This statement is quite easy to prove using logic ( and the constant F = " Falsum" = the proposition equivalent to any logical falsehood).

But I cannot manage to prove it simply with the laws of the algebra of sets. I mean, without analysing the statement in terms of an arbitrary x belonging to the sets involved.

Is it simply the definition of inclusion in the algebra of sets? ( So that it would not be possible to prove it? )

By the laws of the algebra of sets, I mean :


  • $\begingroup$ What is "Inter Complement"? $\endgroup$ – zoli Apr 14 at 6:58
  • $\begingroup$ @zoll. I meant " the Intersection of the set A and of the set Complement of B" $\endgroup$ – Eleonore Saint James Apr 14 at 7:29

You are given that $A \subseteq B$, but in your linked laws of algebra for sets, the only principle involving $\subseteq$ is:

$A \subseteq B \Leftrightarrow B' \subseteq A'$

... meaning that there is no rule in this list that allows you to rewrite the $\subseteq$ into any of the set-algebraic operations you really want to work with, i.e. $\cap$, $\cup$, $'$, and $\emptyset$

So no, with the rules as given, you can't derive $A \cap B' = \emptyset$ from $A \subseteq B$

Well, that's no fun!

OK, let's see how we could rewrite the fact that $A \subseteq B$.

Well, one option is to, as you already say, define $A \subseteq B$ as $A \cap B' = \emptyset$ .. but that's no fun either :P

OK, so how about:

$A \cap B = A$

In fact, I really like that one as a definition of $A \subseteq B$: if $A \subseteq B$, then that means that anything in $A$ is automatically in $B$, meaning that intersecting the set with $B$ does not put any further requirements on the objects we're talking about, and so their intersection should indeed just be $A$

So now let's take this piece of information, and see if we can derive $A \cap B' = \emptyset$:


$A \cap B' \overset{A \cap B = A}{=} (A \cap B) \cap B' \overset{Associative}= A \cap (B \cap B') \overset{Complement}= A \cap \emptyset \overset{Annihilation}= \emptyset$

Yay! (though notice: neither Complement nor Annihilation are in your list of rules ... but they really should be! Those are really, really elementary algebraic principles)


Well, you mean $A\subseteq B\Rightarrow A\cap \overline B=\emptyset$.

[Suppose $x\in A\cap \overline B$. Then $x\in A$ and $x\in \overline B$. Thus by hypothesis $x\in B$ and $x\in\overline B$, i.e., $x\in B\cap\overline B$. But $B\cap \overline B = \emptyset$ and so $A\cap \overline B = \emptyset$. Done.]

We have $A\cap \overline B \subseteq B\cap \overline B$ since by hypothesis $A\subseteq B$ and the intersection operation is monotonous.

Moreover $B\cap \overline B = \emptyset$ and so from above $A\cap \overline B \subseteq \emptyset$. Since also $A\cap \overline B\supseteq \emptyset$, we obtain $A\cap \overline B=\emptyset$. Done.

  • 1
    $\begingroup$ The OP asked for the way to prove it without using this method. $\endgroup$ – Graham Kemp Apr 14 at 7:47
  • $\begingroup$ My question was to know whether the proposition could be derived from some ( hypothetically) more primitive formula, without using any set theoretic tool ( such as the membership relation). In other words, whether it could be proved inside the algebra of sets as a self contained theory). $\endgroup$ – Eleonore Saint James Apr 14 at 7:49

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