# Set Law

(Redirected from Set Axiom)

A Set Law a Constraint in a Set System.

## References

### 2019

Commutative property:
• $\displaystyle{ A \cup B = B \cup A }$
• $\displaystyle{ A \cap B = B \cap A }$
Associative property:
• $\displaystyle{ (A \cup B) \cup C = A \cup (B \cup C) }$
• $\displaystyle{ (A \cap B) \cap C = A \cap (B \cap C) }$
Distributive property:
• $\displaystyle{ A \cup (B \cap C) = (A \cup B) \cap (A \cup C) }$
• $\displaystyle{ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) }$
The union and intersection of sets may be seen as analogous to the addition and multiplication of numbers. Like addition and multiplication, the operations of union and intersection are commutative and associative, and intersection distributes over union. However, unlike addition and multiplication, union also distributes over intersection.

Two additional pairs of properties involve the special sets called the empty set Ø and the universe set $\displaystyle{ U }$ ; together with the complement operator ($\displaystyle{ A^C }$ denotes the complement of $\displaystyle{ A }$ . This can also be written as $\displaystyle{ A' }$ , read as A prime). The empty set has no members, and the universe set has all possible members (in a particular context).

Identity :
• $\displaystyle{ A \cup \varnothing = A }$
• $\displaystyle{ A \cap U = A }$
Complement :
• $\displaystyle{ A \cup A^C = U }$
• $\displaystyle{ A \cap A^C = \varnothing }$
The identity expressions (together with the commutative expressions) say that, just like 0 and 1 for addition and multiplication, Ø and U are the identity elements for union and intersection, respectively.

Unlike addition and multiplication, union and intersection do not have inverse elements. However the complement laws give the fundamental properties of the somewhat inverse-like unary operation of set complementation.

The preceding five pairs of formulae—the commutative, associative, distributive, identity and complement formulae—encompass all of set algebra, in the sense that every valid proposition in the algebra of sets can be derived from them.

Note that if the complement formulae are weakened to the rule $\displaystyle{ (A^C)^C = A }$, then this is exactly the algebra of propositional linear logic.