Hence (A, B, C) refute (32) and (29) is secured.
(A) for allxfor allyfor someznecessarily(x ex Y if-then (z cont x & z emb Y))
(Necessarily, if an i exemplifies an h, these are one in substance.)

(B) for allxfor allyfor allznecessarily((z cont x & z cont y) if-then x = y))
(Necessarily, if i's are one in substance, they are identical.)

(C) for allx(for somey(x ex Y) if-then for someznecessarily(x ex Z))
(If an i exemplifies an h, there is some h it exemplifies necessarily.)

(32) [(x ex X) & ¬necessarily(x ex X) & necessarily((x = x) if-then (x ex X))]
(x exemplifies X contingently but X is essential to x.)

(29) for allx[(x ex X & ¬necessarily(x ex X)) equivalent (x ex X & ¬necessarily(x = x if-then x ex X))]
(x exemplifies X contingently just in case X is accidental to x.)