Cannot prove $T_m(x,y)\leq T_m(x,z)$.

Question

Given $$T_m(x,y)=\min(x,y),$$ for all $x,y\in[0,1]$.

Prove if $y\leq z$ then $T_m(x,y)\leq T_m(x,z)$, for all $x,y,z\in[0,1]$.

Given, \begin{align*} T_m(x,y)&=\min(x,y). \end{align*} For first cases $x\leq y$, \begin{align*} T_m(x,y)&=x. \end{align*} Given that $x\leq y$ and $y\leq z$, so $x\leq z$. We have \begin{align*} T_m(x,y)&\leq z\\ \end{align*} For second cases $x>y$, \begin{align*} T_m(x,y)&=y\\ &\leq z \end{align*}

Now I cannot make form $T_m(x,y)\leq T_m(x,z)$, so how to prove it?

Answer 1

If $x\leq y$ then $T_m(x,y)=x$ and also $T_m(x,z)=x$ since $y\leq z\iff x\leq z$. Hence $T_m(x,y)=T_m(x,z)$ and the condition is satisfied

If $x>y$, we don't know for sure if $x\leq z$ or $x>z$

Supposing $y<x\leq z$, we have $T_m(x,y)=y<x=T_m(x,z)$ so this is satisfied.

Else, supposing $y\leq z<x$, we have $T_m(x,y)=y\leq z=T_m(x,z)$ so this too is satisfied.