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Want $T_{F, G, H} \vdash L1(F, G, H)$, $T_{F, G,
H} \vdash L2(F, G, H)$ and $T_{F, G, H}$ proves
existence of $R$.

Convention: We say ``$T$ is sufficiently strong''
if $T \subseteq \zfc$ is finite and $T$ proves hte
existence of all relevant operations such that
they are absolute for transitive models of $T$.

\begin{proof}
  Observe that by assumption, being an
  ``attempt'' is \gls{abs} for transitive models
  of $T$.

  Let $M \models T$ be transitive.
  \begin{enumerate}[(1)]
    \item To show: If $M \models R(\alpha, \ol{x}) = z$,
      then $R(\alpha, \ol{x}) = z$.

      If $M \models R(\alpha, \ol{x}) = z$, then $M
      \models \exists r, \text{ $r$ is an attempt and
      $r(\alpha, \ol{x}) = z$}$. Without the $\exists
      r$, this would be \gls{abs}. So when we include
      the existential quantifier, we get an \gls{uabs}
      sentence.

      Thus: there is $r$ such that $r$ is an attempt
      and $r(\alpha, \ol{x}) = z$. So $R(\alpha,
      \ol{x}) = z$.
    \item Other direction. Assume $r$ is an
      attempt with  $r(\alpha, \ol{x}) = z$.

      Since $T_{F, G, H} \vdash L2(F, G, H)$, we
      have
      \[
        M \models \exists r', \ub{\text{ $r'$ is an
        attempt and $(\alpha, \ol{x}) \in \dom
        r'$}}_{\text{\gls{abs}}}
      .\]
      Since it is \gls{abs}, $r'$ is a real
      attempt.

      By ???, $r'(\alpha, \ol{x}) = r(\alpha,
      \ol{x})$. Hence $M \models R(\alpha, \ol{x})
      = z$.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{note*}
  This uses the fact that ``$\Delta_1$'' concepts are
  \gls{abs}.
\end{note*}

\begin{fcdefnstar}[Delta1T property]
  \glssymboldefn{delot}%
  A property is called $\Delta_1^T$ if it's both
  $\sigo^T$ and $\pio^T$.
\end{fcdefnstar}

\textbf{Observe:} $\delot$ concepts are \gls{abs}
(upwards from $\sigo$ and downwards from $\pio$).

\subsubsection*{Typical Applications}

 Bounding a quantifier by operation.

 Let $F$ be an operation and $T$ strong enough to
 prove $F$ is an operation and \gls{abs}.
 \begin{align*}
  T
  &\vdash \forall x, \exists z, F(x) = z \\
  T
  &\vdash \forall x, \forall z, \forall z', F(x) =
  z \wedge F(x) = z' \to z = z'
 \end{align*}
 Then the quantifiers $\exists y \in F(x)$ and
 $\forall y \in F(x)$ preserve
 \glsref[abs]{absoluteness}.
 \begin{align*}
   \exists y \in F(x) \psi
   &\iff
   \ub{\exists z \ub{(z = F(x) \wedge \exists y \in
   z\psi)}_{\text{\gls{abs}}}}_{\text{\gls{uabs}}}
   \\
   &\iff \ub{\forall z
   \ub{(z = F(x) \to \exists y \in
   z\psi)}_{\text{\gls{abs}}}
   }_{\text{\gls{dabs}}}
\end{align*}
 
\subsubsection*{Applications}

\begin{enumerate}[(1)]
  \item \glssymboldefn{Fml}%
    Encode formulas as elements of
    $\omega^{<\omega}$.
    \begin{center}
      \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
        \hline
        $\in$
        & $=$
        & $($
        & $)$
        & $\wedge$
        & $\vee$
        & $\neg$
        & $\exists$
        & $\forall$
        & $v_0$
        & $v_1$
        & $v_2$
        & $\cdots$ \\
        \hline
        $0$
        & $1$
        & $2$
        & $3$
        & $4$
        & $5$
        & $6$
        & $7$
        & $8$
        & $9$
        & $10$
        & $11$
        & $\cdots$
        \\
        \hline
      \end{tabular}
    \end{center}
    $\forall v_0 \exists v_1 \neg v_0 \in v_1$
    would be $(8, 9, 7, 10, 6, 8, 0, 10)$.

    $\mathrm{Fml} \subseteq \omega^{<\omega}$.
    So, $\mathrm{Fml}$ is \gls{abs} for some
    (sufficiently strong) finite fragment of
    $\zfc$ (see \es{1}).
  \item If $X$ is any set then 
    \[
      \text{``$X \models \varphi$''}
    \]
    (which means ``$(X, \in) \models \varphi$'')
    is defined by the usual (Tarski) recursion and
    thus also \gls{abs} (\es{1}).
\end{enumerate}

\subsection{The constructible hierarchy}

Fix a set $X$. Define for each $\varphi \in
\Fml$ and each $p \in X^{<\omega}$ (parameter)
\[
  D(\varphi, p, X)
  % \defeq \{a \in X : \forall w, w \in a
  % \leftrightarrow X \models \varphi(p, w)\}
  \defeq \{w \in X : X \models \varphi(p, w)\}
\]
the subset of $X$ defined by $\varphi$ with
parameter $p$.

For a sufficiently strong $T \subseteq \zfc$
finite, we have that $T$ proves that $D$
is an absolute operation (see \es{1}).
\[
  \mathcal{D}(X)
  \defeq \{D(\varphi, p, X) : \varphi \in \Fml, p
  \in X^{<\omega}\}
.\]
This is \gls{abs} for a sufficiently strong theory
(use Replacement to get $\mathcal{D}(X)$).

This $\mathcal{D}(X)$ is sometimes (misleadingly)
called the ``definable power set of $X$'' (it is
misleading because it is more like a ``definable
(by $X$) power set of $X$'').

($\alpha \in \mathcal{D}(X) \iff \exists \varphi
\in \Fml, \exists p \in X^{<\omega}, a =
D(\varphi, p, X)$)

Obvious: $\mathcal{D}(X) \subseteq
\mathcal{P}(X)$. Also: If $X$ is transitive, then
so is $\mathcal{D}(X)$.

\begin{align*}
  L_0
  &\defeq \emptyset \\
  L_{\alpha + 1}
  &\defeq \mathcal{D}(L_\alpha) \\
  L_{\lambda}
  &= \bigcup_{\alpha < \lambda} L_\alpha
\end{align*}
The constructible \gls{hier}.

We usually write $L \defeq \bigcap_{\alpha \in
\Ord} L_\alpha$.

\textbf{Claim:} $L$ is a \gls{hier} (in the sense
of Lecture). See \es{1}.

By closure of \glsref[abs]{absoluteness} under
transfinite recursion, the $L$-\gls{hier} is
\gls{abs} for transitive models of $T \subseteq
\zfc$ where $T$ is strong enough to prove that it
exists.

i.e. if $M \models T$ transitive and $\alpha \in
\Ord \cap M$ and $M \models X = L_\alpha$, then $X
= L_\alpha$. So
\[
  \bigcup_{\alpha \in \Ord \cap M} L_\alpha
  \subseteq M
.\]

The main theorem of next lecture will be:

If $L \models \zf$ and $M \models \zf$ transitive,
then
\[
  \bigcup_{\alpha \in \Ord \cap M} L_\alpha
  \models \zf
.\]
(Minimal $\zf$-model).

\subsubsection*{Some first idea of what the
$L$-hierarchy is like}

Clearly, by induction, $L_\alpha \subseteq
V_\alpha$, and clearly for $n \in \omega$, $L_n =
V_n$. So $L_\omega = V_\omega$.
\[
  L_{\alpha + 1}
  \defeq \bigcup_{\varphi \in \Fml} \bigcup_{p \in
  L_\alpha^{< \omega}} \{D(\varphi, p, L_\alpha)\}
.\]
If $\alpha \ge \omega$, then
\[
  |L_{\alpha + 1}| \le \aleph_0 \cdot
  |L_\alpha^{<\omega}| = \aleph_0 \cdot |L_\alpha|
.\]
Thus $|L_\alpha| = |L_{\alpha + 1}|$.

Therefore $\alpha < \omega_1$, $|L_\alpha| =
\aleph_0$ and $|L_{\omega_1}| = \aleph_1$.

This means: $V_{\omega + 1} \neq L_{\omega + 1}$
(since the first has size $2^{\aleph_0}$, while
the second has size $\aleph_0$).

Note: This does not mean $V \neq L$. ($V = L$
means $\forall x, \exists \alpha, x \in
L_\alpha$).

$V = L$ is called the ``axiom of constructibility''.
