Obfuscation of Abstract Data-Types - Rowan

APPENDIX D. PROVING A TREE ASSERTION 181
Subcase 1 Suppose that z is even:
member 3 p (mktree3 (x : xs))
= {definition of mktree3}
member 3 p (Fork 3 z (mktree3 ys) (mktree3 zs) jt)
= {definition of member 3 , with z even}
z == p ∨ member 3 p (mktree3 zs) ∨ member 3 p (mktree3 ys)
= {induction hypothesis, using (D.4)}
z == p ∨ elem p ys ∨ elem p zs
= {commutativity and associativity of ∨}
elem p ys ∨ (z == p ∨ elem p zs)
= {definition of elem}
elem p ys ∨ elem p (z : zs)
= {Equation (D.5)}
elem p (ys + (z : zs))
= {definition of xs, Equation (D.3)}
elem p xs
Subcase 2 Suppose that z is odd:
member 3 p (mktree3 (x : xs))
= {definition of mktree3}
member 3 p (Fork 3 z kt (mktree3 zs) (mktree3 ys))
= {definition of member 3 , with z odd}
z == p ∨ member 3 p (mktree3 zs) ∨ member 3 p (mktree3 ys)
= {induction hypothesis, using (D.4)}
z == p ∨ elem p ys ∨ elem p zs
= {commutativity and associativity of ∨}
elem p ys ∨ (z == p ∨ elem p zs)
= {definition of elem}
elem p ys ∨ elem p (z : zs)
= {Equation (D.5)}
elem p (ys + (z : zs))
= {definition of xs, Equation (D.3)}
elem p xs
The basic structure of the proof tree is:
(D.12)
Base
even odd
mktree3
member 3
elem