Express boolean function using NAND only?

Question

How can I express $(\bar{x})y + \bar{z}$ using only NAND?

I know the solution is $((x↑x)↑y)↑z$, but I don't understand how to get to there without expanding into an extremely long answer?

The brute force technique given to me was:

1. eliminate addition (using $\bar{a+b} = \bar{a}\bar{b}$)
2. eliminate multiplication (using $ab = (a↑b)↑(a↑b)$)
3. eliminate complement (using $\bar{a} = a↑a$)

Will show my previous work if needed.

Answer 1

$$p ↑ q\equiv\overline{p}+\overline{q}\equiv \neg p\lor\neg q\tag*{aka Nand}$$

And the fully simplified solution should be:

$$(\overline{x}↑y)↑z$$

Basicly, first try to express the statement in form of $\overline{p}+\overline{q}$, to get this we need De Morgan's law:

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Boolean algebra notation:

\begin{align} \overline{x}y+\overline{z}\tag*{$\text{De Morgan's law}$} &\equiv\overline{\overline{\overline{x}}+\overline{y}}+\overline{z}\\ &\equiv\overline{\overline{x}↑y}+\overline{z}\\ &\equiv(\overline{x}↑y)↑z\\ \end{align}

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PL notation:

\begin{align} (\neg x\land y) \lor \neg z\tag*{$\text{De Morgan's law}$} &\equiv\neg(\neg\neg x\lor \neg y)\lor\neg z\\ &\equiv\neg(\neg x↑y)\lor \neg z\\ &\equiv(\neg x↑y)↑z\\ \end{align}

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First use De Morgan's law backward, so the statement is in form of $\overline{p}+\overline{q},$
and the last two line is just substitute the definition.
Therefore we simplified it with Nand.

Answer 2

Since $x \uparrow y = (xy)'$, we just need to get it in that form, and that's easy for this one:

$$x'y+z'=((x'y)'z)'=(((xx)'y)'z)'=((x\uparrow x)\uparrow y)\uparrow z$$