Obfuscation of Abstract Data-Types - Rowan

APPENDIX A. LIST ASSERTION 151
Step Case Suppose that xsp = 〈x 0 ,x 1 〉 b(k) (and so |xsp| b(k)
= |x 0 | + |x 1 |).
Then we have two subcases depending on whether x 1 is empty.
Subcase 1 Suppose that x 0 = a : l, x 1 = [ ] and 〈l, [ ]〉 b(k) satisfies Equation
(A.5). Then
|xsp + b(k) ysp| b(k)
= {definition of xsp}
|〈a : l, [ ]〉 b(k) + b(k) ysp| b(k)
= {definition of + b(k) }
|cons b(k) a (〈l, [ ]〉 b(k) + b(k) ysp)| b(k)
= {Equation (A.6)}
1 + |〈l, [ ]〉 b(k) + b(k) ysp| b(k)
= {induction hypothesis}
1 + |〈l, [ ]〉 b(k) | b(k)
+ |ysp| b(k)
= {definition of | | b(k)
}
1 + |l| + |[ ]| + |ysp| b(k)
= {definition of | |}
|a : l| + |[ ]| + |ysp| b(k)
= {definition of | | b(k)
}
|〈a : l, [ ]〉 b(k) | b(k)
+ |ysp| b(k)
= {definition of xsp}
|xsp| b(k)
+ |ysp| b(k)
Subcase 2 Now suppose that x 0 = a : l, x 1 = r (where r ≠ [ ]) and 〈l +
[head r], tail r〉 b(k) satisfies (A.5). Then
|xsp + b(k) ysp| b(k)
= {definition of xsp}
|〈a : l,r〉 b(k) + b(k) ysp| b(k)
= {definition of + b(k) }
|cons b(k) a (〈l + [head r], tail r〉 b(k) + b(k) ysp)| b(k)
= {Equation (A.6)}
1 + |〈l + [head r], tail r〉 b(k) + b(k) ysp| b(k)
= {induction hypothesis}
1 + |〈l + [head r], tail r〉 b(k) | b(k)
+ |ysp| b(k)
= {definition of | | b(k)
}
1 + |l + [head r]| + |tail r| + |ysp| b(k)
= {Equation (A.1)}