Simplifying Boolean expression?? Help

[VrWagz1]

Is anyone here any good with boolean expressions. Im stuck on this one (A+B).(B+C). It looks like a fairly simply but i just dont seem to be getting it, I have the answer and even the workings but i just dont understand how the workings worked out:S
I fairly new to this so if you could explain a little that would be good

Equation : Z= (A+B).(B+C)
Step 1 : = A.B+A.C+B.B+B.C (this bit i got no worries)
Step 2 : = A.C+B(A+1+C) (How the shizzle did they get this and i cant see where the 1 came from either, if i used boolean laws and identities the only way i can get a 1 is from A+'A and there is no inverted numbers in this equasion at all)

Answer =A.C+B (is the answer which i think i get but if you could explain how to properly get that from step 2 would be great.)

[PaRaDoX]

are you doing this in relation to logic "OR" gates? if you are i think your going about it the wrong way, you need to draw out the circit and work it out as you go. its been a few years since i did digital maths but ill try and help you if i can

[michaelw]

Think of it this way. What needs to be ture for Z to be true.

Z= (A+B).(B+C)

OK so if A is true, either B or C needs to be true for Z to be true.
If B is True, then who cares it is going to be true what ever.
C needs either B or A to be true.

So expand the brackets
Step 1 : = A.B+A.C+B.B+B.C (this bit i got no worries)

Now

Group together

Z = A.C + B.(A+B+C)

No if B is true in the second statement It will also be true in the brackets so you can put in a 1.
Remember B=B.1

So
Z = A.C + B.(A+1+C)

If B is true then who cares about A or C so just drop them

Z = A.C + B

Remember with logic always look for what you can ignore.
Hope this helps.

[VrWagz1]

That is exactly what i needed michaelw thankyou.

In the mean time i have found a law that sort of allows you to cheat. Its the same law that you use to expand in the first step. You use that same law in reverse and its gives you the answer instantly.

the law of Distribution says
A+B.C = (A+B).(A+C)
so in reverse
(A+B).(A+C) = A+B.C

Therefore in our equasion
(A+B).(B.C) = (B+A).(B+C) (Law of commutation)
Answer = B+A.C (Law Of Distribution)

Have learned alot in the last 12 hours but i still got plenty to go, its just starting to click a bit. I think once i start to get a proper handle on this it will come very easy.