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I wrote out Xor operation as $(\lnot x \land y) \lor (x \land \lnot y)$ but don't know how to simplify from here. I don't know if De Morgan's laws will just make things more complicated, and I don't know how to factor anything out when there are NOTS on everything.

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    $\begingroup$ There really isn't any way to "simplifiy" this expression. What do you want to accomplish? $\endgroup$ – naslundx Aug 5 '16 at 8:18
  • $\begingroup$ How do I know when I am done simplifying? $\endgroup$ – Sean Hill Aug 5 '16 at 8:19
  • $\begingroup$ Experience, like most other things. :) Keep practicing! $\endgroup$ – naslundx Aug 5 '16 at 8:25
  • $\begingroup$ If you look up Karnaugh maps you'll see that, at least in terms of AND and OR operations, there are no adjacencies to be simplified. fourier.eng.hmc.edu/e85_old/homework/xor_xnor_4.gif $\endgroup$ – shawnt00 Aug 7 '16 at 3:20
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You can't simplify this expression with boolean algebra as xor is it's own logical connective often denoted with $\oplus$. Thus what you wrote down could be written $x\oplus y$

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What would a simplification look like to you? There aren't much simpler ways to write the exclusive disjunction in terms of $\neg, \land, \lor$. You could write $\neg(x \leftrightarrow y)$, but it's not much cleaner, as usually $x \leftrightarrow y$ is just an abbreviation for $(x \to y) \land (y \to x)$.

(Personally, I find $(x \lor y) \land \neg (x \land y)$ the clearest -- that's how I would describe the exclusive disjunction in words.)

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