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TODO

\subsubsection*{Further Recap}

We proved
\[
  M[G] \models \text{Extensionality + Foundation +
  Pairing + Union}
.\]
Homework was: Think about why power set is not easy.

``Union'' proof was:

collect all natural candidates of names for
elements and assign the natural values.

Problem: If you try to do this for power set,
neither the ``natural candidates for names'' nor
the ``natural values'' are obvious. It'll turn out
that they \emph{are} obvious in the end, but that
requires some assistance.

Note: Separation and Replacement are even worse.

One remaining easy axiom: $\ac$. By well-ordering
theorem, $\ac$ holds if and only if $\forall x,
\exists \alpha, \exists i, i : x \to \alpha$
injection. If $x \in M[F]$, then there is $\sigma
\in \Name$ such that $x = \val(\sigma, F)$.

\begin{notation*}
  I write $\dom(\sigma) \defeq \{\tau : \exists p,
  (\tau, p) \in \sigma\}$.
\end{notation*}

Consider $\dom(\sigma)$. In $M$, I have $i :
\dom(\sigma) \to \alpha$ for some ordinal
$\alpha$. In $M[F]$, define
\[
  y \mapsto \min \{i(\tau) : \val(\tau, F) = y,
  \tau \in \dom(\sigma)\}
.\]
Call this $i^*$. Then $i^* : x \to \alpha$ is an
injection. So, $\ac$ holds in $M[F]$.

\subsubsection*{Forcing}

\begin{fcdefnstar}[Forcing language]
  \glsnoundefn{fl}{forcing language}{forcing
  languages}%
  Fix $M$ a \gls{ctm} of $\zfc$, $\Pbb \in M$ a
  \gls{forcing} poset.

  We call the language
  \[
    \mathcal{L}_{\text{forcing}}
    \defeq \mathcal{L}_{\in} \cup \{\tau :
    \textbf{$\Pbb$-names}\}
  \]
  the \emph{forcing language (over $M$)}.
\end{fcdefnstar}

\begin{fcdefnstar}[Interpretation of forcing language]
  If $G$ is a
  \dgen[\Pbb] over $M$ and $\varphi$ is in the
  forcing language, we say
  \[
    M[G] \models \varphi
  \]
  if and only if
  \[
    M[G], v \models \varphi
  \]
  where $v(\tau) \defeq \val(\tau, G)$.
\end{fcdefnstar}

\begin{fcdefnstar}[Semantic forcing predicate]
  \glssymboldefn{forcing}%
  $p \Vdash_{M, \Pbb} \varphi$: Forall $G$
  \dgen[\Pbb] over $M$ with $p \in G$, $M[G]
  \models \varphi$.

  ``$p$ forces $\varphi$''

  We often omit $M, \Pbb$.
\end{fcdefnstar}

Two theorems at the heart of forcing:
\begin{enumerate}[(1)]
  \item \textbf{Forcing Theorem:} If $G$ is a
    \dgen[\Pbb] over $M$, then
    \[
      M[G] \models \varphi \iff \exists p \in G, p
      \forces \varphi
    .\]
  \item \textbf{Definability Theorem:} ``$p
    \forces \varphi$'' is \gls{abs}ly definable
    for transitive models containing $M$.
\end{enumerate}
We are going to do the following:
\begin{enumerate}[(a)]
  \item Define the \emph{syntactic forcing
    predicate} $\Vdash^*$ that is \gls{abs}ly
    definable.
  \item Prove the Forcing Theorem for $\Vdash^*$.
  \item Derive that $\forces \iff \Vdash^*$.
\end{enumerate}
Lectures 11 and 12.

Observations (under the assumption of the Forcing
Theorem):
\begin{enumerate}[(1)]
  \item If $q \le p$ and $p \forces \varphi$, then
    $q \forces \varphi$.
  \item If $p \forces \exists x, \varphi$, then
    there is a name $\tau$ and $q \le p$ such that
    $q \forces \varphi(\tau)$.

[if $p \forces \exists x, \varphi(x)$ and $p \in
G$, then by definition $M[G] \models \exists x,
\varphi(x)$, so there is a name $\tau$ such that
$M[G] \models \varphi(\tau)$. By Forcing Theorem,
find $r \in G$ such that $r \forces
\varphi(\tau)$. Find $q \le p, r$, $q \in G$. By
(1), $q \forces \varphi(\tau)$.
]
\end{enumerate}

\begin{proof}[Proof of Separation in {$M[G]$}]
  Let $\varphi(x, x_1, \ldots, x_n)$ be an
  $\mathcal{L}_{\in}$-formula (not in the
  \gls{fl}) and $x \in M[G]$. Need a name for
  \[
    A
    \defeq \{z \in x : M[G] \models \varphi(z, \ol{p})\}
  .\]
  [For readability, we now drop parameters $\ol{p}$].

  Fix $\sigma$ such that $\val(\sigma, G) = x$.
  Define
  \[
    \varrho
    \defeq \{(\pi, p) : \pi \in \dom(\sigma)
    \wedge p \forces \pi \in \sigma \wedge
    \varphi(\pi)\}
  .\]
  By the Definability Theorem, $\varrho$ is a name
  in $M$.

  Claim: $\val(\varrho, G) = A$.
  \begin{enumerate}[``$\subseteq$'']
    \item[``$\subseteq$''] If $z \in \val(\varrho,
      G)$, then there is $(\pi, p) \in \varrho$
      such that $z = \val(\pi, G)$, $p \in G$.
      Since $(\pi, p) \in \varrho$, we get $\pi
      \in \dom(\sigma)$, $p \forces \pi \in \sigma
      \wedge \varphi(\pi)$. Together with $p \in
      G$, we get
      \[
        M[G]
        \models \ub{\pi \in \sigma \wedge
        \varphi(\pi)}_{z \in x \wedge \varphi(z)}
      .\]
      Hence $z \in A$.
    \item[``$\supseteq$''] If $z \in A$, then $z
      \in x$ and $M[G] \models \varphi(z)$. So
      there is $\pi \in \dom(\sigma)$, $z =
      \val(\pi, G)$.

      By the Forcing THeorem, $\exists p \in G$,
      $p \forces \varphi(\pi)$. Also, there is $q
      \in G$ and $q \forces \pi \in \sigma$. Find
      $r \in G$, $r \le p, q$ such that $r \forces
      \pi \in r \wedge \varphi(\pi)$.

      Hence $(\pi, r) \in \varrho$, so $z = \val(\pi,
      G) \in \val(\varrho, G)$.
      \qedhere
  \end{enumerate}
\end{proof}
