# Intuition for Absorption and Distributive Laws in Elementary Logic

$P ∧ (Q ∨ R) \equiv (P ∧ Q) ∨ (P ∧ R) \tag{Distributive Law 1}$
$P ∨ (Q ∧ R) \equiv (P ∨ Q) ∧ (P ∨ R) \tag{Distributive Law 2}$
$P ∨ (P ∧ Q) \equiv P \tag{Absorption Law 1}$
$P ∧ (P ∨ Q) \equiv P \tag{Absorption Law 2}$

Wihout any reference to truth tables, Venn diagrams (for the equivalent statements with sets), proof, or any formal argument, how could one intuit the four laws above, in the interest of shunning memorisation?

My attempt : Define the statements: $P :=$ I want Pistachios, $Q :=$ I want Quinoa, and $R :=$ I want Radish.

Then LHS of (DL2) can be construed as: Pistachios or (Quinoa and Radish).
I must want Quinoa and Radish both, (inclusive) or Pistachios. How would I proceed?

What about (AL1) though? Then LHS: I want Pistachios or (Pistachios and Quinoa). How is this equivalent to "I want Pistachios (only)"?

Source: P49 of Mathematical Proofs, 2nd ed. by Chartrand et al, P21 of How to Prove It by Velleman

-
Look at it this way. If you want pistachios, then it doesn’t matter whether you want quinoa or not: the lefthand side is true. If you don’t want pistachios, it still doesn’t matter whether you want quinoa or not: the lefthand side is guaranteed to be false. Is that too much like looking at a truth table, or is it sufficiently intuitive? –  Brian M. Scott Sep 4 '13 at 4:09
I suspect you may be throwing yourself off track with the "or all three" part in both of your attempts. For (AL1), there are only 2 terms. –  luser droog Sep 4 '13 at 4:10
@luserdroog: I suspect that the OP meant all three of Pistachios and Pistachios and Quinoa, but I agree that it sounds rather odd. –  Brian M. Scott Sep 4 '13 at 4:13
@BrianM.Scott: Thank you very much. May I ask to which Law you are referring in your first comment? –  LePressentiment Sep 6 '13 at 4:57
@luserdroog: Thanks. I edited my post. –  LePressentiment Sep 6 '13 at 4:58

I hope it's okay if I use different foods in my explanation, where $P :=$ I have a burger, $Q :=$ I have fish, and $R :=$ I have chips. I'll just explain the $\Rightarrow$ direction of each equivalence for now. In each case, keep in mind that "or" in math means inclusive or, so there is no need to ever say the phrase "or both."

• (DL1) If I have a burger with (ketchup or mustard) then either I have a burger with ketchup, or I have a burger with mustard. My burger must have at least one of the two condiments on it.

• (DL2) If I have a burger or (fish and chips) then statements (a) "I have a burger or fish" and
(b) "I have a burger or chips" must be true. To see this, let's break it down by cases.
Case 1: I have a burger. Then (a) and (b) are both true because of the burger, regardless of the fish and chips.
Case 2: I have fish and chips. Again (a) and (b) are both true because of the fish and the chips respectively, regardless of the burger.

• (AL1) If I have a burger or (a burger with cheese) then in any case I must have a burger. (I can't say for certain whether it has cheese on it, though.)

• (AL2) If I have a burger and (a burger or fish) then I must have a burger. (The fish is just a red herring.)

This is not a complete explanation, so if something still doesn't make sense, please ask.

EDIT: Alternatively, for DL2, note that from "I have a burger or (fish and chips)" we can conclude "I have a burger or fish"; this just weakens the second disjunct (fish and chips) by forgetting the about the chips. Formally, $P \vee (Q \wedge R) \implies P \vee Q$. Likewise we can conclude "I have a burger or chips", while forgetting about the fish. So we can conclude the conjunction: "I have a burger and fish, or I have a burger and chips."

-
I hope that you will not mind my middling element. Also, "the fish is just a red herring" gladdened me. –  LePressentiment Sep 6 '13 at 5:31
Thank you for your helpful answer for which I upvoted. I just have a supplementary on (DL2). Purely intuitively, how did you divine brown from the green, before breaking it down by cases and without any regard to the RHS of (DL2): $\color{green}{\text{I have a burger or (fish and chips)}} \equiv \; \color{brown}{\text{[I have a burger or fish] AND [I have a burger or chips]}}$? –  LePressentiment Sep 6 '13 at 5:36
@LePressentiment Well, sometimes the most intuitive way to derive consequences from a disjunction is to derive consequences from each disjunct separately and then see if there is any overlap. This may seem a bit odd, but keep in mind that most theorems aren't stated with a hypothesis that is a disjunction; it would be more usual to break such a proposition into two separate theorems. Anyway, I'll edit my answer to include an alternative derivation. –  Trevor Wilson Sep 6 '13 at 16:09
Many thanks again. I hope that you will not mind my edit; "fish" appeared twice. Why are most hypotheses in theorems disjunctions instead of conjunctions? For example, I'd reckon that it's easier to work with all the disjuncts of Divergence Theorem at once? If the myriad hypotheses were conjunctions, we'd be forced to anatomise case-by-case EACH hypothesis separately, as well as all the hypotheses together (due to the inclusive "or")? –  LePressentiment Sep 8 '13 at 4:58
@LePressentiment Thanks for the edit. What do you mean be the Divergence Theorem? This: en.wikipedia.org/wiki/Divergence_theorem#Mathematical_statement? In any case I don't understand what you are saying. It would make more sense to me if you switched "conjunction" and "disjunction". In any case, I don't see how the hypothesis being a disjunction (e.g. $P \vee Q$) requires splitting into three cases "$P \wedge \neg Q$", "$Q \wedge \neg P$", and "$P\wedge Q$".) Usually one just considers two cases, $P$ and $Q$. There is no need for the case hypothesis to be mutually exclusive. –  Trevor Wilson Sep 24 '13 at 22:55