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$P ∧ (Q ∨ R) \equiv (P ∧ Q) ∨ (P ∧ R) \tag{1. Distributivity of ∧ over ∨}$
$P ∨ (Q ∧ R) \equiv (P ∨ Q) ∧ (P ∨ R) \tag{2. Distributivity of ∨ over ∧}$
$P ∨ (P ∧ Q) \equiv P \tag{3. ∨ Absorbs ∧}$
$P ∧ (P ∨ Q) \equiv P \tag{4. ∧ Absorbs ∨}$

What are the intuitions for the four Rules of Replacement above? I pursue only intuition;
so please do not rely on or use Truth Tables, Venn Diagrams, or any formal proofs.
This question also enquires about Rule 2 above.

Sources: p 49, Mathematical Proofs, 2nd ed. by Chartrand et al, $\qquad$ p 21, How to Prove It by Velleman.
I apprehended the names of the Absorption Laws herefrom and the Distribution Laws herefrom.

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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
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@LePressentiment, check my questions, it seems i'm in the same predicament that you once were. – seeker Sep 15 '15 at 14:55
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@seeker +1. Thank you; I would not have known about your relevant questions had you not posted here. For posterity's benefit, see math.stackexchange.com/search?q=user%3A45635+[logic]. – LePressentiment Dec 31 '15 at 2:28
up vote 4 down vote accepted

I abbreviate foods as follows 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."

$\huge{1.}$ 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.

$\huge{2.}$ If I have a burger or (fish and chips) then statements
(a) "I have a burger or fish" $\qquad \qquad $ and $\qquad \qquad$ (b) "I have a burger or chips"
must be true. To see this, let's break $(2)$ 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.

Alternatively, 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."

$\huge{3.}$ 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.)

$\huge{4.}$ 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.

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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
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@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

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