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\section{Non-classical Logic}

There are no non-experienced truths -- L.E.J. Brouwer

\subsection{Intuitionistic Logic}

\textbf{Idea:} a proof of $\varphi \to \psi$ is a
``procedure'' that comments a proof of $\varphi$
into a proof of $\psi$.

In particular, $\pnot \pnot \varphi$ is not always
the same as $\varphi$.

\textbf{Fact:} The law of excluded middle
($\varphi \por \pnot \varphi$) is not generally
intuitionistically valid.

Moreover, the Axiom of Choice is incompatible with
intuitionistic set theory.

We take choice to mean that any family of
inhabited sets admits a choice function.

\begin{fcthm}[Diaconescu]
  % Theorem 1.1.1
  The law of excluded middle can be
  intuitionistically deduced from the Axiom of
  Choice.
\end{fcthm}

\begin{proof}
  Let $\varphi$ be a proposition. By the Axiom of
  Separation, the following are sets (i.e. we can
  construct a proof that they are sets):
  \[ A \defeq \{x \in \{0, 1\} : \varphi \por (x =
  0)\} \qquad B \defeq \{x \in \{0, 1\} : \varphi
  \por (x = 1)\} .\]
  As $0 \in A$ and $1 \in B$, we have that $\{A,
  B\}$ is a family of inhabited sets, thus admits
  a choice function $f : \{A, B\} \to A \cup B$ by
  the Axiom of Choice. This satisfies $f(A) \in A$
  and $f(B) \in B$ by definition.

  Thus we have
  \[ (f(A) = 0 \por \varphi) \pand (f(B) = 1 \por
  \varphi) \]
  and $f(A), f(B) \in \{0, 1\}$. Now $f(A) \in
  \{0, 1\}$ means that $(f(A) = 0) \por (f(A) = 1)$
  and similarly for $f(B)$.

  We can have the
  following:
  \begin{enumerate}[(1)]
    \item We have a proof of $f(A) = 1$, so
      $\varphi \por (1 = 0)$ has a proof, so we
      must have a proof of $\varphi$.
    \item We have a proof of $f(B) = 0$, which
      similarly gives a proof of $\varphi$.
    \item We have $f(A) = 0$ and $f(B) = 1$, in
      which case we can prove $\pnot \varphi$:
      given a proof of $\phi$, we can prove that
      $A = B$ (by Extensionality), in which case
      $0 = f(A) = f(B) = 1$, a contradiction.
  \end{enumerate}
  So we can always specify a proof of $\varphi$
  or a proof of $\varphi$ or a proof of $\pnot
  \varphi$.
\end{proof}

Why bother?
\begin{itemize}
  \item Intuitionistic maths is more general: we
    assume less.
  \item Several notions that are conflated in
    classical maths are genuinely different
    constructively.
  \item Intuitionistic proofs have a computable
    content that may be absent in classical
    proofs.
  \item Intuitionistic logic is the internal
    logic of an arbitrary topos.
\end{itemize}
Let's try to formalise the BHK interpretation of
logic.

We will inductively define a provability relation
by enforcing rules that implement the BHK
interpretation.

We will use the notation $\Gamma \syn \varphi$
to mean that $\varphi$ is a consequence of the
formulae in the set $\Gamma$.
