Let $R$ be a total order on set $S$. Prove that if $S$ has a minimal element, than the minimum element is unique.
I have difficulties with proofs. I know any graph of a total order is a straight line, which clearly has a minimal element. How doI tell when something is mathematically proven?
EDIT: this is how I would prove it
Let $x, y \in S$ such that x and y are minimal. Since S is a total order $x$ and $y$ are comparable and either
case 1: one comes after the other and therefore one isn't the minimal case 2: both are the same and $x=y$
I know this isn't a mathy argument and could use help making it more presentable.
If $S$ is a partially ordered set in which is $s$ is a minimal element, then $\forall x\in S(x\leq s\implies x=s)$.
Since $R$ is a total order and $s$ is minimal, it follows that $s$ is in fact a minimum.
Now to answer your question.
Hint: Suppose $s'\in S$ is also a minimum. Relate it do $s$ and try to conclude that $s=s'$.
What you wrote is good enough for an answer, but you're asking for a 'more symbolic' answer, so here it is.
Full answer:
As stated above, since $R$ is a partial order with a minimum element $s\in S$, then $\forall x\in S(x\leq s\implies x=s)$.
But $R$ is a total order, thus $\forall x\in S(x\leq s\lor s\leq x)$, therefore, since $s$ is minimal $\forall x\in S(x=s\lor s\leq x)$, i.e., $\forall x\in S(s\leq x)$, that is, $s$ is a minimum of $S$ with respect to the order $R$.
Now suppose $s'$ is also a minimum of $S$ with respect to $R$, then, by definition of $s'$ being a minimum, it holds that $s'\leq s$. On the other hand, $s$ is a minimum too, so $s\leq s'$. From $s'\leq s\land s\leq s'$ it follows that $s=s'$. This proves that all minimum elements are the same, that is, there is a unique minimum element.
It's probably good to note that the fact that $R$ is a total order is only necessary to prove that a minimum exists.
On any poset, if there is a minimum, the reasoning above proves it is unique.