Definition of XOR, and how to remember its expansion in disjunctive normal form for N variables

As you know, you use the upper part of the structure (from Vout to Vcc) to set the 1's and the lower part of the structure (from Vout to Vss) to set the 0's.

You can do this directly from the function expression. For example, let say your function is F=A·B*+C. I've chosen a different (and shorter function) for a simpler illustration of the method.

The upper part is easy (A and B* in series for AND) and C in parallel (for the OR part). The variables enter negated into the PMOS gate, because the PMOS will turn ON for 0 V (and not 5 V).

^{simulate this circuit – Schematic created using CircuitLab}

Then the lower part must implement the 0's. Recalling the De Morgan Theorem, by negating F, AND becomes OR and viceversa. Applying to our function: we'll get F*=(A*+B)·C*

Therefore, in the lower part we have A* and B that are in parallel while C is in series. The variables doesn't enter negated into the NMOS gates because NMOS turn ON for 5V.

This method can be applied also to your original function, it just needs much more space!

Actally, it is very simple:

f = (y XOR z) XOR x

^{simulate this circuit – Schematic created using CircuitLab}

This just by looking at the thruth table and realzing that the bottom half is the inverse of thetop half.

Here's the math using boolean calculus.

We know x ^ y = x'y + xy' (^ is the XOR operator). Then:

g = a'b'c + abc + ab'c' + a'bc'
g = (a'b' + ab)c + (ab' + a'b)c'
     ^using a for X and b' for Y
g = (a ^ b')c + (a ^ b)c'

Now we have (a ^ b') and (a ^ b), which may seem not be possible to match, unless you realise that due to the nature of the XOR, (a ^ b') = (a ^ b)' [I'll leave that as an exercise to the reader]. Thus

g = (a ^ b)'c + (a ^ b)c'
g = (a ^ b) ^ c

This is only possible due to the special nature of the original equation. If you need an AND or OR you will find it impossible to reduce the function like this.