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I know this was answered before but I'm having one particular problem on the proof that I'm not getting.

My Understanding of the distribution law on the absorption law is making me nuts, by the answers of the proof it should be like this.

A∨(A∧B)=(A∧T)∨(A∧B)=A∧(T∨B)=A∧T=A

This should prove the Absoption Law but on the Step (A ^ (T v B)), I'm not getting how they get to it.

If (A ^ T) v (A ^ B) will be distributed by my understand of this, the following is the answer (A v A) ^ (A v B) ^ (T v A) ^ (T v B) That we can go to A ^ (A v B) ^ T I'm getting lost on something here, because it looks to me we will enter a loop on it as: A ^ T ^ (A v B) will be distributed again and I will go back to A ^ B if I distribute with A but if I distribute with T it will be B ^ T, nothing usefull also is it?

Can anyone help on this? Thanks in advance.

Bram28
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DGabardo
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  • As noted below, distribution is a law of equivalence; it can go both ways. $$A\wedge(\top\vee B)~\iff~(A\wedge \top)\vee(A\wedge B)\$$ So of course $$(A\wedge \top)\vee(A\wedge B)~\iff~A\wedge(\top\vee B)$$ You should know that in common algebra, by "distributing out the common factor" we have: $$a\cdot x+a\cdot y=a\cdot(x+y)$$ This is the same principle; we just distribute out the common conjunct. – Graham Kemp Oct 29 '19 at 22:45

2 Answers2

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Please note that Distribution is an equivalence!

That is, Distribution says:

$P\lor (Q \land R) \Leftrightarrow (P \lor Q) \land (P \lor R)$

but since this is an equivalence, we can just as well say that:

$(P \lor Q) \land (P \lor R) \Leftrightarrow P \lor (Q \land R)$

In other words, we can go from left to right, but we can also go from right to left.

Now, I realize that going from $(P \lor Q) \land (P \lor R)$ to $P \lor (Q \land R)$ doesn't feel like 'Distribution' ... it feels like that should be called 'Reverse Distribution' or 'Factoring' ... but it is nevertheless a correct application of Distribution.

So, this is what is going on with your $(A \land \top) \lor (A \land B)$: when told to use Distribution, you were only able to see this as going in the 'normal' left-to-right' direction, but they key was to recognize that this expression can also be seen as the result of distributing $A$ over $(\top \lor B)$. That is, applying 'Reverse Distribution', we get $A \land (\top \lor B)$

Bram28
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  • I see, thank you very much for this, when you told me about the Reverse Distribution I could understand way better, I haven't noticed that. Thanks a lot. – DGabardo Oct 31 '19 at 13:32
  • @DGabardo You're welcome! This is a very common thing to miss, and it's really because of the name. Glad I could help. :) – Bram28 Oct 31 '19 at 13:51
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The distributive law is being applied "in reverse." This becomes more obvious if you start with $A \wedge (\top \vee B)$ and then apply the distributive law as you are familiar with doing, from-left-to-right.

$A \wedge (\top \vee B) \Leftrightarrow (A \wedge \top) \vee (A \wedge B)$

Since the two statements are logically equivalent, we can convert back and forth, from left-to-right or from right-to-left, as desired. Hence,

$A \vee (A \wedge B)$

$\Leftrightarrow (A \wedge \top) \vee (A \wedge B)$ by the identity law

$\Leftrightarrow A \wedge (\top \vee B)$ by the distributive law

$\Leftrightarrow A \wedge \top $ by the domination law

$\Leftrightarrow A $ by the identity law

RyRy the Fly Guy
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