De Morgan's Laws (Set Theory)

Set Difference
Let $$S, T_1, T_2$$ be sets.

Then:


 * $$S \setminus \left({T_1 \cap T_2}\right) = \left({S \setminus T_1}\right) \cup \left({S \setminus T_2}\right)$$

where:
 * $$S \setminus \left({T_1 \cup T_2}\right) = \left({S \setminus T_1}\right) \cap \left({S \setminus T_2}\right)$$
 * $$T_1 \cap T_2$$ denotes set intersection;
 * $$T_1 \cup T_2$$ denotes set union.

Relative Complement
Further, if $$T_1, T_2$$ are both subsets of $$S$$, we can use the notation of the relative complement:


 * $$\complement_S \left({T_1 \cap T_2}\right) = \complement_S \left({T_1}\right) \cup \complement_S \left({T_2}\right)$$


 * $$\complement_S \left({T_1 \cup T_2}\right) = \complement_S \left({T_1}\right) \cap \complement_S \left({T_2}\right)$$

Set Complement
When $$T_1, T_2$$ are understood to belong to a universe $$\mathbb U$$, the notation of the set complement can be used:


 * $$\overline {T_1 \cap T_2} = \overline T_1 \cup \overline T_2$$


 * $$\overline {T_1 \cup T_2} = \overline T_1 \cap \overline T_2$$

It is arguable that this notation is easier to follow:


 * $$\complement \left({T_1 \cap T_2}\right) = \complement \left({T_1}\right) \cup \complement \left({T_2}\right)$$


 * $$\complement \left({T_1 \cup T_2}\right) = \complement \left({T_1}\right) \cap \complement \left({T_2}\right)$$

General Result
Let $$S$$ and $$T$$ be sets.

Let $$\mathcal P \left({T}\right)$$ be the power set of $$T$$.

Let $$\mathbb T \subseteq \mathcal P \left({T}\right)$$.

Then:
 * $$S \setminus \bigcap \mathbb T = \bigcup_{T' \in \mathbb T} \left({S \setminus T'}\right)$$


 * $$S \setminus \bigcup \mathbb T = \bigcap_{T' \in \mathbb T} \left({S \setminus T'}\right)$$

where:
 * $$\bigcap \mathbb T \ \stackrel {\mathbf {def}} {=\!=} \ \left\{{x: \forall T' \in \mathbb T: x \in T'}\right\}$$

i.e. the intersection of $$\mathbb T$$;


 * $$\bigcup \mathbb T \ \stackrel {\mathbf {def}} {=\!=} \ \left\{{x: \exists T' \in \mathbb T: x \in T'}\right\}$$

i.e. the union of $$\mathbb T$$.

When $$T$$ is a subset of $$S$$:


 * $$\complement_S \left({\bigcap \mathbb T}\right) = \bigcup_{T' \in \mathbb T} \complement_S \left({T'}\right)$$


 * $$\complement_S \left({\bigcup \mathbb T}\right) = \bigcap_{T' \in \mathbb T} \complement_S \left({T'}\right)$$

In the context of set complement:


 * $$\complement \left({\bigcap \mathbb T}\right) = \bigcup_{T' \in \mathbb T} \complement \left({T'}\right)$$


 * $$\complement \left({\bigcup \mathbb T}\right) = \bigcap_{T' \in \mathbb T} \complement \left({T'}\right)$$

Set Difference

 * $$S \setminus \left({T_1 \cap T_2}\right) = \left({S \setminus T_1}\right) \cup \left({S \setminus T_2}\right)$$:

$$ $$ $$ $$ $$

So $$S \setminus \left({T_1 \cap T_2}\right) = \left({S \setminus T_1}\right) \cup \left({S \setminus T_2}\right)$$.


 * $$S \setminus \left({T_1 \cup T_2}\right) = \left({S \setminus T_1}\right) \cap \left({S \setminus T_2}\right)$$:

$$ $$ $$ $$ $$

So $$S \setminus \left({T_1 \cup T_2}\right) = \left({S \setminus T_1}\right) \cap \left({S \setminus T_2}\right)$$.

Relative Complement
Let $$T_1, T_2 \subseteq S$$.

Then:
 * $$T_1 \cap T_2 \subseteq S$$ from Intersection Subset and Subsets Transitive;
 * $$T_1 \cup T_2 \subseteq S$$ from Union Smallest.

So we can talk about $$\complement_S \left({T_1 \cap T_2}\right)$$ and $$\complement_S \left({T_1 \cup T_2}\right)$$.

Hence the following results are defined:

$$ $$ $$

$$ $$ $$

Set Complement
$$ $$ $$

$$ $$ $$

General Proof
It is necessary only to prove:


 * $$S \setminus \bigcap \mathbb T = \bigcup_{T' \in \mathbb T} \left({S \setminus T'}\right)$$


 * $$S \setminus \bigcup \mathbb T = \bigcap_{T' \in \mathbb T} \left({S \setminus T'}\right)$$

... as the others follow directly from the definition of relative complement and set complement.

First result
To prove that:
 * $$S \setminus \bigcap \mathbb T = \bigcup_{T' \in \mathbb T} \left({S \setminus T'}\right)$$:

Suppose:
 * $$x \in S \setminus \bigcap \mathbb T$$

Note that by Set Difference Subset we have that $$x \in S$$ (we need this later).

Then:

$$ $$ $$ $$ $$ $$

Therefore:
 * $$S \setminus \bigcap \mathbb T = \bigcup_{T' \in \mathbb T} \left({S \setminus T'}\right)$$

Second result
To prove that:
 * $$S \setminus \bigcup \mathbb T = \bigcap_{T' \in \mathbb T} \left({S \setminus T'}\right)$$

Suppose:
 * $$x \in S \setminus \bigcup \mathbb T$$

Note that by Set Difference Subset we have that $$x \in S$$ (we need this later).

Then:

$$ $$ $$ $$ $$ $$

Therefore:
 * $$S \setminus \mathbb T = \bigcap_{T' \in \mathbb T} \left({S \setminus T'}\right)$$

Caution: General Case
It is tempting to set up an argument to prove the general case using induction. While this works, and is a perfectly valid demonstration for an elementary student in how such proofs are crafted, such a proof is inadequate as it is valid only when $$\mathbb T$$ is countable.

The proof as given here relies only upon De Morgan's laws as applied to predicate logic. Thus the uncountable case has been reduced to a result in logic.

However, for better or worse, the following is an example of how one might achieve this result using induction.

Proof by Induction
Let $$\mathbb T = \left\{{T_i: i \in I}\right\}$$, where each $$T_i$$ is a set and $$I$$ is some indexing set.

Then:


 * $$S \setminus \bigcap_{i \in I} T_i = \bigcup_{i \in I} \left({S \setminus T_i}\right)$$
 * $$S \setminus \bigcup_{i \in I} T_i = \bigcap_{i \in I} \left({S \setminus T_i}\right)$$

Strictly speaking, these are not the actual laws he devised, but an application of those laws in the context of set theory.

This result is known by some authors, for example, as the duality principle.