How do I acquire an intuitive understanding of the distributive law over a disjunction?

Question

The distributive law over a disjunction is given to be: $ P \vee (Q \wedge R) \equiv (P \vee Q) \wedge (P \vee R) $

I want an intuitive understanding of this statement, in order to do I tend to write these logical statements out in words and use concrete examples.

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Statements

$P=$ I will eat chocolate ice cream
$Q=$ I will eat vanilla ice cream
$R=$ I will eat strawberry

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Stating $ P \vee (Q \wedge R)$ in words:

$ P \vee (Q \wedge R)$ = EITHER [I will eat chocolate ice cream] AND/OR [BOTH vanilla and strawberry ice cream].

Thus there are three distinct possibilities that will ensure this statement is true:

1. $P$: I will eat chocolate ice cream
2. $(Q\wedge R)$: I wont eat chocolate ice cream but I will eat BOTH vanilla and strawberry ice cream.
3. $P \wedge (Q \wedge R)$: That you eat all three, i.e. [I will eat chocolate ice cream] AND [BOTH vanilla AND strawberry ice cream].

I believe these have exhausted the possibilities of what makes this statement true.

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Stating $(P \vee Q) \wedge (P \vee R) $ in words:

[EITHER I will eat chocolate OR vanilla ice] AND [EITHER I will eat chocolate or strawberry ice cream]

That was very inelegantly phrased, and from this phrasing I can't see from that why the two statements are equivalent. I would appreciate it greatly if someone could write $(P \vee Q) \wedge (P \vee R) $ in words (using the above statements) conducive to an intuitive understanding, that will allow me to see the parallels to its equivalent statement $ P \vee (Q \wedge R)$.

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EDIT: Truth Table

I seem to have noticed something in the truth table that is confusing me:

\begin{array} \hline P &Q &R &(Q \wedge R) & (P \vee Q) &(P\vee R) &P \vee (Q \wedge R) &(P \vee Q) \wedge (P \vee R)\\ \hline F &F &F &F &F &F &F &F \\ \hline F &F &T &F &F &T &F &F\\ \hline F &T &T &T &T &T &T &T\\ \hline F &T &F &F &T &F &F &F \\ \hline T &F &T &F &T &T &T &T\\ \hline T &F &F &F &T &T &T &T \\ \hline T &T &F &F &T &T &T &T \\ \hline T &T &T &T &T &T &T &T \\ \end{array}

From the truth table it seems that both statements are equivalent because one is true when the other is true. However it seems certain outcomes are only exclusive to one statement. If both statements are equal how can they have different outcomes?

If you're confused here are the 5 instances when both these statements are true and their outcomes:

1. When $P=$False, $Q=$True, $R=$True
   

   • $P \vee (Q \wedge R)$: $(Q \wedge R)$
   • $(P \vee Q) \wedge (P \vee R)$: $(Q \wedge R)$

2. When $P=$True, $Q=$False, $R=$True
   

   • $P \vee (Q \wedge R)$: $P$
   • $(P \vee Q) \wedge (P \vee R)$: $P \wedge (P \wedge R)= P \wedge R$

3. When $P=$True, $Q=$False, $R=$False
   

   • $P \vee (Q \wedge R)$: $P$
   • $(P \vee Q) \wedge (P \vee R)$: $P \wedge P = P$

4. When $P=$True, $Q=$True, $R=$False
   

   • $P \vee (Q \wedge R)$: $P$
   • $(P \vee Q) \wedge (P \vee R)$: $(P \vee Q) \wedge P= P \wedge Q$

5. When $P=$True, $Q=$True, $R=$True
   

   • $P \vee (Q \wedge R)$: $P \wedge Q \wedge R$
   • $(P \vee Q) \wedge (P \vee R)$: $P \wedge Q \wedge R$

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My Questions

1. How do I phrase $(P \vee Q) \wedge (P \vee R) $ as a sentence, that will allow me to see how this statement is similar to $P \vee (Q \wedge R)$?
2. How is $ P \vee (Q \wedge R) \equiv (P \vee Q) \wedge (P \vee R) $ when they both have distinctly different outcomes (as seen above)?

   EDIT: I've just gone through the book and it seems i'm also stumped by the absorption law.

3. How does $P \vee (P \wedge Q) \equiv P$, using my example this is equivalent to saying: Either I will eat chocolate OR/"OR BOTH" [chocolate and vanilla ice cream] is the same as saying I will eat chocolate ice cream.
4. How does $P \wedge (P \vee Q) \equiv P$, using my example this is equivalent to saying: I will eat chocolate ice cream AND [chocolate OR/"OR BOTH" vanilla ice cream] is the same as saying I will eat chocolate ice cream.

Answer 1

I'll either eat chocolate or vanilla and either chocolate or strawberry. This is the same as eating chocolate or both vanilla and strawberry.

Answer 2

As you stated, $(P \vee Q) \wedge (P \vee R) $ =
[EITHER I will eat chocolate OR vanilla] AND [EITHER I will eat chocolate or strawberry]

There are also three distinct possibilities that will ensure this statement is true:

$\Large{1.} \; P$ : If you eat chocolate ice cream both $(P \vee Q)$ and $(P \vee R)$ are true. So the proposition is true.

$\Large{2.} \; Q \wedge R$ : If you don't eat chocolate ice cream, you have to eat both vanilla ice cream (to make $(P \vee Q)$ true) and strawberry ice cream (to make $(P \vee R)$ true) .

$\Large{3.} \; P \wedge (Q \wedge R)$ : If you eat all three flavours, then the proposition is obviously true.

You may remind yourself that when we use and ($\wedge$), we mean the entire proposition is true if and only if both propositions connected by this are true. Propositions or expressed by $\vee$ (or) are not exclusive, i.e. you only need to ensure one of the propositions is true.

Answer 3

Firstly, English (or any other natural language) is not designed for logic and it is often counter-productive to attempt solving logic problems using such a poor tool.

The English word "and" is not the same as ∧, and "or" is definitely not a good stand-in for ∨ (as it doesn't distinguish between inclusive or exclusive). They are only similar. Moreover, the use of 'either' may be problematic in some cases.

But let's go through it anyway.

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P∨(Q∧R)

has two cases, of which only one has to be true

case 1: P (will eat choco), regardless of what else i may eat
case 2: Q∧R (will eat both vanilla and strawberry), regardless of what else i may eat

In English:

[It's true that] Either I will eat chocolate ice cream, or I will have strawberry and vanilla -- or all three

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(P∨Q)∧(P∨R)

has two conditions that must be met (which each have 2 cases)

condition 1: P∨Q - (will eat choco or vanilla)
condition 2: P∨R - (will eat choco or strawberry)

In English:

[It's true that] I will have either chocolate or vanilla ice cream -- or both. [It's] also [true that] I will have either chocolate or strawberry ice cream -- or both.

Both sentences (or conditions) can be met immediately if i eat choco P. So, let's call that case 1.

Now, Let's ignore P and assume i didn't eat choco ~P. How can i still satisfy both conditions?

For condition 1, i would have to eat vanilla Q; and for condition 2, i would have to eat strawberry R. So to get both conditions, I'd have to eat both vanilla and strawberry Q∧R. And we can call that case 2, and we'll also note that it doesn't matter if i additionally ate choco P in this case.

We can summarize these two cases like this:

case 1: P (will eat choco), regardless of what else i may eat
case 2: Q∧R (will eat both vanilla and strawberry), regardless of what else i may eat

But that's the same cases we found for P∨(Q∧R)

Therefore, (P∨Q)∧(P∨R) = P∨(Q∧R).

(note: this is not a rigorous proof, but rather an aid for intuition)

Answers

1. How do I phrase (P∨Q)∧(P∨R) as a sentence? See above.
2. How is P∨(Q∧R)≡(P∨Q)∧(P∨R) when they both have distinctly different outcomes (as seen above)? They don't! You have only partially substituted the truth values into the expressions. Finish the substitutions, and you'll find they are still equal.

3. P∨(P∧Q)≡P: if Q is false, P∧Q is false; so only P can make this true. If Q is true, then P∧Q is only true when P is. So Q is irrelevant.

   I will eat choco or I will eat choco and vanilla.

   So we can say for sure you're going to eat choco P...

4. P∧(P∨Q)≡P: if P is false, the whole statement is false because ∧. if P is true, P∨Q will be true no matter what Q is, and so the whole statement will be true. Q is again irrelevant.

   [It is true that] I will eat choco and [it's also true] I will eat choco and/or vanilla.

   So, we can say for sure, you're going to eat choco P...

Answer 4

2. When P= True, Q= False, R= True

   • P∨(Q∧R) : P
   • (P∨Q)∧(P∨R) : P∧(P∧R)=P∧R

I see what you're doing is setting the false variables to false and leaving the true variables as variables, and then after elimination of the false values claiming the results can't be equal because that would mean $P= P\wedge R$, which isn't generally true

However as, in this specific case, $P=\top$ and $R=\top$, then it is so that $P=P\wedge R$ because $\top=\top\wedge\top$.