I was reading over the following solution for the sum of all natural numbers using the well-ordering principle.
Let Fact 1 be:
$$\forall n:\ (1+2+3+...n)=\frac{n(n+1)}{2}$$
By contradiction, assume that Fact 1 is false. Then, some nonnegative integers serve as counterexamples to it. Let’s collect them in a set:
$$C= \left\{ n \in \mathbb{N} \mid (1+2+3+...n) \neq \frac{n(n+1)}{2} \right\}$$
Assuming there are counterexamples, $C$ is a nonempty set of nonnegative integers. So, by the well-ordering principle, $C$ has a minimum element, which we’ll call $c$. That is, among the nonnegative integers, $c$ is the smallest counterexample to Fact 1.
Since $c$ is the smallest counterexample, we know that Fact 1 is false for $n=c$ but true for all nonnegative integers $n<c$. But Fact 1 is true for $n=0$, so $c>0$. This means $c - 1$ is a nonnegative integer, and since it is less than $c$, Fact 1 is true for $c - 1$. That is,
$$(1+2+3+...c-1)=\frac{c(c-1)}{2}$$
But then, adding $c$ to both sides, we get
$$(1+2+3+...+(c-1)+c)=\frac{c(c-1)}{2} + c = \frac{c(c+1)}{2}$$
which means that (Fact 1) does hold for $c$, after all! This is a contradiction.
I am confused when the proof states "Since $c$ is the smallest counterexample, we know that Fact 1 is false for $n=c$ but true for all nonnegative integers $n<c$." How can we assume this? All we know is that Fact 1 is false when $n=c$, it does not say anything about it being true. I am a beginner when it comes to discrete math so keep that in mind during an explanation. Thanks!