Obfuscation of Abstract Data-Types - Rowan

CHAPTER 4. SPLITTING HEADACHES 80
Case 2 Suppose that k > 0. We prove Equation (4.20) by structural induction
on xs.
Base Case Suppose that xs = [ ]. Then
unsplit b(k) (split b(k) [ ])
= {definition of split b(k) }
unsplit b(k) 〈[ ], [ ]〉 b(k)
= {definition of unsplit b(k) }
[ ] + [ ]
= {definition of +}
[ ]
Step Case We now suppose that xs = t : ts and let ts ❀ 〈l,r〉 b(k−1) . For
the induction hypothesis, we suppose that ts satisfies Equation (4.20), i.e. ts =
l + r. Then
unsplit b(k) (split b(k) (t : ts))
= {definition of split b(k) and refinement of ts}
unsplit b(k) 〈t : l,r〉 b(k)
= {definition of unsplit b(k) }
(t : l) + r
= {definition of +}
t : (l + r)
= {induction hypothesis}
t : ts
so
We can consider unsplit b(k) to be an abstraction function for this split and
xs ❀ 〈l,r〉 b(k) ⇔
xs = unsplit b(k) 〈l,r〉 b(k) ∧ ((|r| = 0 ∧ |l| < k) ∨ (|l| = k))
(4.21)
We would like a function cons b(k) that satisfies:
split b(k) (a : xs) = cons b(k) a (split b(k) xs)
and so we define:
cons b(k) a 〈l,r〉 b(k) =
where the functions init and last satisfy:
xs = init xs + [last xs] where xs ≠ [ ]
{
〈a : l,r〉b(k) if |l| < k
〈a : (init l), (last l) : r〉 b(k) otherwise
From earlier in this chapter, by using Theorem 1, we can define
map b(k) f 〈l,r〉 b(k) = 〈map f l, map f r〉 b(k)