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\begin{module}[id=bbt-size]
\importmodule[balanced-binary-trees]{balanced-binary-trees}
\importmodule[\KWARCslides{dmath/en/cardinality}]{cardinality}

\begin{frame}
  \frametitle{Size Lemma for Balanced Trees}
  \begin{itemize}
  \item
    \begin{assertion}[id=size-lemma,type=lemma] 
    Let $G=\tup{V,E}$ be a \termref[cd=binary-trees]{balanced binary tree} 
    of \termref[cd=graph-depth,name=vertex-depth]{depth}$n>i$, then the set
     $\defeq{\livar{V}i}{\setst{\inset{v}{V}}{\gdepth{v} = i}}$ of
    \termref[cd=graphs-intro,name=node]{nodes} at 
    \termref[cd=graph-depth,name=vertex-depth]{depth} $i$ has
    \termref[cd=cardinality,name=cardinality]{cardinality} $\power2i$.
   \end{assertion}
  \item
    \begin{sproof}[id=size-lemma-pf,proofend=,for=size-lemma]{via induction over the depth $i$.}
      \begin{spfcases}{We have to consider two cases}
        \begin{spfcase}{$i=0$}
          \begin{spfstep}[display=flow]
            then $\livar{V}i=\set{\livar{v}r}$, where $\livar{v}r$ is the root, so
            $\eq{\card{\livar{V}0},\card{\set{\livar{v}r}},1,\power20}$.
          \end{spfstep}
        \end{spfcase}
        \begin{spfcase}{$i>0$}
          \begin{spfstep}[display=flow]
           then $\livar{V}{i-1}$ contains $\power2{i-1}$ vertexes 
           \begin{justification}[method=byIH](IH)\end{justification}
          \end{spfstep}
          \begin{spfstep}
           By the \begin{justification}[method=byDef]definition of a binary
              tree\end{justification}, each $\inset{v}{\livar{V}{i-1}}$ is a leaf or has
            two children that are at depth $i$.
          \end{spfstep}
          \begin{spfstep}
           As $G$ is \termref[cd=balanced-binary-trees,name=balanced-binary-tree]{balanced} and $\gdepth{G}=n>i$, $\livar{V}{i-1}$ cannot contain
            leaves.
          \end{spfstep}
          \begin{spfstep}[type=conclusion]
           Thus $\eq{\card{\livar{V}i},{\atimes[cdot]{2,\card{\livar{V}{i-1}}}},{\atimes[cdot]{2,\power2{i-1}}},\power2i}$.
          \end{spfstep}
        \end{spfcase}
      \end{spfcases}
    \end{sproof}
  \item 
    \begin{assertion}[id=fbbt,type=corollary]	
      A fully balanced tree of depth $d$ has $\power2{d+1}-1$ nodes.
    \end{assertion}
  \item
      \begin{sproof}[for=fbbt,id=fbbt-pf]{}
        \begin{spfstep}
          Let $\defeq{G}{\tup{V,E}}$ be a fully balanced tree
        \end{spfstep}
        \begin{spfstep}
          Then $\card{V}=\Sumfromto{i}1d{\power2i}= \power2{d+1}-1$.
        \end{spfstep}
      \end{sproof}
    \end{itemize}
  \end{frame}
\begin{note}
  \begin{omtext}[type=conclusion,for=binary-tree]
    This shows that balanced binary trees grow in breadth very quickly, a consequence of
    this is that they are very shallow (and this compute very fast), which is the essence of
    the next result.
  \end{omtext}
\end{note}
\end{module}

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           As $G$ is \termref[cd=balanced-binary-trees,name=balanced-binary-tree]{balanced} and $\gdepth{G}=n>i$, $\livar{V}{i-1}$ cannot contain
 
\begin{module}[id=bbt-size]\importmodule[balanced-binary-trees]{balanced-binary-trees}\importmodule[\KWARCslides{dmath/en/cardinality}]{cardinality}​\begin{frame}  \frametitle{Size Lemma for Balanced Trees}  \begin{itemize}  \item    \begin{assertion}[id=size-lemma,type=lemma]     Let $G=\tup{V,E}$ be a \termref[cd=binary-trees]{balanced binary tree}     of \termref[cd=graph-depth,name=vertex-depth]{depth}$n>i$, then the set     $\defeq{\livar{V}i}{\setst{\inset{v}{V}}{\gdepth{v} = i}}$ of    \termref[cd=graphs-intro,name=node]{nodes} at     \termref[cd=graph-depth,name=vertex-depth]{depth} $i$ has    \termref[cd=cardinality,name=cardinality]{cardinality} $\power2i$.   \end{assertion}  \item    \begin{sproof}[id=size-lemma-pf,proofend=,for=size-lemma]{via induction over the depth $i$.}      \begin{spfcases}{We have to consider two cases}        \begin{spfcase}{$i=0$}          \begin{spfstep}[display=flow]            then $\livar{V}i=\set{\livar{v}r}$, where $\livar{v}r$ is the root, so            $\eq{\card{\livar{V}0},\card{\set{\livar{v}r}},1,\power20}$.          \end{spfstep}        \end{spfcase}        \begin{spfcase}{$i>0$}          \begin{spfstep}[display=flow]           then $\livar{V}{i-1}$ contains $\power2{i-1}$ vertexes            \begin{justification}[method=byIH](IH)\end{justification}          \end{spfstep}          \begin{spfstep}           By the \begin{justification}[method=byDef]definition of a binary

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