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\begin{enumerate}[(1)]
  \setcounter{enumi}{2}
  \item
    \begin{flashcard}[empty-ax]
      \glsnoundefn{Emp}{(Emp)}{N/A}
      \prompt{\cloze{(3) }}\textbf{\cloze{Empty
      set axiom} (Emp):} \cloze{
      \[ (\pexists x)(\pforall y)(\pnot y \zin x) \]
      By \gls{Ext}, this set is unique which we
      denote by $\emptyset$. Formally, we add a
      constant $\emptyset$ to the language with
      the sentence $(\pforall y)(\pnot y \zin
      \emptyset)$.
      }
    \end{flashcard}
  \item
    \begin{flashcard}[pair-set-ax]
      \glsnoundefn{Pair}{(Pair)}{N/A}
      \prompt{\cloze{(4) }}\textbf{\cloze{Pair
      set axiom} (Pair):} \cloze{
      ``We can form unordered pairs''.
      \[ (\pforall x)(\pforall y)(\pexists
      z)(\pforall t) ((t \zin z) \pimpl (t = x
      \por t = y)) .\]
      Unique by \gls{Ext}. We denote this set $z$
      by $\{x, y\}$. Define singletons as $\{x, x\}$.

      The following an be proved:
      \[ (\pforall x)(\pforall y)(\{x, y\} = \{y,
      x\}) .\]
      We can use \gls{Pair} to define ordered
      pairs: for $x, y$ the ordered pair $(x, y) =
      \{\{x\}, \{x, y\}\}$. One can then prove
      that:
      \[ (\pforall x)(\pforall y)(\pforall
      t)(\pforall z) ((x, y) = (t, z) \iff (x = t
      \pand y = z)) .\]
      }
    \end{flashcard}
    
    We introduce abbreviations:
    \begin{itemize}
      \item ``$x$ is an ordered pair'' for 
        $(\pexists y)(\pexists z)(x = (y, z))$.
      \item ``$f$ is a function'' for
        \[ (\pforall
        x)(x \zin f \pimpl x \text{ is an ordered
        pair}) \]
        \[ \pand (\pforall x)(\pforall
        y)(\pforall z)(((x, y) \zin f \pand (x, z)
        \zin f) \pimpl (y = z)) \]
      \item ``$x = \dom f$'' for
        \[ \text{`$f$ is a function'} \pand
        (\pforall y)(y \zin x \iff (\pexists z)((y
        z) \zin f)) \]
      \item ``$f$ is a function from $x$ to $y$''
        for
        \[ (x = \dom f) \pand (\pforall t)
        ((\pexists z)(z, t) \zin f \pimpl t \zin y) \]
    \end{itemize}
  \item
    \begin{flashcard}[union-set-ax]
      \glsnoundefn{Un}{(Un)}{N/A}
      \prompt{\cloze{(5) }}\textbf{\cloze{Union
      axiom} (Un):} \cloze{
      \[ (\pforall x)(\pexists y)(\pforall z)(z
      \zin y \iff (\pexists t)(t \zin x \pand z
      \zin t)) .\]
      Denote this set $y$ by $\bigcup x$.
      }
    \end{flashcard}
    \begin{example*}
      \vspace{0.5em}
      For $x, y$, $t \zin \{x, y\} \iff (t \zin x
      \por t \zin y)$. We also write $\bigcup \{x,
      y\} = x \cup y$.
    \end{example*}
    \begin{remark*}
      \vspace{0.5em}
      No new axiom eeded for intersection as this
      can be formed by \gls{Sep}. So the following
      line follows from axioms so far:
      \[ (\pforall x)(\pnot x = \emptyset \pimpl
      (\pexists y)(\pforall z)(z \zin y \iff
      (\pforall t)(t \zin x \pimpl z \zin t)) .\]
      Denote the set $y$ by $\bigcap x$. To prove
      this, given $x$, form
      \[ y = \{z \zin \bigcup x : (\pforall t)(t
      \zin x \pimpl z \zin t)\} \]
      by \gls{Sep}. Check that
      \[ (\pforall z)(z \zin y \iff (\pforall t)(t
      \zin x \pimpl z \zin t)) .\]
      Given $x, y$, denote $\bigcap \{x, y\}$ by
      $x \cap y$.
    \end{remark*}
  \item
    \begin{flashcard}[power-set-ax]
      \glsnoundefn{Pow}{(Pow)}{N/A}
      \prompt{\cloze{(6) }}\textbf{\cloze{Power
      set axiom} (Pow):} \cloze{
      \[ (\pforall x)(\pexists y)(\pforall z)(z
      \zin y \iff z \subset x) \]
      where $z \subset x$ is an abbreviation for
      $(\pforall t)(t \zin z \pimpl t \zin x)$. We
      denote $y$ by $\PP x$.

      We can now form Cartesian product $x \times
      y$ for sets $x, y$: an element of $x \times
      y$ is an ordered pair $(s, t)$ where $s \zin
      x$, $t \zin y$. Note that
      \[ (s, t) = \{\{x\}, \{x, y\}\} \zin \PP
      \PP(x \cup y) ,\]
      so by \gls{Sep} we can form
      \[ \{z \zin \PP\PP(x \cup y) : (\pexists
      s)(\pexists t)(s \zin x \pand t \zin y \pand
      z = (s, t)\} .\]

      We can also form, from sets $x, y$
      \[ y^x = \{f \zin \PP(x \times y) : (f : x
      \to y)\} ,\]
      which is the set of all functions from $x$
      to $y$.
      }
    \end{flashcard}
  \item
    \begin{flashcard}[ax-of-infty]
      \glsnoundefn{Inf}{(Inf)}{N/A}
      \prompt{\cloze{(7) }}\textbf{\cloze{Axiom
      of infinity} (Inf):} \cloze{
      From axioms so far, any model $V$ will be
      infinite, for example
      \[ \emptyset, \PP \emptyset, \PP\PP
      \emptyset, \ldots \]
      are all distincnt elements of $V$.

      For a set $x$ define the successor of $x$ as
      $x^+ = x \cup \{x\}$. Then
      \[ \emptyset, \emptyset^+, \emptyset^{++},
      \ldots \]
      are distinct elements of $V$:
      \[ \emptyset^+ = \{\emptyset\},
      \emptyset^{++} = \{\emptyset,
      \{\emptyset\}\}, \emptyset^{+++} =
      \{\emptyset, \{\emptyset\}, \{\emptyset,
      \{\emptyset\}\}\}, \ldots \]
      We write $0 = \emptyset$, $1 = \emptyset^+$,
      $2 = \emptyset^{++}, \ldots$. We have a copy
      of $\NN_0$ in $V$. From the outside, $V$ is
      infinite. From the inside, $V$ is not a set:
      $\pnot (\pexists x)(\pforall y)(y \zin x)$
      (Russell's paradox).

      Abbreviate ``$x$ is a successor set'':
      \[ \emptyset \zin x \pand (\pforall y)(y
      \zin x \pimpl y^+ \zin x) .\]
      Axiom (Inf) says:
      \[ \boxed{(\pexists x)(\text{$x$ is a
      succcessor set})} .\]
      The intersection of successor sets is a
      successor set. So we can construct
      ``smallest'' successor set, i.e. we can
      prove
      \[ (\pexists x)((x \text{ is a successor
      set}) \pand (\pforall y)(\text{$y$ is a
      successor set} \pimpl x \subset y)) \]
      (Pick any sucessor set $z$, let $x = \bigcap
      \{y \zin \PP z \st \text{$y$ is a sucessor
      set}\}$. $x$ is then a successor set, and if
      $y$ is any successor set then $x \subset (y
      \cap z)$.) We denote the smallest successor
      set by $\omega$.

      If $x \subset \omega$ is a successor set
      then $x = \omega$, i.e.
      \[ (\pforall x)(((x \subset \omega) \pand
      (\emptyset \zin x) \pand (\pforall y)(y \zin
      x \pimpl y^+ \zin x)) \pimpl x = \omega) .\]
      This is true induction.

      We can prove by induction:
      \begin{itemize}
        \item $(\pforall x)(x \zin \omega
          \pimpl \pnot x^+ = \emptyset)$
        \item $(\pforall x)(\pforall y)(((x \zin
          \omega) \pand (y \zin \omega) \pand (x^+
          = y^+) \pimpl x = y))$.
      \end{itemize}
      We can define abbreviations:
      \begin{itemize}
        \item ``$x$ is finite'' for $(\pexists
          y)((y \zin \omega) \pand (\pexists f)(f
          : x \to y \pand \text{$f$ is a
          bijection}))$.
        \item ``$x$ is countable'' for $(\pexists
          f)(f : x \to \omega \pand \text{$f$ is
          injective})$
      \end{itemize}
      }
    \end{flashcard}
\end{enumerate}
