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\begin{proof}[Proof of power set]
  \es{3}.
\end{proof}

\begin{visiblefc}{3e98ca0d0301}
\begin{proof}[Proof of Replacement]
  \divcloze{
  Since we already have Separation, it's enough to
  show the following:
  \begin{quote}
    if $\varphi$ is a functional formula and $x
    \in M[G]$, then there is $R \in M[G]$ such
    that
    \[
      M[G] \models \forall y \in x, \exists z \in
      R, \varphi(y, z, \cancel{\ol{p}})
      \tag{$*$}
      \label{lec11eq1}
    \]
    (as before, we suppress parameters for
    notational ease).
  \end{quote}
  We work in $M$ and identify a name $\rho$ for
  $R$. Fix $\sigma$ such that $x = \val(\sigma,
  G)$. Find $\alpha$ large enough such that
  $\dom(\sigma) \subseteq V_\alpha$. Consider
  $\psi(p, \pi) \defeq \exists \mu, p \forces
  \varphi(\pi, \mu)$ ($p \in \Pbb$, $\pi$ a name).
  By Definability Theorem,
  this is an $\mathcal{L}_\in$ formula.

  By \nameref{lrt}, find $\theta > \alpha$ such
  that $\psi$ is \gls{abs} between $V_\theta$ and
  $V = M$. Define
  \[
    \rho \defeq \{(\mu, \mathbbm{1}) : \mu \in
    V_\theta\}
  \]
  and $R \defeq \val(\rho, G)$.

  Now we verify \eqref{lec11eq1} holds: Fix $y \in
  x$, $y = \val(\pi, G)$. Since $\varphi$ is
  functional, we know that $\varphi(\pi, \mu)$
  holds in $M[G]$ for some $\mu$. By Forcing
  Theorem, there is $p \in G$ such that $p \forces
  \varphi(\pi, \mu)$. So $M \models \psi(p, \pi)$.

  By \gls{abs}ness, $V_\theta \models \psi(p,
  \pi)$. This means $\exists \mu \in V_\theta$
  such that $p \forces \varphi(\pi, \mu)$.

  Thus: $\val(\mu, G) \in \val(\rho, G) = R$.
  }
\end{proof}
\end{visiblefc}

\begin{fcdefnstar}[Dense below $p$]
  \glsadjdefn{dbel}{dense below}{filter}%
  $D \subseteq \Pbb$ is called \emph{dense below
  $p$} if $\forall q \le p, \exists r \le q, r \in
  D$.
\end{fcdefnstar}

\begin{fclemma}[]
  \label{lec11lemmalots}
  Assuming:
  - $G$ is \dgen[\Pbb] over $M$
  - $E \subseteq \Pbb$
  - $E \in M$
  Thens:[(i)]
  - If $E$ is \gls{dbel} $p$, $q \le p$, then $E$
  is \gls{dbel} $q$.
  - If $\{r : \text{$E$ is \gls{dbel}
  $r$}\}$ is \gls{dbel} $p$, then $E$ is
  \gls{dbel} $p$
  - Either $G \cap E \neq \emptyset$ or $\exists q
  \in G$, $\forall r \in E, r \perp q$.
  - If $p \in G$, $E$ is \gls{dbel} $p$, then
  $G \cap E \neq \emptyset$.
\end{fclemma}

\noproof

\begin{proof}
  \es{3}
\end{proof}

\subsubsection*{Definition of the syntactic
forcing relation}

$p \forces^* \varphi(\ol{\tau})$.

Two recursions:

First ``$\forces^* =$'' by recursion on
$\Name_\alpha$.

Then ``$\forces^* \in$'' without recursion.

Then the rest by recursion on formula complexity.

$p \forces^* \tau_0 = \tau_1$: if and only if
\begin{align*}
  \forall (\pi_0, s_0) \in \tau_0
  & \\
  &\{q \le p : q \le s_0 \to \exists (\pi_1, s_1)
  \in \tau_1, (q \le s_1 \wedge q \forces^* \pi_0
  = \pi_1)\} \\
  &\text{is \gls{dbel} $p$}
\end{align*}
and
\begin{align*}
  \forall (\pi_1, s_1) \in \tau_1
  & \\
  &\{q \le p : q \le s_1 \to \exists (\pi_0, s_0)
  \in \tau_0, (q \le s_0 \wedge q \forces^* \pi_0
  = \pi_1)\} \\
  &\text{is \gls{dbel} $p$}
\end{align*}

\begin{remark*}
  This is a recursion on $\Name_\alpha$.
\end{remark*}

$p \forces^* \tau_0 \in \tau_1$: if and only if
\[
  \{q \le p : \exists (\pi, s) \in \tau_1, (q \le
  s \wedge q \forces^* \pi = \tau_0)\}
\]
is \gls{dbel} $p$.

\begin{remark*}
  No recursion involved, just ``$\forces^* =$''.
\end{remark*}

Recursion on complexity of formulas:
\begin{itemize}
  \item $p \forces^* \varphi \wedge \psi$: if and
    only if $p \forces^* \varphi$ and $p \forces^*
    \psi$.
  \item $p \forces^* \neg \varphi$: if and only if
    $\forall q \le p$, $q \not \forces^* \varphi$.
  \item $p \forces^* \exists x, \varphi(x)$: if
    and only if $\{r : \exists \sigma, r \forces^*
    \varphi(\sigma)\}$.
\end{itemize}

\begin{remark*}
  These definitions remind us of Kripke semantics
  for intuitionistic logic.
\end{remark*}

\begin{fclemma}[]
  The following are equivalent:
  \begin{enumerate}[(i)]
    \item $p \forces^* \varphi$.
    \item $\forall r \le p, r \forces^* \varphi$.
    \item $\{r : r \forces^* \varphi\}$ is
      \gls{dbel} $\rho$.
  \end{enumerate}
\end{fclemma}

\begin{proof}
  If this is true for $\varphi$ of the form
  $\tau_0 = \tau_1$ and of the form $\tau_0 \in
  \tau_1$, then it's true for all formulas.

  For $\varphi$ atomic, we get that (ii)
  $\implies$ (i) and (ii) $\implies$
  (iii) are trivial.

  (i) $\implies$ (ii) follows from
  \cref{lec11lemmalots}(i).

  (iii) $\implies$ (i) follows from
  \cref{lec11lemmalots}(ii).
\end{proof}

\textbf{Strategy:}

\begin{fcthm}[Syntactic Forcing Theorem]
  \label{synFT}
  \begin{iffc}
    \lhs
    $M[G] \models \varphi$
    \rhs
    $\exists p \in G, M \models r \forces^* \varphi$.
  \end{iffc}
\end{fcthm}

\noproof

\begin{fccoro}[Definability Theorem]
  \label{lec11coro1}
  \begin{iffc}
    \lhs
    $p \forces \varphi$
    \rhs
    $p \forces^* \varphi$.
  \end{iffc}
\end{fccoro}

\noproof

($p \forces_M \varphi \iff M \models p \forces^*
\varphi$)

\begin{fccoro}[Forcing Theorem]
  \begin{iffc}
    \lhs
    $M[G] \models \varphi$
    \rhs
    $\exists p \in G, p \forces \varphi$.
  \end{iffc}
\end{fccoro}

\noproof

This corollary is immediate from combining the two
previously stated results.

\begin{proof}[Proof of Definability Theorem from
  Syntactic Forcing Theorem]
  \phantom{}
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Leftarrow$] This is just
      \nameref{synFT}.
    \item[$\Rightarrow$] Suppose $p \forces
      \varphi$. By \cref{lec11lemmalots}, we need
      to show
      \[
        \{r \le p : r \forces^* \varphi\}
      \]
      is \gls{dbel} $p$.

      Suppose not. Then I find $q \le p$ such that
      $\forall r \le q$, $r \not \forces^*
      \varphi$. So $q \forces^* \neg\varphi$.

      If $q \in G$, then $p \in G$. Since $p
      \forces \varphi$, $M[G] \models \varphi$.
      Since $q \forces^* \neg \varphi$, $M[G]
      \models \neg \varphi$. Contradiction!
      \qedhere
  \end{enumerate}
\end{proof}

\begin{proof}[Proof of the Syntactic Forcing
  Theorem]
  The Theorem for $=$ can be proved by induction on
  $\Name_\alpha$.

  The Theorem for $\in$ can be proved once the
  Theorem has been established for $=$.

  Rest is induction on formula complexity.

  Let's do $\neg$ and leave the rest to \es{3}.
  Assume we have proved it for $\varphi$. We will
  prove it for $\neg \varphi$.

  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] $M[G] \forces \neg \varphi$. Consider
      \[
        D
        \defeq \{p : p \forces^* \varphi \text{ or } p
        \forces^* \neg \varphi\}
      .\]
      By definition of $\forces^* \neg$, this is
      \gls{dense}. Find $p \in G \cap D$.

      By assumption, $p \not\forces^* \varphi$, so $p
      \forces^* \neg \varphi$.
    \item[$\Leftarrow$] Let $p \in G$ such that $p
      \forces^* \neg \varphi$. By definition,
      $\forall q \le p$, $q \not\forces^*
      \varphi$. If $M[G] \not \models \neg
      \varphi$, thus $M[G] \models \varphi$. By
      induction hypothesis, find $r \in G$, $r
      \forces^* \varphi$. So $q \le r, p, q \in
      G$.

      By Lemma, $q \forces^* \varphi$,
      contradiction to $\forall q \le p, q
      \not\forces^* \varphi$.
  \end{enumerate}
\end{proof}
