Converting a boolean expression to nand gates

Question

Let f(x,y) = x.y + x.ycomplement. I want to convert this into nand gates.
Although this is simply x I am not sure how to convert this into a NAND gate.
which is generally done using DeMorgan's law

Answer 1

Given it is just $x$ as you note, to write it via nand gates: first note that for any $Y$ we have $Y|Y$ is equivalent to the usual "not $Y$." (I'm using in general $U|V$ as the nand gate formed from inputs $U$ and $V$.) Then to get $x$ itself as the result, use that $x$ is equivalent to "not(not x)" so to get $x$ in nand gates it is $(x|x)|(x|x).$