De Morgan's Laws (Set Theory)

Set Difference

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

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


 * $$\mathcal{C}_S \left({T_1 \cap T_2}\right) = \mathcal{C}_S \left({T_1}\right) \cup \mathcal{C}_S \left({T_2}\right)$$
 * $$\mathcal{C}_S \left({T_1 \cup T_2}\right) = \mathcal{C}_S \left({T_1}\right) \cap \mathcal{C}_S \left({T_2}\right)$$

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:


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

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:


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

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


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

In the context of set complement:


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

Set Difference

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

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


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

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

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


 * $$\mathcal{C}_S \left({T_1 \cap T_2}\right) = \mathcal{C}_S \left({T_1}\right) \cup \mathcal{C}_S \left({T_2}\right)$$:


 * $$\mathcal{C}_S \left({T_1 \cup T_2}\right) = \mathcal{C}_S \left({T_1}\right) \cap \mathcal{C}_S \left({T_2}\right)$$:

Set Complement

 * $$\mathcal{C} \left({T_1 \cap T_2}\right) = \mathcal{C} \left({T_1}\right) \cup \mathcal{C} \left({T_2}\right)$$:


 * $$\mathcal{C} \left({T_1 \cup T_2}\right) = \mathcal{C} \left({T_1}\right) \cap \mathcal{C} \left({T_2}\right)$$:

Generalized Proof
It is necessary only to prove:


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

... 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 $$I \cong \mathbb{N}_n$$ and we can express the propositions:


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

as:


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

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


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

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


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


 * $$P(2)$$ is the case $$S - \left({T_1 \cap T_2}\right) = \left({S - T_1}\right) \cup \left({S - T_2}\right)$$ 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:

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

Then we need to show:

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

This is our induction step:

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

Therefore $$\overline {\bigcap_{i = 1}^n T_i} = \bigcup_{i = 1}^n \overline {T_i}$$, i.e. $$\overline {\bigcap_{i \in I} T_i} = \bigcup_{i \in I} \overline {T_i}$$.

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


 * $$P(1)$$ is true, as this just says $$\overline {T_1} = \overline {T_1}$$.


 * $$P(2)$$ is the case $$\overline {T_1 \cup T_2} = \overline {T_1} \cap \overline {T_2}$$ 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:

$$\overline {\bigcup_{i = 1}^k T_i} = \bigcap_{i = 1}^k \overline {T_i}$$

Then we need to show:

$$\overline {\bigcup_{i = 1}^{k+1} T_i} = \bigcap_{i = 1}^{k+1} \overline {T_i}$$

This is our induction step:

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

Therefore $$\overline {\bigcup_{i = 1}^n T_i} = \bigcap_{i = 1}^n \overline {T_i}$$, i.e. $$\overline {\bigcup_{i \in I} T_i} = \bigcap_{i \in I} \overline {T_i}$$.

QED