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I'm trying to prove that $(xyz)' = x'+y'+z'$ using theorems/axioms.

I tried this but I just want to make sure if its the correct route or if I've done something "illegal"/wrong.

(xyz)' = [(xy)z]' by associativity 
       = [(x*y)'+z'] by DeMorgan's Law
       = [(x'+y') + z'] by DeMorgan's Law
       = [(x'+z')+(y'+z')] by Distribution
       = x'+y'+z' by simplifying redundant z' terms.

Is this the correct method?

user1766888
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    How did you apply distribution? You already had what you wanted to prove... you just needed to drop parentheses. – Doug Spoonwood Jun 28 '13 at 04:00
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    It looks OK, but I guess it would be good if you had included the set of axioms/theorems that you are allowed to use (so that we can check according to them). For example, one can write $(xyz)' = x'+y'+z'$ (by DeMorgan) and that would be perfectly fine for course XYZ, but maynot be OK for course ABC. Regardless, I highly doubt that you have an axiom that says "you may simplify the redundant $z$ terms," so you may want to formalize that step. – Lord Soth Jun 28 '13 at 04:00
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    Doug is correct, don't apply distribution just apply associativity again. – A.E Jun 28 '13 at 04:01
  • Ok. But is distribution of the z' a valid step, albeit unnecessary? – user1766888 Jun 28 '13 at 04:02
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    I am hesitant to say no outright, but typically you would only see distribution applied in the case a+(bc)=(a+b)(a+c) or a(b+c)=ab+ac... That "distributive" step you did is more confusing than anything else. – A.E Jun 28 '13 at 04:06

1 Answers1

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I wouldn't say you did anything "illegal", but distribution is usually used in the following manner:

$\quad (x + y)z = xz + yz\quad $ or $\quad(xy)+ z = (x+z)(y+ z)$

or the "flip side"

$\quad x(y+z)=xy + xz \quad $ or $\quad x +(yz)=(x+ y)(x+z)$

Your work was done, essentially, when you after your second application of DeMorgan's. Then, we simply use associativity again:

$$[(x'+y') + z'] = x' + y' + z'\tag{by associativity}$$

amWhy
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