I need to express these with $\downarrow$
$x+ y + z$
This one I think I can do, I guess at it and copy the wikipedia page since my book has no explanation on how to do this I get
$(x \downarrow y \downarrow z) \downarrow (x \downarrow y \downarrow z) $
which makes sense I think when I run through it.
$(x + z)y$
copying wikipedia is much more difficult now since they don't work with three variables I don't know what to do on this one and there is no example in my book.
$x$
This one should be easy x seems right, no need for a nor but I am guessng this is wrong.
The last one is hard to type in latex but it is x times the complement of y
$\;x\overline{y}$
I have no idea how to do this.
To express any given boolean formula with $\def\nor{\downarrow}$$\nor$, we will try to express $\bar{}$ (not), $+$ (or) and $\cdot$ (and) with it. You can use these for your two given formulas. Let's start with not. As $x \nor y$ is $\overline{x+y}$, we can - using that $x+x \equiv x$ - rewrite not as $$ \overline x \equiv \overline{x+x} \equiv x \nor x $$ Now for or, we have, as not is idempotent, that $$ x+y \equiv \overline{\overline{x+y}} \equiv \overline{x\nor y} $$ As we already know that $\overline{p} \equiv p \nor p$ for a formula $p$, we get $$ x+y \equiv \overline{x\nor y} \equiv (x \nor y) \nor (x\nor y) $$ To express and, recall that $xy \equiv \overline{\bar x + \bar y}$, so (using again what we know for not) $$ xy \equiv \overline{\bar x +\bar y} \equiv \bar x \nor \bar y \equiv (x\nor x)\nor (y \nor y)$$ Using these we have \begin{align*} (x+z)y &\equiv\bigl((x \nor z) \nor (x\nor z)\bigr) y\\ &\equiv \Bigl(\bigl(x \nor z) \nor (x\nor z)\bigr) \nor \bigl(x \nor z) \nor (x\nor z)\bigr)\Bigr)\nor (y \nor y) \end{align*} This will work, but produces lengthy formulas. You can of course also follow the ideas we used to express the basic connectives, giving \begin{align*} (x+z)y &\equiv \overline{\overline{x+z} + \bar y }\\ &\equiv (x\nor z) \nor(y\nor y) \end{align*}