%! TEX root = LC.tex
% vim: tw=50
% 06/11/2024 11AM

\begin{proof}[Proof (sketch)]
  Let $S$ be the set of all \gls{primef}
  \glspl{filt} of $H$, ordered by inclusion. We
  write $F \forces p$ if and only if $v(p) \in
  F$ for primitive propositions $p$.

  We prove by induction that $F \forces \varphi$
  if and only if $v(\varphi) \in F$ for arbitrary
  propositions.

  For the implication case, say that $F \forces
  (\psi \to \psi')$ and $v(\psi \to \psi') =
  [v(\psi) \himplies v(\psi')] \notin F$. Let $G'$
  be the least \gls{filt} containing $F$ and
  $v(\psi)$. Then
  \[
    G' = \{b : (\exists f \in F) (f \meet v(\psi)
    \le b)\}
  .\]
  Note that $v(\psi') \notin G'$, or else $f \meet
  v(\psi) \le v(\psi')$ for some $f \in F$, whence
  $f \le v(\psi \to \psi')$ and so $v(\psi \to
  \psi') \in F$ (as $F$ is a terminal segment).

  In particular, $G'$ is \gls{propf}. So let $G$ be a
  \gls{primef} \gls{filt} extending $G'$ that does
  not contain $v(\psi')$ (exists by Zorn's lemma).

  By the induction hypothesis, $G \forces \psi$,
  and since $F \forces (\psi \to \psi')$ and $G'$
  (this $G$) contains $F$, we have that $G \forces
  \psi'$. But then $v(\psi') \in G$,
  contradiction.

  This settles that $F \forces (\psi \to \psi')$
  implies $v(\psi \to \psi') \in F$.

  Conversely, say that $v(\psi \to \psi') \in F
  \subseteq G \forces \psi$. By the induction
  hypothesis, $v(\psi) \in G$, and so $v(\psi)
  \himplies v(\psi) \in G$ (as $F \subseteq G$).
  But then $v(\psi') \ge v(\psi) \meet (v(\psi)
  \himplies v(\psi')) \in G$, as $G$ is a
  \gls{filt}.

  So the induction hypothesis gives $G \forces
  \psi'$, as needed.

  The cases for the other connectives are easy
  ($\por$ needs primality). So $(S, \le, \forces)$
  is a \gls{km}. Want to show that $v \hmodels
  \varphi$ if and only if $S \forces \varphi$, for
  each $\varphi$.

  Conversely, say $S \forces
  \varphi$, but $v \not\hmodels \varphi$. Since
  $v(\varphi) \neq \top$, there must be a
  \gls{propf} \gls{filt} that does not contain it.
  We can extend it to a \gls{primef} \gls{filt}
  $G$ that does not contain it, but then $G
  \not\forces \varphi$, contradiction.
\end{proof}

\begin{fcthm}[Completeness of the Kripke semantics]
  % Theorem 1.4.20
  Assuming:
  - $\varphi$ a proposition
  Then:
  \begin{iffc}
    \lhs
    $\Gamma \syn_{\text{\gls{ipc}}} \varphi$
    \rhs
    for all \glspl{km} $(S, \le, \forces)$,
    the condition $S \forces \Gamma$ implies $S
    \forces \varphi$.
  \end{iffc}
\end{fcthm}

\begin{proof}
  \textbf{Soundness:} induction over the
  complexity of $\varphi$.

  \textbf{Adequacy:} Say $\Gamma \not
  \syn_{\text{\gls{ipc}}} \varphi$. Then $v
  \hmodels \Gamma$ but $v \not \hmodels \varphi$
  for some \gls{ha} $H$ and \gls{hvalt} $v$
  (\cref{thm:1.4.12}). But then
  \cref{lemma:1.4.19} applied to $H$and $v$
  provides a \gls{km} $(S, \le, \forces)$ such
  that $S \forces \Gamma$, but $S\not\forces
  \varphi$, contradicting the hypothesis on every
  \gls{km}.
\end{proof}

\subsection{Negative translations}

\begin{fcdefn}[Double-negation translation]
  % Definition 1.5.1
  We recursively define the
  $\pnot\pnot$-translation $\varphi^N$ of a
  proposition $\varphi$ in the following way:
  \begin{citems}
    \item If $p$ is a primitive proposition, then
      $p^N \defeq \pnot \pnot p$;
    \item $(\varphi \pand \psi)^N \defeq \varphi^N
      \pand \psi^N$
    \item $(\varphi \to \psi)^N \defeq \varphi^N \to
      \psi^N$
    \item $(\varphi \por \psi)^N \defeq \pnot
      (\pnot \varphi^N \pand \pnot \psi^N)$
    \item $(\pnot \varphi)^N \defeq \pnot
      \varphi^N$
  \end{citems}
\end{fcdefn}

\begin{fclemma}[]
  % Lemma 1.5.2
  Assuming:
  - $H$ a \gls{ha}
  Then:
  the map $\pnot \pnot : H \to H$ preserves $\pand$
  and $\himplies$.
\end{fclemma}

\begin{proof}
  Example Sheet 2.
\end{proof}

\begin{fclemma}[Regularisation]
  % Lemma 1.5.3
  Assuming:
  - $H$ a \gls{ha}
  Thens:[(1)]
  - The subset $H_{\pnot \pnot} \defeq \{x
  \in H : \pnot \pnot x = x\}$ is a \gls{ba};
  - For every \gls{hh} $g : H
  \to B$ into a \gls{ba}, there is a unique
  map of \glspl{ba} $g_{\pnot \pnot}
  :H_{\pnot \pnot} \to B$ such that $g(x) =
  g_{\pnot \pnot}(\pnot \pnot x)$ for all $x
  \in H$.
\end{fclemma}
