How do I prove this logic rule with Boolean algebra?

[Mason98]

Homework Statement

Boolean algebra help

Relevant Equations

A.(a+b) = a

Hello, can anyone help how i prove this logic rule? I am not sure whether i have to draw a digital circuit or something. if someone could help me solve it showing the steps they took i'd appreciate thanks

 

[cnh1995]

What do you get when you use the distributive property here?

PS: Forum rules require you to show your attempt at a solution. We can only help by providing hints and pointing mistakes in your work.

 

[Mason98]

What do you get when you use the distributive property here?

PS: Forum rules require you to show your attempt at a solution. We can only help by providing hints and pointing mistakes in your work.

I thought it could possibly be A.B + A.C but I am not sure tbh

 

[cnh1995]

I thought it could possibly be A.B + A.C but I am not sure tbh

Where is C in that expression?

 

[Mason98]

A.A + A.B i mean sorry, thanks for the help, i tried multiplying the brackets out

 

[cnh1995]

A.A + A.B i mean sorry

Right.

You need to use the laws of boolean algebra to simplify this expression. The distributive property was one of these laws. How will you reduce this further?
Can you find anything in your lecture notes?

 

[Mason98]

Right.

You need to use the laws of boolean algebra to simplify this expression. The distributive property was one of these laws. How will you reduce this further?
Can you find anything in your lecture notes?

Would i now use the Idempotent Law, which would change A.A to just A which would leave me with A +A.B and the A.B would change to B.A, so A + B.A ?

 

[cnh1995]

Would i now use the Idempotent Law, which would change A.A to just A which would leave me with A +A.B?

Right.
Then?

 

[cnh1995]

and the A.B would change to B.A, so A + B.A ?

No need to change A.B to B.A.

Hint (if you haven't got the next step yet): A=A.1

 

[Mason98]

No need to change A.B to B.A.

Hint (if you haven't got the next step yet): A=A.1

Hmm thanks for the help by the way appreciate it :), I'm thinking it could be, A.1 + A.B?

 

[Mark44]

Hmm thanks for the help by the way appreciate it :), I'm thinking it could be, A.1 + A.B?

Now factor the expression, using the reverse of the distributive law. What do you get?

 

[Mason98]

Now factor the expression, using the reverse of the distributive law. What do you get?

A.(1+B)?

 

[Mark44]

A.(1+B)?

Right, and how can $1 + B$ be simplified? Remember that + is used for OR, so in terms of sets, this would be $U \cup B$, where U is the universal set.

 

[Mason98]

1 + B is basically 1 or B? So, it can be simplified down to just 1? I'm so confused

 

[Master1022]

1 + B is basically 1 or B? So, it can be simplified down to just 1? I'm so confused

Yes it can. If you think about the logic table for the expression $1 + B$, what is the outcome if $B = 1$? What if $B = 0$? Does it matter what $B$ is? That should help prove to you the identity $X + 1 = 1$ for a boolean variable $X$.

Another method when facing these problems is to use a Karnaugh map. These help when the logical expressions become more complex.

 

[FactChecker]

Do you know how to make a logic table. If so, you should make one and if you have trouble, show your work.