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DataAbstraction chapter: proofreading

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Adam Chlipala 2017-02-20 17:06:06 -05:00
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frap_book.tex
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@ -600,8 +600,9 @@ As a result, client code (and its correctness proofs) can use this fancy code, e
Why did we need to go through the trouble of introducing custom equivalence relations? Why did we need to go through the trouble of introducing custom equivalence relations?
Consider the following two queues. Consider the following two queues.
Are they equal? Are they equal?
(We write $\pi_1$ for the function that projects out the first element of a pair.)
\begin{eqnarray*} \begin{eqnarray*}
\mt{enqueue}(\mt{empty}, 2) &\stackrel{?}{=}& \mt{dequeue}(\mt{enqueue}(\mt{enqueue}(\mt{empty}, 1), 2)) \mt{enqueue}(\mt{empty}, 2) &\stackrel{?}{=}& \pi_1(\mt{dequeue}(\mt{enqueue}(\mt{enqueue}(\mt{empty}, 1), 2)))
\end{eqnarray*} \end{eqnarray*}
No, they aren't equal! The first expression reduces to $([2], [])$, while the second reduces to $([], [2])$. No, they aren't equal! The first expression reduces to $([2], [])$, while the second reduces to $([], [2])$.
@ -668,7 +669,7 @@ We implement our optimized set type like so, assuming an operation $\mt{fromRang
\mt{t} &=& \mt{Empty} \mid \mt{Range}(\mathbb N \times \mathbb N) \mid \mt{AdHoc}(\mt{t}_0) \\ \mt{t} &=& \mt{Empty} \mid \mt{Range}(\mathbb N \times \mathbb N) \mid \mt{AdHoc}(\mt{t}_0) \\
\mt{empty} &=& \mt{Empty} \\ \mt{empty} &=& \mt{Empty} \\
\mt{add}(\mt{Empty}, k) &=& \mt{Range}(k, k) \\ \mt{add}(\mt{Empty}, k) &=& \mt{Range}(k, k) \\
\mt{add}(\mt{Range}(n_1, n_2), k) &=& s\textrm{, when $n_1 \leq k \leq n_2$} \\ \mt{add}(\mt{Range}(n_1, n_2), k) &=& \mt{Range}(n_1, n_2)\textrm{, when $n_1 \leq k \leq n_2$} \\
\mt{add}(\mt{Range}(n_1, n_2), n_1-1) &=& \mt{Range}(n_1-1, n_2)\textrm{, when $n_1 \leq n_2$} \\ \mt{add}(\mt{Range}(n_1, n_2), n_1-1) &=& \mt{Range}(n_1-1, n_2)\textrm{, when $n_1 \leq n_2$} \\
\mt{add}(\mt{Range}(n_1, n_2), n_2+1) &=& \mt{Range}(n_1, n_2+1)\textrm{, when $n_1 \leq n_2$} \\ \mt{add}(\mt{Range}(n_1, n_2), n_2+1) &=& \mt{Range}(n_1, n_2+1)\textrm{, when $n_1 \leq n_2$} \\
\mt{add}(\mt{Range}(n_1, n_2), k) &=& \mt{AdHoc}(\mt{add}_0(\mt{fromRange}(n_1, n_2), k))\textrm{, otherwise} \\ \mt{add}(\mt{Range}(n_1, n_2), k) &=& \mt{AdHoc}(\mt{add}_0(\mt{fromRange}(n_1, n_2), k))\textrm{, otherwise} \\