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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

pensee
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  • Why would you want to do this when the function is just $f(x,y) = x$? I don't see the point. – Stephen Donovan May 01 '21 at 01:14
  • But NAND is a universal function and every function should be expressible in terms of nand gate. So shouldn't there be some of expressing this in terms of NAND gates. But what if we think of this as x.y+x.ycomplement – pensee May 01 '21 at 01:23

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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 $(x|x)|(x|x).$

coffeemath
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