%! TEX root = FC.tex
% vim: tw=50
% 04/03/2025 12PM

\begin{note*}
  The Forcing Theorem is very useful! The inner
  details of the proof are not very important
  though. We won't really discuss these details
  once we have proved the theorem.
\end{note*}

Continuing the proof of \nameref{synFT}. Assume
\nameref{synFT} for ``$=$'' and prove if for
``$\in$''.

Want to show:

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

\begin{iffproof}
  \rightimpl
  Assume that $M[G] \models \tau_0 \in \tau_1$,
  i.e. $\val(\tau_0, G) \in \val(\tau_1, G)$.
  Thus, there is $(\pi, s) \in \tau_1$ such that
  $\val(\pi, G) = \val(\tau_0, G)$ ($\iff M[G]
  \models \pi = \tau_0$) and $s \in G$.
  So by \nameref{synFT} for $=$, find $r \in G$
  such that $r \forces^* \pi = \tau_0$.

  Find $p \le r, s$ such that $p \in G$.

  Claim that $\mathcal{D}$ is \gls{dbel} $p$. Let
  $q \in D$ and pick the above $(\pi, s)$. Then we
  have $q \le p \le s$ and $q \forces^* \pi =
  \tau_0$ (since $q \le p \le r$).

  \leftimpl
  Assume $p \in G$, $p \forces^* \tau_0 \in
  \tau_1$, i.e. $D$ is \gls{dbel} $p$. By
  \cref{lec11lemmalots}(iv), find $q \in G \cap D$,
  thus there is $(\pi, s) \in \tau_1$ such that $q
  \le s$, $q \forces^* \pi = \tau_0$. Since $q \in
  G$, $s \in G$, we have $\val(\pi, G) \in
  \val(\tau_1, G)$. By \nameref{synFT} for
  ``$=$'', we have $\val(\pi, G) = \val(\tau_0,
  G)$. So $\val(\tau_0, G) \in \val(\tau_1, G)$.
\end{iffproof}

Now we prove it for ``$=$''.

We prove this by induction on name rank, so assume
that \nameref{synFT} for ``$=$'' is true for names
$\pi_0, \pi_1$ with smaller name rank.

Reminder: we need to show $M[G] \models \tau_0 =
\tau_1$ if and only if $\exists p \in G$, $p
\forces^* \tau_0 = \tau_1$.

TODO: double check the below

\begin{iffproof}
  \leftimpl
  Assume $p \in G$ such that $p \forces^* \tau_0 =
  \tau_1$. Need to prove $\val(\tau_0, G) =
  \val(\tau_1, G)$.

  We'll show that the fact that $D_0$ is
  \gls{dbel} $p$ implies $\val(\tau_0, G)
  \subseteq \val(\tau_1, G)$. The other direction
  is the same proof but with $0$ and $1$ flipped.

  Proof of this: Let $x \in \val(\tau_0, G)$, so
  find $(\pi_0, s_0) \in \tau_0$ such that $s_0
  \in G$ and $x = \val(\pi_0, G)$. So, the
  corresponding $D_0$ is \gls{dbel} $p$. Find $q
  \le s_0, p$ such that $q \in G$. Then $D_0$ is
  \gls{dbel} $q$. So find $r \le q$ such that $r
  \in G \cap D_0$. (Note that $r \le q \le s_0$,
  so $r \le s_0$).

  Since $r \in D_0$ and $r \le s_0$, find $(\pi_1,
  s_1) \in \tau_1$ such that $r \le s_1 \wedge r
  \forces^* \pi_0 = \pi_1$. By induction
  hypothesis, $r \in G$ and $r \forces^* \pi_0 =
  \pi_1$, which implies $x = \val(\pi_, v) =
  \val(\pi_1, G)$ (since $(\pi_1, s_1) \in \tau_1$
  and $s_1 \in G$).

  \rightimpl
  Assume $\val(\tau_0, G) = \val(\tau_1, G)$.
  Consider the set
  \begin{align*}
    D
    &\defeq \{r : r \forces^* \tau_0 = \tau_1
    \text{ or } \\
    &\Phi_r^0 ~~\exists (\pi_0, s_0) \in \tau_0 (r
    \le s_0 \wedge \forall (\pi_1, s_1) \in
    \tau_1, \forall q((q \le s_1 \wedge q
    \forces^* \pi_0 = \pi_1) \to q \perp r) \\
    &\Phi_r^1 ~~\exists (\pi_1, s_1) \in \tau_1 (r
    \le s_1 \wedge \forall (\pi_0, s_0) \in
    \tau_0, \forall q ((q \le s_0 \wedge q
    \forces^* \pi_0 = \pi_1) \to q \perp r) \\
    &\}
  \end{align*}
  Claim: $D$ is dense. If $p \forces^*$, then $p
  \in D$, so nothing to show. If $p \not\forces^*
  \tau_0 = \tau_1$, then there is some $D_0 / D_1$
  that fails to be \gls{dbel} $p$.

  We'll show: if $D_0$ is not \gls{dbel} $p$, then
  we can find $r \le p$ such that $\Phi_r^0$
  holds.

  (Other proof: $D_1$ not \gls{dbel} $p$ $\to
  \Phi_r^1$ is flipping $0$ and $1$).

  Suppose $p \not\forces^* \tau_0 = \tau_1$ and
  there is $(\pi_0, s_0) \in \tau_0$ such that
  $D_0$ is not \gls{dbel} $p$. This means: there
  is $r \le p$ such that
  \[
    \forall q \le r (q \le s_0 \wedge \forall
    (\pi_1, s_1) \in \tau_1 \neg (q \le s_1 \wedge
    q \forces^* \pi_0 = \pi_1)
  .\]
  So, we get
  \[
    r \le s_0 \wedge \forall (\pi_1, s_1) \in
    \tau_1
  \]
  for any $q$ that satisfies $q \le s_1 \wedge q
  \forces^* \pi_0 = \pi_1$. It can't be compatible
  with $r$ (finishes the proof of the claim that
  $D$ is \gls{dense}). 

  Summary: We now have that $D$ is \gls{dense}.
  \[
    D
    = \{r : r \forces^* \tau_0 = \tau_1 \text{ or
    } \Phi_r^0 \text{ or } \Phi_1^r\}
  .\]
  Claim 2: If $r \in G$, then neither $\Phi_r^0$
  nor $\Phi_r^1$ holds (again, just do $\Phi_r^0$
  and then flip $0$ and $1$ for $\Phi_r^1$).

  If $\Phi_r^0$ holds, then find $(\pi_0, s_0) \in
  \tau_0$ such that $r \le s_0$ and the
  incompatibility statement holds. $r \in G \to
  s_0 \in G$, so $\val(\pi_0, G) \in \val(\tau_0,
  G)$. If $(\pi_1, s_1) \in \tau_1$ such that TODO

  Put everything together: since $D$ is dense by
  Claim 1, find $r \in D \cap G$.
  Therefore, by Claim 2, $\Phi_r^0$, $\Phi_r^1$ do
  not hold.

  Thus $r \forces^* \tau_0 = \tau_1$.
\end{iffproof}
