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I have to reduce this boolean algebra :

((x⊕y)⊕(y⊕z))⊕(x⊕(1⊕((y⊕0)⊕z)))

I found out that

  1. ((x⊕y)⊕(y⊕z)) = x⊕z
  2. y⊕0 = y
  3. 1⊕(y⊕z) = negative(y⊕z)

looking at the solution it must be reducable to negative(y), but i don't know how to get from (x⊕z)⊕(x⊕negative(y⊕z)) to negative(y)

epoyraz
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1 Answers1

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Next, note that we have $x\oplus \lnot(y\oplus z) = \lnot(x\oplus(y\oplus z))$. In fact, this $\lnot$ can be moved all the way to the front: $$ \lnot\big((x\oplus z)\oplus(x\oplus(y\oplus z))\big) $$ Lastly, note that $\oplus$ is both associative and commutative, so we can take away the parentheses and rearrange the terms to get $$ \lnot (x\oplus x \oplus y \oplus z \oplus z) = \lnot (0\oplus y \oplus 0) = \lnot y $$

Arthur
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  • Wow, how did you get from the question that you can take the negation out like that? And how did you know that the operator is associative and commutative? – Bram28 Oct 23 '16 at 20:38
  • Thanks. I got it ! The note was really helpful :) – epoyraz Oct 23 '16 at 20:40
  • Huh! I still don't get it ... – Bram28 Oct 23 '16 at 20:47
  • @Bram28 $\oplus$ is a common notation for the XOR operation. I assumed that that's what it is, since also $y \oplus 0 = y$ and $1 \oplus (y \oplus z) = \lnot (y\oplus z)$. – Arthur Oct 23 '16 at 22:40
  • @Arthur Oh! OK, I thought it was some given mystery operator. Got it, thanks! – Bram28 Oct 24 '16 at 03:07