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I'm looking at some class examples on basic postulates and I can't figure out how the 2nd part is reduced (see below). Could someone explain it to me?



this is the part I do not understand, and could use some explanation using the formulas below.

= 'a('c+c) + a('c+c)

The example uses theses formulas:

[p6a] x*y = y*x
[p8a] x(y+z) = x*y + x*z

and to finish it all off: (which I understand)

= 'a*1 + a*1
= 'a+a
= 1
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You are also using commutativity of addition, right? (x+y=y+x) And associativity, perhaps? ((a+b)+c=a+(b+c)) – Andrés E. Caicedo Feb 9 '11 at 0:08
looking at… those might be [p6b] and [p7b] – Henry Feb 9 '11 at 0:25
@Henry, I agree as well. In the handout's it states p6a and p8a however I have a feeling that may have been a typo. I'll run it by the prof next class – cdnicoll Feb 9 '11 at 5:51
up vote 1 down vote accepted

To be completely explicit: we begin with the formula $$X = a' c' + ac + a'c+ac'$$ (where I have written 'c as $c'$ in order to make the latex easier). We then switch the second and third terms:

$$X = a' c' + a' c + ac + ac'.$$

We can now apply p8a (in "reverse", as it were):

$$X = a' (c' + c) + a (c + c').$$

Finally, we reorder the terms inside the second set of parentheses:

$$X = a' (c' + c) + a (c' + c),$$

and you can take it from here.

Note that this derivation doesn't use p6a (the commutativity of multiplication), but it does use the commutativity of addition.

Edit: As Henry pointed out in the comments above, your class might be calling the commutativity of addition p6b.

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Thank you, this makes sense now! I think what I had confusing me was x+y = y+x, in that I could only move a single variable and not a term. – cdnicoll Feb 9 '11 at 0:28

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