Two total orders Let $\mathcal{R}$  and $\mathcal{S}$ be two total orders on $E$ such that $\forall(x,y) \in E\times{E}, x\mathcal{R}y \Rightarrow x\mathcal{S}y$. Show that $\forall(x,y) \in E\times{E}, x\mathcal{R}y\Leftrightarrow x\mathcal{S}y$.
I need some help with this problem. I know that a total order is a set where every pair of elements in a relation such that anti-symmetry and transitivity holds. However I am having a problem from transferring from a relation $\mathcal{R}$ to a relation $\mathcal{S}$. Could you please give me some hints?

We are going to use the converse of 
$$x\mathcal Ry \Rightarrow x\mathcal S y$$
which is
$$\neg (x\mathcal Ry) \Rightarrow \neg(x\mathcal S y).$$
The order is total, so if we have $\neg (x\mathcal Ry)$ we have $y\mathcal R x$.
By the first property it gives us
$$y\mathcal S x.$$
By anti-symetry we can conclude that $x=y$ or $\neg (x\mathcal Sy)$.
But $x=y$ can not be because $\neg(x\mathcal Ry)$ (and you always have $x\mathcal R x$).
So $\neg (x\mathcal Sy)$.
Finally, 
$$x\mathcal Ry \iff x\mathcal S y.$$
