Boolean algebra - distribution

[Rectifier]

The problem
I am trying to show that $a'c' \vee c'd \vee ab'd$ is equivalent to $(a \vee c')(b' \vee c')(a' \vee d)$

The attempt
$(a \vee c')(b' \vee c')(a' \vee d) \\ (c' \vee (ab'))(a' \vee d)$
The following step is the step I am unsure about. I am distributing the left parenthesis over the right.
$(c' \vee (ab'))(a' \vee d) \\ c'a \vee c'd \vee ab'a' \vee ab'd$

I am almost there but the term $ab'a'$ differs. I suspect that you can remove that term since it does not afect the output because it is always false. What do you say about that?

 

[micromass]

Can you show $ab'a' = 0$?

 

[Rectifier]

Unfortunately, I cant, but I know it is true for all values of 1 and 0 for a and b. Since $a$ is the opposite of $a'$ this means that $a \wedge a'$ will always be 0 and it does not matter which value b has . I can write a truth table but I am on my phone so its a bit hard.

 

[micromass]

What are your axioms for Boolean algebra?

 

[Rectifier]

Are you thinking about $x \cdot x' = 0$ ?

 

[micromass]

I don't know, how did you define a boolean algebra?

 

[Rectifier]

In boolean algebra variables can only have two values.

 

[micromass]

Either you tell me how you defined a boolean algebra, or I'm out of this thread.

 

[Rectifier]

I am sorry that my answer didn't fit you but I can't come up with anything better. In any case, I am thankful for the help I have received so far.

 

[micromass]

How are you supposed to prove anything about a Boolean algebra if you're not given its definition or anything??

 

[Rectifier]

Well, let's suppose that we use the following axioms(?) then:
https://wikimedia.org/api/rest_v1/media/math/render/svg/f3b7cce1f6dba36569346955c3c1e27d31522d06

 

[micromass]

Those are axioms for a structure that is called a bounded lattice. It's not the axioms for a Boolean algebra. Usually, there are 10 axioms for a Boolean algebra. They can be found here: https://en.wikipedia.org/wiki/Boolean_algebra_(structure)#Definition

Note that in this case, it is an axiom that $x\wedge x' = 0$. This is why I asked for your definition, since you might adopt different (but equivalent) ways of doing things.

 

[Rectifier]

I didnt think they were different :). Thank you for you help kind stranger!

So now I can basically remove that ab'a'. Halleluja, I am done with that problen.