Smallest possible 6-bit XOR (even/odd parity)

Question

I have come across an interesting problem that is kind of boggling my mind: It's about implementing a 6-bit XOR or even/odd parity with as few logic gates as possible.

Obviously this would mean calculating Z = ((A XOR B) XOR (C XOR D)) XOR (E XOR F).

The caveat is that this should be done with NO xor gates, so only AND/OR/NOT. I'm trying to get this down to as few of these gates as possible, and I have a solution that needs 13 gates so far (not counting NOT):

X1 = (ABC + !A!BC + !AB!C + A!B!C)
X2 = (DEF + !D!EF + !DE!F + D!E!F)
Z = (X1 + X2) !(X1 X2)

However, supposedly this is possible with only 12 gates (again, not counting NOT). I feel like I'm missing some rather obvious boolean algebra optimization step here (not my strong suit), is there something I can reduce here?

EDIT: I'm ok with using n-bit OR/AND gates, e.g. X1 uses only 1 OR gate.

Answer 1

Figured out how to do it:

X1 = A + B + C
X2 = A + !B + !C
X3 = !A + B + !C
X4 = !A + !B + C
Y1 = D + E + F
Y2 = D + !E + !F
Y3 = !D + E + !F
Y4 = !D + !E + F
Z1 = X1 * X2 * X3 * X4 * Y1 * Y2 * Y3 * Y4
OUT = (-Z1 * X1 * X2 * X3 * X4) + (-Z1 * Y1 * Y2 * Y3 * Y4)

So basically I switched to using CNF instead of DNF, and to using the 4-NAND XOR for combining them. This way I can combine AND gates.