Fundamental Principle of Counting

Theorem
Let $$A$$ be a set.

Let $$\left \langle {B_n} \right \rangle$$ be a sequence of distinct finite subsets of $$A$$ which form a partition of $$A$$.

Let $$p_k = \left|{B_k}\right|$$ for each $$k \in \left[{1 \,. \, . \, n}\right]$$.

Then $$\left|{A}\right| = \sum_{k=1}^n p_k$$.

That is, the sum of the numbers of elements in the subsets of a partition of a set is equal to the total number of elements in the set.

Proof
Let $$r_0 = 0$$, and let:

$$\forall k \in \left[{1 \,. \, . \, n}\right]: r_k = \sum_{j=1}^k {p_j}$$

Then $$r_{k-1} + p_k = r_k$$, so $$r_{k-1} < r_k$$.

Thus by Isomorphism to Closed Interval, $$\left[{r_{k-1} \,. \, . \, r_k}\right]$$ has $$r_k - r_{k-1} = p_k$$ elements.

As a consequence, there exists a bijection $$\sigma_k: B_k \to \left[{r_{k-1} \,. \, . \, r_k}\right]$$ for each $$k \in \left[{1 \,. \, . \, n}\right]$$.

Let $$\sigma: A \to \N$$ be the mapping that satisfies:

$$\forall x \in B_k: \forall k \in \N: \sigma \left({x}\right) = \sigma_k \left({x}\right)$$

By Strictly Increasing Sequence on Ordered Set, $$\left \langle {r_k} \right \rangle_{0 \le k \le n}$$ is a strictly increasing sequence of natural numbers.

Thus by Strictly Increasing Sequence induces Partition, $$\sigma: A \to \left[{1 \,. \, . \, r_n}\right]$$ is a bijection.

By Isomorphism to Closed Interval, $$\left[{1 \,. \, . \, r_n}\right]$$ has $$r_n$$ elements.

Hence the result.