Is $p \lor (q \land r)\equiv p \land (q \lor r)$?

Question

Before writing, I'm not good at English. As I developed the law of distribution, I felt strange.

$$\begin{align} p \land (q \lor r)& \equiv (p \land q) \lor (p \land r)\\ &\equiv X \lor (p\land r)\tag{$X=p \land q$}\\ &\equiv (X \lor p) \land (X \lor r)\\ &\equiv [(p \land q) \lor p] \land [(p \land q) \lor r]\\ &\equiv (p \lor p) \land (p \lor q) \land (p \lor r) \land (q \lor r)\\ &\equiv p \land (q \lor r) \land (p \lor q) \land (p \lor r)\\ &\equiv [p \land (q \lor r)] \land p \lor (q \land r)\end{align} $$

When called $p \land (q \lor r)=A,\ p \lor (q \land r)=B.$ I have never heard of $A≡A \land B$. (When they are not the same) So I developed $p \lor (q \land r)$

$$\begin{align} p \lor (q \land r) &\equiv (p \lor q) \land (p \lor r)\\ &\equiv Y \land (p \lor r)\ \ \ \ (let\ Y=p \lor q)\\ &\equiv (Y \land p) \lor (Y \land r)\\ &\equiv [(p \lor q) \land p] \lor [(p \lor q) \land r]\\ &\equiv (p \land p) \lor (p \land q) \lor (p \land r) \lor (q \land r)\\ &\equiv [p \lor (q \land r)] \lor [p \land (q \lor r)]\end{align} $$

$A\equiv A \land B,\ B\equiv A \lor B$

I don't know if they are equal or if I made a mistake.

Answer 1

$p \vee (q \wedge r)$ and $p \wedge (q \vee r)$ are not equivalent. For instance, let $p$, $q$, $r$ be the statements $$ \begin{aligned} p\colon \quad \text{$0 < 1\phantom{.}$}\\ q\colon \quad \text{$1 < 0\phantom{.}$}\\ r\colon \quad \text{$2 < 0.$}\\ \end{aligned} $$ Since $q$ and $r$ are false, both $q \wedge r$ and $q \vee r$ are false.

But then since $p$ is true, $p \vee (q \wedge r)$ is true while $p \wedge (q \vee r)$ is false.

Answer 2

At first you make a mistake in third rows from down to up.

Then it's true that in general $A \equiv A \land B$ isn't true for independent $A,B$.

But in your case they are dependent in value!

Answer 3

Consider when $p$ is true but both $q$ and $r$ are false. Here is a link to a useful online tool.

Answer 4

$p$ or ($q$ and $r$) is obviously not equivalent to $p$ and ($q$ or $r$).

$p = FALSE; q=TRUE;r=TRUE$ will make $p$ or ($q$ and $r$)= TRUE but $p$ and ($q$ or $r$)=FALSE. (This is clear as [$p$ AND M] requires $p$ to be true but [$p$ OR K]$ does not.

When yous said $A = p$ or ($q$ and $r$)$ and $B = p$ and ($q$ or $r$). The $A$ and $B$ are NOT independent.

Notice that $B\implies A$. It is not possible for $B$ to be true and $A$ to be false. But for all other cases $A$ and ($A$ and $B$) have the same truth values.

If you have $B\implies A$ then, yes, $A$ and ($A$ and $B$) have the same truth values.

If you discover that $[A] \equiv [A] \land [B]$, you can not conclude $[A] \equiv [B]$. What you can conlclude is that $[B] \implies [A]$

So, no, $p\lor (q \land r) \not \equiv p\land(q\lor r)$

But $p\land(q\lor r) \implies p\lor (q \land r)$

Answer 5

This kind of thing can be easier to understand if you produce a real-world example. On paper, $p\lor(q\land r)$ is just abstract symbols. But when you translate it into something practical, a different part of your brain can work on it in a different way.

Our brains are good at thinking about social rules. So let's imagine that you live in a place where a person is allowed to drink alcohol in a restaurant if:

• they are over the age of 18 ($p$), OR
• their parent is present ($q$) AND their parent gives permission ($r$).

This rule is $p\lor(q\land r)$ and seems plausible.

Now let's see if $p\land(q\lor r)$ is the same thing. This new rule says that a person is allowed to drink alcohol in a restaurant if

1. they are over the age of 18 ($p$) AND
2. their parent is present ($q$) OR has given permission ($r$)

This means that even a 60-year-old customer must get permission from their 80-year-old parent. Or, if they can't get permission, they have to bring their parent into the restaurant.

Perhaps it is clear that, while the original rule was reasonable, this second rule is absurd. If so, it should be clear that the two rules can't mean the same thing.