Welcome to Digital Electronics lecture series. I, Professor Hitesh Dholakia is going to explain you Consensus Theorem in this session. So there were two theorems which was there by consensus.
Here in this session, I will state those theorems and we will prove that by comparing LHS and RHS. So first theorem is, now let us prove this consensus theorem. So here if I consider LHS over here then as per Boolean equations or Boolean rules we don't see anything which we can resolve here. But see here to provide some common data out of this two from this all we can do is we can say over here BC into A plus.
a bar right then what will happen is like see this b a that we can take it common from this and a bar c that we can take it common from it right so here let us simplify this further so a b plus a bar c plus b c a plus b c a bar where now you can see this AB and this AB is common right so we can take common of that two terms so AB common 1 plus C plus here now you can see a bar And c that is common. So we can take common out of those two terms. So a bar c into 1 plus b.
Right. Now see 1 plus c that is 1 and 1 plus b that is even 1 as per Boolean rules. Right. So you can see my that session based on Boolean rules. You will get to know like how to apply all those rules.
Right. So 1 plus C is 1 and 1 plus B is 1. So this is AB plus A bar C. And you can see that is our RHS. Now I'll state second theorem and we will prove it. So you can see it is similar to first theorem only.
Here only we replace plus with dot and dot with plus. Right. But still I'll prove this by having Boolean equations.
So let us consider LHS over here first. Now see in this LHS let us multiply this two. Now here in this bracket we can see a a bar.
So a a bar that is 0. So we need to cancel. So here we can say simplify. Now let us simplify this further by taking this bracket off. So, ACB plus BA bar B plus BCB plus ACC plus BA bar C plus BCC.
Right. Now, here. we can see see this is what b b so that is b this b b that is even b this c c that is even c and this c c that is even c so i need to rewrite this a b c that is this term plus b a bar that is this term plus b c that is this term plus a c that is this term plus a bar b c that is this term plus b c that is this term. Now here you can see this b c plus b c so that is b c only right.
So we don't need to write 2 b c over here. BC plus BC is only single. You can take BC common. So let me simplify this further by removing one BC over here. So ABC plus B bar A plus BC plus AC plus A bar BC.
Now you can see BC is common over here, over here as well as over here. Right. So I can take BC common out of this three term.
So BC in bracket A plus 1 plus A bar plus this BA bar plus AC. So A plus 1 plus A bar. So that is 1. Right.
So we can say this is BC plus BA bar plus AC. Now I don't see any further simplification with this. Right. So what I'm doing is like see I'm just taking this as LHS. Now let us take RHS.
So let us take off this bracket. So A A bar plus AC plus B A bar plus BC. So see this a a bar that we can say that is 0 right.
So we can say this is ac plus b a bar plus b c and you can see LHS and RHS both are equal. You can see LHS and RHS both are equal. So we have proved this consensus second theorem right. So usually by having Boolean algebra rules we can solve any Boolean expression and even consensus theorem we can prove it over here you can see. So in next few sessions I'll explain you some competitive examination questions based on Boolean equations and I'll resolve that by applying Boolean rules.
You can see my previous session based on Boolean rules which will be helpful to you to resolve. examples based on Boolean equations. I hope that you are getting all those things which I am teaching here.
Thank you so much for watching this video. Please do give your valuable suggestions definitely based on your suggestions in future I will make