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I'm trying to prove that XYZ + XYZ' + XY'Z + X'YZ = XY+ XZ + YZ

I have gotten so far i think. I don't know if I am on the right track but i keep getting stuck when i end up with this xy + xy’z + x’yz....

here is what i have done:

F(X,Y,Z) = XYZ + XYZ'+XY'Z + X'YZ

XYZ + XYZ’ + ZXY’ + XYX’ (Commutative)

XY(Z) + XY(Z’) + ZXY’ + ZYX’ (associative)

XY(Z+Z’) + ZXY’ + ZYX’ (distributive)

XY(1) + ZXY’ + ZYX’ (inverse)

XY + ZXY’ + ZYX’ (identity)

and as you can see... i get stuck here... cause i need to have XY + ZX + ZY but i can't get rid of the two last negatives..

tiinkx
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    I'm voting to close this question as off-topic because it is about mathematics (Boolean algebra), not programming. – Pang Apr 07 '15 at 01:18

1 Answers1

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I suggest:

XYZ + XYZ' + XY'Z + X'YZ
= (XYZ + XYZ + XYZ) + XYZ' + XY'Z + X'YZ
= (XYZ + XYZ') + (XYZ + XY'Z) + (XYZ + X'YZ)
= XY(Z + Z') + X(Y + Y')Z + (X + X')YZ
= XY + XZ + YZ

ruakh
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  • Nice and sweet. Informally it is also quite obvious: "Either all three are correct, or two are correct and one false" is the same as "At least two of these three must be true". – Amadan Apr 06 '15 at 01:10
  • how did you get that first line??? the rest makes perfect sense and thanks so much, but I'm at a loss with how i can add two more xyz? i just want to know for future so i can learn where i was going wrong. – tiinkx Apr 06 '15 at 02:42
  • @tiinkx: For any *foo*, *foo* = *foo* + *foo*. Or, if you prefer, *foo* = *foo* · 1 = *foo* · (1 + 1) = *foo* · 1 + *foo* · 1 = *foo* + *foo*. – ruakh Apr 06 '15 at 02:46
  • so is that like a law?? @ruakh – tiinkx Apr 06 '15 at 04:09
  • @tiinkx: I'm not sure "law" is the right word, but -- yes. + is defined such that 0 + 0 = 0 and 1 + 1 = 1. Since 0 and 1 are the only two possible values, we have that *foo* + *foo* = *foo* for both possible values. – ruakh Apr 06 '15 at 14:36
  • As ruakh said, it is formally defined to be so; but informally you can see for yourself it works. "A + B" in logic means "A or B". If you have "A", then you also have "A + A + A" because... see here: if you know "sun is orange", then you also know that "sun is orange, or sun is orange, or sun is orange". Makes sense? – Amadan Apr 09 '15 at 04:36