Anyone good at Boolean algebra?

[Looneybomber]

I have a problem in class that I didn't get very far on.

Just so you know what I'm putting down
x=x
x'= x not.

Here's the problem. I have to prove that the left side equals the right side.

x'z + xyz' + x'y + xy' = y'z + xz' + yz' + x'yz

I eliminated the Y in the 3 variable term on the left side using the distributive property, and I'm sure I can get rid of it on the right, but that gets me no where.

x'z + x'y + x(yz'+y')
x'z + x'y + x[(y'+y)(y'+z')]
x'z + x'y + x(y'+z')
x'z + x'y + xz' + xy' = right hand side.

Not sure where to go from here in order to get closer to the right side.

[4cefed]

I forget, is (x not) = 1 or -x? Whenever I got stuck on these I had to start multiplying both sides by some crazy version of 1. Like (x'+yz')/(x'+yz'). If no one figures this out I'll work on it tonight.

[C-Kwik]

I never formally learned this and had some very limited use of it in an internship last summer. But, if you are just looking to make both sides equal, you can multiply each term that has only 2 variables by the equivalent of 1 (A + A' = 1) where you add in the missing variable (in the case of, x'z, use x'z(y + y')). You'll end up with 7 terms on each side, but a pair of terms on each side will reduce using A + A = A, leaving 6 terms on each side. Assuming I understand this correctly and my math was right, both sides will now be equal. I'll leave out the actual steps so you can try it yourself, but let me know if anything is unclear about what I said.

[Eikon]

but let me know if anything is unclear about what I said.

All of it.

[Looneybomber]

I forget, is (x not) = 1 or -x?

It's binary based, so if x=0 x'=1 | x=1 x'=0

Whenever I got stuck on these I had to start multiplying both sides by some crazy version of 1. Like (x'+yz')/(x'+yz'). If no one figures this out I'll work on it tonight.

I will try that tomorrow at work. Usually I like to combine things like (x+y) = a. I can distribute "a" into the problem and then substitute (x+y) for "a" later, but I don't think my substitution method will work in this case.

Side note: In boolean algebra, we have "and gates" (*) and "or gates" (+). No division symbols or subtraction. Well there's also "exclusive or", but we're not using that.

Ckwik: I will try that too tomorrow.
The teacher wants us to use algebra, but so far all the folks in class that I've talked to couldn't do it and used a truth table instead. He bribed us with candy

[Jesda]

C-Kwik is smarter than the average pony.

[nissangirl74]

C-Kwik is smarter than damn near all of us. That problem is way above my pay grade but I sure would like to know the answer when you figure it out. I know just enough to be interested in how you get it to work. Good luck.

[headhunt3r]

There's no chance that the second last term isn't YX' instead of YZ' right...? Because I got as far as:

X'ZY + Y'Z + XZ' + X'Y = Y'Z + XZ' + YZ' + X'YZ

so the only terms that aren't equal is X'Y on the left and YZ' on the right... It would be stupid if your teach wrote the question wrong.

I can show work too:
X'Z + XYZ' + X'Y + XY' = Y'Z + XZ' + YZ' + X'YZ
X'ZY + X'ZY' + XYZ' + X'ZY + X'YZ' + XZY' + XZ'Y' = right side
1 2 3 4 5 6 7
X'ZY + Y'Z(X'+X) + XZ'(Y+Y') + X'Y(Z+Z') = RS
1 2and6 3and7 4and5
X'ZY = Y'Z = XZ' = YX' = right side

which leaves me the YX' = YZ' to solve, which seems funny

Edit: gah my spacing got destroyed. I wonder if there's a code tag I can use.

[Looneybomber]

There's no chance that the second last term isn't YX' instead of YZ' right...? Because I got as far as:

X'ZY + Y'Z + XZ' + X'Y = Y'Z + XZ' + YZ' + X'YZ

so the only terms that aren't equal is X'Y on the left and YZ' on the right... It would be stupid if your teach wrote the question wrong.

I can show work too:
X'Z + XYZ' + X'Y + XY' = Y'Z + XZ' + YZ' + X'YZ
X'ZY + X'ZY' + XYZ' + X'ZY + X'YZ' + XZY' + XZ'Y' = right side

You missed a NOT symbol. Putting terms in alphabetical order I will copy yours and then put mine right below it.
X'ZY + X'ZY' + XYZ' + X'ZY + X'YZ' + XZY' + XZ'Y' = right side
x'yz + x'y'z + xyz' + x'yz' + xy'z + xy'z' = RHS.

[Looneybomber]

you can multiply each term that has only 2 variables by the equivalent of 1 (A + A' = 1) where you add in the missing variable (in the case of, x'z, use x'z(y + y')). You'll end up with 7 terms on each side, but a pair of terms on each side will reduce using A + A = A, leaving 6 terms on each side.

I tried this and I end up with,
y'z + xz' + x'y = y'z + xz' + x'y

Looks like I will get candy in class. Ckwik, if you pay shipping I'll let you have it

Well that was 7a, time to do 7b, 7c.

[PoorManQ45]

Easy enough, the answer is in between the left and right side.

x'z + xyz' + x'y + xy' = y'z + xz' + yz' + x'yz

[gwoods]

oh man I loved this stuff!!! But sadly I took it 9 years and 3 kids ago and can't remember anything. Are ya going to school for computer programming?

[PoorManQ45]

oh man I loved this stuff!!! But sadly I took it 9 years and 3 kids ago and can't remember anything. Are ya going to school for computer programming?

Why would you love this stuff?

It is not used anywhere.

When you have a formula you plug in the real values that you are given or have figured out. There is no where that xyz = zyx is the answer to a real question.

[headhunt3r]

When you're designing a simple circuit you can't give a formula "real values" and solve for the answer. You need to simplify your design to use as few gates as possible, and that's where boolean algebra comes in.

These exercises are so you get comfortable doing that.

[gwoods]

LOL at you man, Boolean Algebra was part of Abstract Algebra I & II and part of the BS in CS degree at UMKC when I attended. You may not ever use the math but you will use this type of logic/thinking in programming and engineering. If your going into either your gonna need it on your job.

I HATE math in general but for some reason loved Abstract Algebra...

[szh]

Why would you love this stuff?

It is not used anywhere.

Um, not correct. Boolean logic is used in programming, circuit design, and a number of other similar fields.

Basically, if you want to have a good engineering career, then this is fundamental knowledge to learn.

Z

[C-Kwik]

Looks like I will get candy in class. Ckwik, if you pay shipping I'll let you have it

No worries. I have pretzel M&M's here so I'm golden.

Um, not correct. Boolean logic is used in programming, circuit design, and a number of other similar fields.

Basically, if you want to have a good engineering career, then this is fundamental knowledge to learn.

Z

I saw it in code for a balancing robot (like a segway). I have a feeling I may see some of this in controls.

[Looneybomber]

oh man I loved this stuff!!! But sadly I took it 9 years and 3 kids ago and can't remember anything. Are ya going to school for computer programming?

Shoot, I'm 29 and have 2 kids...and NOW I decide to finish my BSEE. However, you gotta know some CS stuff if going into EE.

Looks like I will get candy in class. Ckwik, if you pay shipping I'll let you have it

No worries. I have pretzel M&M's here so I'm golden.

I didn't even get candy. What a gyp.