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?
eq:
X='a'c+ac+'ac+a'c
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
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.