# Commutative

## Definition 2

Let $$(S, \circ)$$ be an algebraic structure. Then $$\circ$$ is commutative on $$S$$ if and only if

$$\forall x,y \in S : x \circ y = y \circ x$$
THEOREM

Let $$(S, \circ)$$ be an algebraic structure. Then $$\circ$$ is not commutative on $$S$$ if and only if

$$\exists x,y \in S : x \circ y \ne y \circ x$$

Proof available

THEOREM

Conjunction is commutative.

$$P \wedge Q \dashv\vdash Q \wedge P$$

Proof available

THEOREM

Natural number multiplication is commutative.

$$\forall x,y \in \mathbb{N} : x * y = y * x$$

Proof available

THEOREM

Disjunction is commutative.

$$P \vee Q \dashv\vdash Q \vee P$$

Proof available

THEOREM

Union is commutative.

$$A \cup B = B \cup A$$

Proof available

THEOREM

$$\forall x,y \in \mathbb{N} : x + y = y + x$$

Proof available

THEOREM

$$\forall x,y \in \mathbb{C} : x + y = y + x$$

Proof available

THEOREM

The biconditional operation is commutative.

$$P \Leftrightarrow Q \dashv\vdash Q \Leftrightarrow P$$

Proof available

THEOREM

Intersection is commutative.

$$A \cap B = B \cap A$$

Proof available

THEOREM

$$\forall x,y \in \mathbb{Z} : x + y = y + x$$