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    \item[($\pand$-E)] By the induction
      hypothesis, we have $v(\bigwedge \Gamma) \le
      v(B \pand C) = v(B) \wedge v(C) \le v(B),
      v(B)$.
    \item [($\to$-E)] We know that $v(A \to B) =
      (v(A) \himplies v(B))$. From $v(A \to B) \le
      v(A) \himplies v(B)$, we derive $v(A) \wedge
      v(A \to B) \le v(B)$ by definition of
      $\himplies$. So if $v(\bigwedge \Gamma) \le
      v(A \to B)$ and $v(\bigwedge \Gamma) \le
      v(A)$, then $v(\bigwedge \Gamma) \le v(B)$,
      as needed.
    \item[($\por$-E)] By induction hypothesis:
      $v(A \por \bigwedge \Gamma) \le v(C)$,
      $v(B \por \bigwedge \Gamma) \le v(C)$ and
      $v(\bigwedge \Gamma) \le v(A \por B) = v(A)
      \vee v(B)$. This last fact means that
      $v(\bigwedge \Gamma) \wedge (v(A) \vee
      v(B)) = v(\bigwedge \Gamma)$. Now this is
      the same as $(v(\bigwedge \Gamma) \wedge
      v(A)) \vee (v(\bigwedge \Gamma) \wedge
      v(B))$ as \glspl{ha} are \gls{dist}
      \glspl{latt} (see Example Sheet 2), and this
      is $\le v(C)$ by the first two inequalities
      of this paragraph.
    \item[($\bot$-E)] If $v(\bigwedge \Gamma) \le
      v(\false) = \bot$, then $v(\bigwedge \Gamma) =
      \bot$, in which case $v(\bigwedge \Gamma)
      \le v(A)$ for any $A$ by minimality of
      $\bot$ in $H$. \qedhere
  \end{enumerate}
\end{proof}

\begin{example}
  % Example 1.4.8
  The Law of Excluded Middle is not
  intuitionistically valid. Let $p$ be a
  primitive proposition and consider the \gls{ha}
  given by the topology $\{\emptyset, \{1\}, \{1,
  2\}\}$ on $\{1, 2\}$.

  We can define a \gls{valt} $v$ with $v(p) =
  \{1\}$, in which case $v(\pnot p) = \pnot
  \{1\} = \interior(X \setminus \{1\}) = \emptyset$.

  So $v(p \por \pnot p) = \{1\} \vee \emptyset =
  \{1\} \neq \top$. Thus \nameref{hssoundness}
  implies that $\not\syn_{\text{\gls{ipc}}} p \por
  \pnot p$.
\end{example}

\begin{example}
  % Example 1.4.9
  Peirce's Law $((p \to q) \to p) \to p$ is not
  intuitionistically valid.

  Take the valuation on the usual topology of
  $\Rbb^2$ that maps $p$ to $\Rbb^2 \setminus \{(0,
  0)\}$ and $q$ to $\emptyset$.
\end{example}

Classical completeness: $\Gamma \syn_{\text{CPC}}
A$ if and only if $\Gamma \models_2 A$.

Intuitionistic completeness: no single finite
replacement for $2$.

\begin{fcdefnstar}[Lindenbaum-Tarski algebra]
  \glsnoundefn{ltalg}{Lindenbaum-Tarski
  algebra}{Lindenbaum-Tarski algebras}%
  \glssymboldefn{F}%
  Let $Q$ be a logical doctrine (CPC, \gls{ipc}, etc), $L$
  be a propositional language, and $T$ be an
  $L$-theory. The \emph{Lindenbaum-Tarski algebra}
  $F^Q(T)$ is built in the following way:
  \begin{citems}
    \item The underlying set of $F^Q(T)$ is the
      set of equivalence classes $[\varphi]$ of
      propositions $\varphi$, where $\varphi \sim
      \psi$ when $T, \varphi \syn_Q \psi$ and $T,
      \psi \syn_Q \varphi$;
    \item If $\bowtie$ is a logical connective in
      the fragment $Q$, we set $[\varphi] \bowtie
      [\psi] \defeq [\varphi \bowtie \psi]$
      (should check well-defined: exercise).
  \end{citems}
\end{fcdefnstar}

We'll be interested in the case $Q = \text{CPC}$,
$Q = \text{\gls{ipc}}$, and $Q = \text{\gls{ipc}} \setminus
\{\to\}$.

\begin{fcprop}[]
  % Proposition 1.4.10
  The \gls{ltalg} of any theory in $\text{\gls{ipc}}
  \setminus \{\to\}$ is a \gls{dist} \gls{latt}.
\end{fcprop}

\begin{proof}
  Clearly, $\pand$ and $\por$ inherit
  associativity and commutativity, so in order for
  $F^{\text{\gls{ipc}} \setminus \{\to\}} (T)$ to be a
  \gls{latt} we need only to check the absorption
  laws:
  \begin{align*}
    [\varphi] \por [\varphi \pand \psi]
    &=
    [\varphi] \tag{$\alpha$} \label{alpha9} \\
    [\varphi] \pand [\varphi \por
    \psi]
    &= [\varphi] \tag{$\beta$} \label{beta9}
  \end{align*}
  Equation \eqref{alpha9} is true since $T,
  \varphi \syn_{\gls{ipc} \setminus \{\to\}}
  \varphi \por (\varphi \pand \psi)$ by
  ($\por$-I), and also $T, \varphi \por (\varphi
  \pand \psi) \syn_{\text{\gls{ipc}} \setminus
  \{\to\}} \varphi$ by ($\por$-E). Equation
  \eqref{beta9} is similar.

  Now, for distributivity: $T, \varphi \pand (\psi
  \por \chi) \syn (\varphi \pand \psi) \por
  (\varphi \pand \chi)$ by ($\pand$-E) followed by
  ($\por$-E):
  \begin{logicproof}
    \varphi \pand (\psi \por \chi)
    \\($\pand$-E)
    \varphi \qquad \psi \por \chi
    \\($\por$-E)
    (\varphi \pand \psi) \por (\varphi \pand \chi)
  \end{logicproof}
  Conversely, $T, ((\varphi \pand \psi) \por
  (\varphi \pand \chi)) \syn \varphi \pand (\psi
  \por \chi)$ by ($\vee$-E) followed by
  ($\pand$-I).
\end{proof}
