Boolean Laws, and Theorem

What's up wonder community of math wizards. I was hoping you could check a problem for me. For every step I need to pick which numbered law from the list was used. HERE IS THE LIST.

Given =$\displaystyle \overline{(A\overline{BC}+\overline{AC})(\overline {A+B)}}$

19 ===$\displaystyle \overline{A\overline{BC}+\overline{AC}}+\overline{ \overline{A}+\overline{B}}$

20 ===$\displaystyle \overline{A\overline{BC}+\overline{AC}}+\overline{ \overline{AB}}$

9 ====$\displaystyle \overline{A\overline{BC}+\overline{AC}}+AB$

20 ===$\displaystyle \overline{(A\overline{BC}})\overline{(\overline{AC )}}+AB$

9 ====$\displaystyle \overline{(A\overline{BC}})(AC)+AB$

19 ===$\displaystyle (\overline {A}+\overline{\overline{BC}})(AC)+AB$

9 ====$\displaystyle (\overline{A}+BC)(AC)+AB$

15 ===$\displaystyle \overline {A}AC+ABCC+AB$

7 ====$\displaystyle 0\bullet C+ABCC+AB$

3 ====$\displaystyle 0+ABCC+AB$

1 ====$\displaystyle ABCC+AB$

5 ====$\displaystyle ABC+AB$

18 ===$\displaystyle AB$

I think I might of got it right. #15 I'm not feeling good about.

Re: Boolean Laws, and Theorem

Quote:

Originally Posted by

**Jarod_C** What's up wonder community of math wizards. I was hoping you could check a problem for me. For every step I need to pick which numbered law from the list was used.

HERE IS THE LIST.
Given =$\displaystyle \overline{(A\overline{BC}+\overline{AC})(\overline {A+B)}}$

<snip>

9 ====$\displaystyle (\overline{A}+BC)(AC)+AB$

15 ===$\displaystyle \overline {A}AC+ABCC+AB$

this should be $\displaystyle \overline {A}AC+BCAC+AB$, yes they are the same but you have to use one of your laws to get there.

$\displaystyle \overline {A}AC+BCAC+AB$ --> $\displaystyle \overline {A}AC+BACC+AB$ --> $\displaystyle \overline {A}AC+ABCC+AB$ both by 10

Quote:

7 ====$\displaystyle 0\bullet C+ABCC+AB$

3 ====$\displaystyle 0+ABCC+AB$

1 ====$\displaystyle ABCC+AB$

5 ====$\displaystyle ABC+AB$

18 === $\displaystyle AB$

you sort of skipped a couple steps here.

I'd say $\displaystyle ABC+AB$ --> $\displaystyle AB(C+1)$ by 14, --> $\displaystyle AB1$ by 2, --> $\displaystyle AB$ by 4

Re: Boolean Laws, and Theorem

That's how it was on my paper, and I had to list the steps. I didn't really even notice that, it's good you said something. For a school paper you'd figure they would put it together better, now I can see why some of it was hard to make sense of. I'm going to use what you pointed out for food for thought. Thanks.

Re: Boolean Laws, and Theorem

I've been looking at it now for a little bit, and I'm really glad you brought the missing steps up. It's going to help me in class. In digital it always made sense to me when looking at the gates, but when I get to math, and I don't get to see it first hand it's a little hard to put together. Like the missing step in $\displaystyle ABC+AB=AB$ In the circuit you can see it, but on just paper it needs to be explained like you showed. Just had to say thanks again.

Re: Boolean Laws, and Theorem

Re: Boolean Laws, and Theorem

I really liked this concept and also the logic that was used to solve this problem. Nice..!!

----------

TOMS Shoes Australia Online - Hype DC