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I'm sure this is an easy one but I'm struggling. From my notes, there's this example on how to simplify a proposition using proposition laws:

p $\lor$ (p$\land$ q)
$\equiv$ (p $\land$ t) $\lor$ (p $\land$ q) identity law
$\equiv$ p $\land$ (t $\lor$ q) distribution law
$\equiv$ p $\land$ t identity law
$\equiv$ p identity law

I just can't get my head round how the identity law works. Reading up, I found that:

p $\land$ T ≡ p
p $\lor$ F ≡ p

p $\lor$ T ≡ T domination law
p $\land$ F ≡ F domination law

However, I have been unable to find what T is. Does it stand for Truth and F false? If so, why does p AND T = p? What is the value of p? What is happening in the example then? Please could someone explain this to me I also included the domination law since it uses T and F again which is alien to me).

Also, on the second line, how does p $\land$ (t $\lor$ q) $\equiv$ p $\land$ t. What happens to q? Thanks in advance

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"T" (or $\top$) stands for an arbitrary tautology, while "F" (or $\bot$) stands for an arbitrary contradiction.

So the identity law, $p \wedge T \equiv p$, means that the conjunction of any sentence $p$ with an arbitrary tautology T will always have the same truth value as $p$ (i.e., will be logically equivalent with $p$). For, $p$ is either true or false; if it is true, then $p \wedge T$ will also be true (because both conjuncts are true, then); if it is false, then $p \wedge T$ will also be false (because one of the conjuncts is false). So whatever truth value $p$ has, $p \wedge T$ will have the same.

Similar reasoning applies to the second identity law you mention.

Now, let's look at the first domination law, $p \vee T \equiv T$. It means that the disjunction of any sentence $p$ with an arbitrary tautology $T$ will always be true (will itself be a tautology). This is because if you have a disjunction and know that one of the disjuncts is true, then you don't need to know the truth value of the other disjunct anymore, to know that the whole disjunction is true. For if $p$ were true, then the disjunction $p \vee T$ would also be true; but even if $p$ were false, the disjunction would again be true, because the truth of the tautology $T$ suffices to make the disjunction true.

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  • $\begingroup$ such a great answer! Thank you! $\endgroup$ – user152810 Mar 4 at 0:38
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$\top \wedge p \equiv p\wedge \top \equiv p$ is called the "identity law" because $\top$ acts like an identity in a monoid. Compare this to the similar properties telling you, that: $0 + x = x + 0 = x$, $1\cdot x = x\cdot 1 = x$ or $id_A \circ f = f\circ id_A = f$ (where $f$ is a function $A \to A$).

It's true because $p\wedge q$ is true, if and only if $p$ is true AND $q$ is true. But since $\top$ is automatically true, the truth value of $\top \wedge p$ (or $p\wedge \top$) only depends on $p$, so $\top \wedge p \equiv p\wedge \top \equiv p$.

To understand $p \vee \top \equiv \top$ and $p\wedge \bot \equiv \bot$, comparise this to:

$$\max(x, \infty) = \infty, \min(x, -\infty) = -\infty$$

The "biggest" proposition is $\top$, because it is implied be everything ($p\to \top$). The "smallest" is $\bot$, because from contradiction everything follows ($\bot \to p$). Similarly $\infty$ is bigger then all real numbers ($x\leq \infty$) and $-\infty$ smaller then all real numbers ($-\infty\leq x$). (No, this similarity is NOT a coincedence, look up "bounded lattices"). The actual reason for the laws is this: If $q$ is false, so is $p\wedge q$, so $p\wedge \bot$ is always false. Similarly, if $q$ is true, then so is $p\vee q$, because then $p$ is true OR $q$ is true. So $p\vee \top$ is always true.

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