Product of Cardinals is Associative

Theorem
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.

Then:
 * $\mathbf a \paren {\mathbf {b c} } = \paren {\mathbf {a b} } \mathbf c$

where $\mathbf {a b}$ denotes the product of $\mathbf a$ and $\mathbf b$.

Proof
Let $\mathbf a = \card A$, $\mathbf b = \card B$ and $\mathbf c = \card C$ for some sets $A$, $B$ and $C$.

By definition of product of cardinals:


 * $\mathbf a \paren {\mathbf {b c} }$ is the cardinal associated with $A \times \paren {B \times C}$.

Consider the mapping $f: A \times \paren {B \times C} \to \paren {A \times B} \times C$ defined as:
 * $\forall a \in A, b \in B, c \in C: \map f {a, \tuple {b, c} } = \tuple {\tuple {a, b}, c}$

Let $a_1, a_2 \in A, b_1, b_2 \in B, c_1, c_2 \in C$ such that:
 * $\map f {a_1, \tuple {b_1, c_1} } = \map f {a_2, \tuple {b_2, c_2} }$

Then:

Thus $f$ is an injection.

Thus $f$ is a surjection.

So $f$ is both an injection and a surjection, and so by definition a bijection.

Thus a bijection has been established between $A \times \tuple {B \times C}$ and $\tuple {A \times B} \times C$.

It follows by definition that $A \times \tuple {B \times C}$ and $\tuple {A \times B} \times C$ are equivalent.

The result follows by definition of cardinal.