Showing that $x \sim _{0=1} y$ is an equivalence relation

Question

Let $x \sim _{0=1} y \iff x=y \text{ or } x,y\in \{0,1\}$ be a relation on $[0,1]$. I have to show that this is indeed an equivalence relation. I think that's not really hard If my argumentation is correct:

it's reflexive, because $x\sim_{0=1} x \iff x=x \text{ or } x\in \{0,1\}$ is true since only one (either $x=x$ or $x \in \{0,1\}$ has to be true) and $x=x$ is true using the fact that "$=$" is reflexive.

Furthermore, the relation is symmetric, since $x\sim_{0=1} y \implies y \sim_{0=1}$ also because $x=y \implies y=x$.

Lastly, it is transitive, also because $x=y \, \land y=z \implies x=z$.

Answer 1

Symmetric: let $x\sim y$ then $x=y$ or $x,y\in \{0,1\}$. If $x=y$ then $y=x\checkmark$. If $x,y\in\{0,1\}$ then $y,x\in\{0,1\}\checkmark$

Transitive: let $x\sim y$ and $y\sim z$. Then we have four possibilities:

1. If $x=y$ and $y=z$ then $x=z\checkmark$
2. If $x=y$ and $y,z\in\{0,1\}$ then $x,z\in\{0,1\}\checkmark$
3. If $x,y\in\{0,1\}$ and $y=z$ then $x,z\in\{0,1\}\checkmark$
4. If $x,y\in\{0,1\}$ and $y,z\in\{0,1\}$ then $x,z\in\{0,1\}\checkmark$