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In boolean algebra I found the rules like the redundancy theorem and de morgan's law, a little unintuitive.

Although the truth table shows it all, I wonder if the rules were made by experimenting like this in the first place, and if not, how do I prove such rules without using the truth table?

JRE
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Mahmoud Salah
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  • By using logic. It's really not clear what your question is – there's not really many axioms you need for boolean logic, and all are as intuitive as math gets. – Marcus Müller Apr 04 '20 at 19:07
  • @Marcus How something like the redundancy theorem looks intuitive to you ? , either I am missing something or it's not. – Mahmoud Salah Apr 04 '20 at 20:34
  • Well, intuitivity is something subjective, I'll agree. But the redundancy law is super intuitive: $XY + X\bar Y = X$: It doesn't matter what $Y$ is, if the same term appears in conjunction with both $Y$ and its complement $\bar Y$. Doesn't get much more intutive than that? – Marcus Müller Apr 04 '20 at 20:37
  • It's also a direct consequence of one of the only 3 axioms that boolean algebra has, so the proof is literally a one liner: Because of distributivity, $XY + X\bar Y = X(Y+\bar Y)=X$ – Marcus Müller Apr 04 '20 at 20:39
  • @Marcus What I'm actually facing is something more confusing like this ( XY + X'Z + YZ ) and what the source says is that this last term is redundant which I can't relate to your example – Mahmoud Salah Apr 04 '20 at 20:50
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    It's not redundant. – Marcus Müller Apr 04 '20 at 20:50
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    @MarcusMüller YZ is redundant. It's not obvious, but it's true. YZ=1 with (X,Y,Z )= (0,1,1) and (X,Y,Z) = (1,1,1) But XY+X'Z is already =1 with these 2 combinations, so YZ gives nothing. –  Apr 06 '20 at 07:28
  • @user287001 ah you're right, I should have done a truth table in my head. – Marcus Müller Apr 06 '20 at 07:30
  • @MahmoudSalah Hey, you have got a good suggestion. User analogsystemsrf proposed Venn Diagrams. They visualize 1...4 variable rules with low effort. For ex the redundancy of YZ in XY + X'Z + YZ can be seen easily simply by coloring at first areas XY and X'Z. Try it. –  Apr 06 '20 at 07:45
  • @user287001 Actually it will make much more sense if it's proved this way , but I'm not sure – Mahmoud Salah Apr 08 '20 at 20:28

2 Answers2

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We can find the opposite to be impossible i.e. show that assuming for ex. De Morgan's rule false contains a contradiction.

An example: Let's assume (AB)' in some case to be different than A'+B'. That means there's at least one pair A,B which make the following exclusive or ((AB)' xor (A'+B')) =1. We remember that X xor Y = X'Y+XY'.

Thus with some A,B there must be ((AB)(A'+B') + (AB)'(A'+B')')=1

The left half cannot be 1 because it's ABA'+ABB' which both are zero for every A,B.

Thus (AB)'(A'+B')' must be =1. That means (AB)' and (A'+B')' must both be =1.

(AB)'=1 means that AB=0, so either A or B or both are =0. Then A' or B' or both are =1.

In that case A'+B'=1, (A'+B')'=0. Just a couple of lines above this (A'+B')' was 1. That's the needed contradiction. So De Morgan's rule is true.

I guess the truth table is much more easier way.

Another way is to perform direct algebraic manipulation by catching 1 as a multiplier or zero as added, both not affecting the logic value. An example:

X'+Y= 1(X'+Y)=(X+X')(X'+Y)= XX'+XY+X'X'+X'Y= 0+XY+X'+X'Y= X'+X'Y+XY= X'(1+Y)+XY=X'+XY

We have proved that X'+XY = X'+Y. It's possible that one makes this calculation without anybody asking and really by accident founds a theorem, but as well someone can have asked a proof.

  • I didn't get how this could make something like the redundancy theorem more intuitive – Mahmoud Salah Apr 04 '20 at 20:37
  • This is a proof, it doesn't affect the theorem. A theorem is true or false as it is, no matter does someone write a proof or not. Someone can have so high capacity brain that a theorem is intuitive for him, but not for me or someone else who has as ineffective flesh between his ears. I asked a chess player how he could calculate the result several turns earlier. He said "I saw the situation and the consequences as one indivisible block with all details, I didn't split it to a sequence an check the possibilities in every phase, all was there at the same time" Intuitive sight is unexplained. –  Apr 04 '20 at 20:53
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Since de Morgan did his work at a time when "logic" was in upheaval, we might consider what he used: Venn Diagrams.

analogsystemsrf
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