# How to prove “ If (A is included in B) then (A Intersection Complement of B is equal to the null set) ” using only set algebra laws?

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 :

https://www.math-only-math.com/laws-of-algebra-of-sets.html

• What is "Inter Complement"? – zoli Apr 14 at 6:58
• @zoll. I meant " the Intersection of the set A and of the set Complement of B" – 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 $$A \cap B' = \emptyset$$:

Well:

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

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.

• The OP asked for the way to prove it without using this method. – Graham Kemp Apr 14 at 7:47
• 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). – Eleonore Saint James Apr 14 at 7:49