De Morgan's Laws (Set Theory)

Set Difference

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

Relative Complement
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:


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

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


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

When each of $$T_i$$ are subsets of $$S$$:


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

In the context of set complement:


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

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
$$ $$ $$

$$ $$ $$

Generalized Proof
It is necessary only to prove:


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

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

Let the cardinality $$\left|{I}\right|$$ of the indexing set $$I$$ be $$n$$.

Then by the definition of cardinality, it follows that $$I \cong \N^*_n$$ and we can express the propositions:


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

as:


 * $$S \setminus \bigcap_{i = 1}^n T_i = \bigcup_{i = 1}^n \left({S \setminus T_i}\right)$$
 * $$S \setminus \bigcup_{i = 1}^n T_i = \bigcap_{i = 1}^n \left({S \setminus T_i}\right)$$

The proof of these is more amenable to proof by Principle of Mathematical Induction.

First result

 * $$S \setminus \bigcap_{i = 1}^n T_i = \bigcup_{i = 1}^n \left({S \setminus T_i}\right)$$:

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition: $$S \setminus \bigcap_{i = 1}^n T_i = \bigcup_{i = 1}^n \left({S \setminus T_i}\right)$$.


 * $$P(1)$$ is true, as this just says $$S \setminus T_1 = S \setminus T_1$$.


 * $$P(2)$$ is the case $$S \setminus \left({T_1 \cap T_2}\right) = \left({S \setminus T_1}\right) \cup \left({S \setminus T_2}\right)$$ which has been proved. This is our basis for the induction.


 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:


 * $$S \setminus \bigcap_{i = 1}^k T_i = \bigcup_{i = 1}^k \left({S \setminus T_i}\right)$$

Then we need to show:


 * $$S \setminus \bigcap_{i = 1}^{k+1} T_i = \bigcup_{i = 1}^{k+1} \left({S \setminus T_i}\right)$$

This is our induction step:

$$ $$ $$ $$

So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$S \setminus \bigcap_{i = 1}^n T_i = \bigcup_{i = 1}^n \left({S \setminus T_i}\right)$$, i.e. $$S \setminus \bigcap_{i \in I} T_i = \bigcup_{i \in I} \left({S \setminus T_i}\right)$$.

Second result

 * $$S \setminus \bigcup_{i = 1}^n T_i = \bigcap_{i = 1}^n \left({S \setminus T_i}\right)$$:

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition: $$S \setminus \bigcup_{i = 1}^n T_i = \bigcap_{i = 1}^n \left({S \setminus T_i}\right)$$.


 * $$P(1)$$ is true, as this just says $$S \setminus T_1 = S \setminus T_1$$.


 * $$P(2)$$ is the case $$S \setminus \left({T_1 \cup T_2}\right) = \left({S \setminus T_1}\right) \cap \left({S \setminus T_2}\right)$$ which has been proved. This is our basis for the induction.


 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:

$$S \setminus \bigcup_{i = 1}^k T_i = \bigcap_{i = 1}^k \left({S \setminus T_i}\right)$$

Then we need to show:

$$S \setminus \bigcup_{i = 1}^{k+1} T_i = \bigcap_{i = 1}^{k+1} \left({S \setminus T_i}\right)$$

This is our induction step:

$$ $$ $$ $$

So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$S \setminus \bigcup_{i = 1}^n T_i = \bigcap_{i = 1}^n \left({S \setminus T_i}\right)$$, i.e. $$S \setminus \bigcup_{i \in I} T_i = \bigcap_{i \in I} \left({S \setminus T_i}\right)$$.

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