Principle of Counting

Theorem
Let $$T$$ be a set such that $$T \sim \N_n$$.

Then $$\forall m \in \N: n \ne m \Longrightarrow T \nsim \N_m$$.

Proof
The result follows directly from Equality of Natural Numbers and the fact that set equivalence is transitive.

Comment
Suppose we have a set of objects which we want to count.

That is, we want to know how many objects there are in our set.

What are doing is trying to match up our set with one of the sets $$\N_n$$.

That's what we (effectively) do when we count. We pick a set $$\N_n$$ and compare it in size to see whether it matches the one we're counting. If it doesn't, we pick another one and try again.

In fact, what we do is test them in order. We give each $$n \in \N$$ a name, and we learn what order they go in and which name matches which $$n$$. Then we separate the set we are counting into two subsets: the "ones counted" and the "ones still to be counted". As we move each object one by one from the latter into the former, we recite the name of the element of $$n$$ which follows the previous one we have just counted.

When there are no more elements in the "ones still to be counted" subset, the $$n$$ we have reached is the name of the $$\N_n$$ which is equivalent to the "ones counted" subset.

Note
This is (confusingly) not to be confused with the Fundamental Principle of Counting, otherwise known as the "counting principle".