not obviously
\gls{abs}. In particular, $z = \mathcal{P}(x)$ is
not \gls{abs}.

Consider $\mathbb{L}_{\omega + 1}$: we have
$\omega \in \mathbb{L}_{\omega + 1}$ and
$\mathcal{P} \cap \mathbb{L}_{\omega + 1}$ is
countable.

In $\mathbb{L}_{\omega + 2}$, we find
\[
  \{a \in \mathbb{L}_{\omega + 1} : a
  \subseteq \omega\}
\]
which is the best possible answer to the question
``what is the power set of $\omega$?'' that
$\mathbb{L}_{\omega + 1}$ can give, but unlikely
to be the correct answer.

Consider instead $\mathcal{P}(\omega) \cap
\mathbb{L} \eqdef P$ and define
\[
  \Omega
  \defeq \{\rho_{\mathbb{L}}(a) : a \in P\}
.\]
(reminder: $\rho_{\mathbb{L}}(a)$ is the least
$\alpha$ such that $a \in \mathbb{L}_{\alpha
+ 1}$)

By Replacement, $\Omega$ is a set of ordinals, so
find $\alpha > \Omega$. Then $P \subseteq
\mathbb{L}_\alpha$.

Therefore $P = \{a \in \mathbb{L}_\alpha : a \subseteq
\omega\} \in \mathbb{L}_{\alpha + 1}$, so $P \in
\mathbb{L}$.

Separation:

\[
  \forall \ol{p}, \forall x, \exists s, \forall w,
  (w \in s \leftrightarrow \ub{w \in x \wedge \varphi(w,
  \ol{p})}_{\varphi'(w, x, \ol{p})})
\]
If $x \in \mathbb{L}_\alpha$, then
\begin{align*}
  D(\varphi', x, L_\alpha)
  &\defeq \{w \in L_\alpha : L_\alpha \models
  \varphi'(w, x, \ol{p})\} \\
  &= \{w \in L_\alpha : L_\alpha \models w \in x
  \wedge \varphi(w, \ol{p})\} \\
  &\stackrel{?}{=} \{w \in
  \ub{\mathbb{L}}_{\text{not a problem}} :
  \ub{\mathbb{L}}_{\text{this is a problem}} \models w \in x
  \wedge \varphi(w, \ol{p})\}
\end{align*}
If $\varphi$ is not \gls{abs} between $L_\alpha$
and $L$, this won't work.

\nameref{lrt} to the rescue: $\forall \varphi,
\forall \alpha, \exists \theta > \alpha$ such that
$\varphi$ is \gls{abs} between $L_\theta$ and $L$.

Thus: form
\begin{align*}
  D(\varphi', x, L_\theta)
  &= \{w \in L_\theta : L_\theta \models w \in x
  \wedge \varphi(w, \ol{p})\} \\
  &\stackrel{\text{\gls{abs}}}{=} \{w \in L_\theta
  : L \models w \in x \wedge \varphi(w, \ol{p})\}
\end{align*}

Replacement:

This will be on \es{2}. The proof is a combination
of the ideas from power set and separation.

\begin{fccoro}[Minimality]
  Assuming:
  - $T$ is a transitive model of $\zf$
  Then:
  for all $\alpha \in T \cap \Ord$,
  $\mathbb{L}_\alpha \subseteq T$.

  Axiom of TODO.
\end{fccoro}

\begin{remark*}
  Remark on the Axiom of Choice.

  G\"odel (1938): $\Con(\zf) \to \Con(\zfc)$.

  Note first that everything we did so far only
  needed $\zf$ in the universe. We will sketch
  that $\mathbb{L} \models \ac$. In fact a strong
  version of $\ac$ known as GLOBAL CHOICE: there
  is an absolutely definable bijective operation
  between $\mathbb{L}$ and $\Ord$.

  Sketch: Recursive construction of bijections
  $\pi_\alpha : \mathbb{L}_\alpha \to \eta_\alpha$
  for some ordinal $\eta_\alpha$, and such that
  for $\beta < \alpha$ we have $\pi_\alpha
  |_{\mathbb{L}_\beta} = \pi_\beta$.

  If $\lambda$ is a limit and $\pi_\alpha$ is
  defined for $\alpha < \lambda$, then let
  \[
    \pi_\lambda(x) \defeq \pi_\alpha(x)
  \]
  if $x \in \mathbb{L}_\alpha$.

  Suppose $\alpha = \beta + 1$ and $\pi_\beta$ is
  given by $\pi_\beta : \mathbb{L}_\beta \to
  \eta_\beta$.

  Consider $\Fml \times L_\beta^{<\omega}$.
  Well-order it in order type $\eta_\beta'$ via
  the induced $\pi_\beta$ well-order. Then if $x
  \in L_\alpha$, say
  \[
    \pi_\alpha(x)
    \defeq \begin{cases}
      \pi_\beta(x) & \text{if $x \in L_\beta$} \\
      \eta_\beta \xi & \text{if $\xi$ is the ordinal
      corresponding to the least $(\varphi,
      \ol{p})$ such that $x = D(\varphi, \ol{p},
      TODO)$}
    \end{cases}
  \]
\end{remark*}