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\begin{fcdefn}[Forcing relation]
  % Definition 1.4.15
  Let $(S, \le, \forces)$ be a \gls{km} for a
  propositional language. We define the extended
  \gls{forcing} relation inductively as follows:
  \begin{citems}
    \item There is no $s \in S$ with $s \forces
      \false$;
    \item $s \forces \varphi \pand \psi$ if and
      only if $s \forces \varphi$ and $s \forces
      \psi$;
    \item $s \forces \varphi \por \psi$ if and
      only if $s \forces \varphi$ or $s \forces
      \psi$;
    \item $s \forces (\varphi \to \psi)$ if and
      only if $s' \forces \varphi$ implies $s'
      \forces \psi$ for every $s' \ge s$.
  \end{citems}
\end{fcdefn}

It is easy to check that the \gls{pers}
property extends to arbitrary propositions.

Moreover:
\begin{itemize}
  \item $s \forces \pnot \varphi$ if and only if $s'
    \not\forces \varphi$ for all $s' \ge s$.
  \item $s \forces \pnot \pnot \varphi$ if and only if for every
    $s' \ge s$,
    there exists $s'' \ge s'$ with $s'' \forces \varphi$.
\end{itemize}

\begin{notation*}
  \glssymboldefn{wforces}%
  $S \Vdash \varphi$ for $\varphi$ a proposition
  if all worlds in $S$ \gls{force} $\varphi$.
\end{notation*}

\begin{example}
  % Example 1.4.16
  Consider the following \glspl{km}:
  \begin{enumerate}[(1)]
    \item
      \begin{picmath}
        \begin{tikzcd}
          s' \forces p \ar[d, no head] \\
          s
        \end{tikzcd}
      \end{picmath}
    \item
      \begin{picmath}
        \begin{tikzcd}
          s' \ar[rd, no head]
          &
          & s'' \forces p \ar[ld, no head] \\
          & s
        \end{tikzcd}
      \end{picmath}
    \item
      \begin{picmath}
        \begin{tikzcd}
           s' \forces p, \forces q
           \ar[d, no head] \\
           s
        \end{tikzcd}
      \end{picmath}
  \end{enumerate}
  In (1), we have $s \not\forces \pnot p$, since
  $s' \ge s$ and $s' \forces p$. We also know that
  $s \not\forces p$, thus $s \not\forces p \por
  \pnot p$.

  It is also the case that $s \forces \pnot \pnot
  p$, yet $s \not \forces p$, so
  $s\not\forces(\pnot \pnot p \to p)$ either.

  In (2), $s \not\forces \pnot \pnot p$ since $s'
  \ge s$ can't access a world that \glspl{force}
  $p$. Also $s \not \forces \pnot p$ either, as
  $s'' \ge s$ \glspl{force} $p$. So $s \not\forces
  \pnot \pnot p \por \pnot p$.

  In (3), $s\not\forces (p \to q) \to (\pnot p
  \por q)$. All \glspl{world} \gls{force} $p \to
  q$, and $s \not\forces q$. So to check the claim
  we just need to verify that $s\not\forces \pnot
  p$. But that is the case, as $s' \ge s$ and $s'
  \forces p$.
\end{example}

\begin{fcdefn}[Filter]
  \glsnoundefn{filt}{filter}{filters}%
  % Definition 1.4.17
  A \emph{filter} $F$ on a \gls{latt} $L$ is a
  subset of $L$ with the following properties:
  \begin{citems}
    \item $F \neq \emptyset$
    \item $F$ is a \gls{tseg} of $L$ (i.e., if $f
      \le x$ and $f \in F$, then $x \in F$)
    \item $F$ is closed under finite meets
  \end{citems}
\end{fcdefn}

\begin{example}
  % Example 1.4.18
  \phantom{}
  \begin{enumerate}[(1)]
    \item
      \glsadjdefn{princf}{principal}{\gls{filt}}%
      Given an element $j \in I$ of a set $I$,
      then the family $F_j$ of all subsets of $I$
      containing $j$ is a \gls{filt} on
      $\mathcal{P}(I)$. Such a \gls{filt} is
      called a \emph{principal filter}.
    \item The family of all cofinite subsets of
      $I$ is a filter on $\mathcal{P}(I)$, the
      Fr\'echet \gls{filt}.

      Exercise: a maximal \gls{propf} \gls{filt} (known as an
      \emph{ultra filter}) is not \gls{princf} if and
      only if it contains the Fr\'echet \gls{filt}.
    \item The family of all subsets of $[0, 1]$
      with Lebesgue measure $1$ is a \gls{filt}.
  \end{enumerate}
\end{example}

\glsadjdefn{propf}{proper}{\gls{filt}}%
A \gls{filt} is \emph{proper} if $F \neq L$.

\glsadjdefn{primef}{prime}{\gls{filt}}%
A \gls{filt} $F$ on a \gls{ha} is \emph{prime} if
it is \gls{propf} and satisfies: whenever $(x \por
y) \in F$, we can conclude that $x \in F$ or $y
\in F$.

If $F$ is a \gls{propf} \gls{filt} and $x \notin
F$, then there is a \gls{primef} \gls{filt}
extending $F$ that still doesn't contain $x$ (by
Zorn's Lemma).

\begin{fclemma}[]
  \label{lemma:1.4.19}
  % Lemma 1.4.19
  Assuming:
  - $H$ a \gls{ha}
  - $v$ a \gls{hvalt}
  Then:
  \begin{iffc}
    \lhs
    there is a \gls{km} $(S, \le, \forces)$ such
    that $v \hmodels \varphi$
    \rhs
    $S \wforces \varphi$, for every proposition
    $\varphi$.
  \end{iffc}
\end{fclemma}