TITLE: Showing that $t\mapsto t \cup \{t\}$ is injective.
QUESTION [5 upvotes]: Assuming the axiom of regularity of ZFC, we can show several (maybe) useful lemmas:
$\forall x : x\notin x$
$\forall x,y: \neg(x\in y \wedge y \in x)$
Now, I would like to show that the map defined by : $t\mapsto t \cup \{t\}$ is injective.
My go:
Suppose not, that is, $t\cup \{t\} = s \cup\{s\}$, and $t\neq s$. That is, WLOG there is $x$ s.t. $x\in t $ and $x\notin s$, we have:
$$
x\in t \Rightarrow x\in t \cup \{t\} \Rightarrow x \in s\cup \{s\} \Rightarrow x\in \{s\} \Rightarrow x=s
$$
firstly, we used the definition of $\cup$, secondly, we used the assumption, next we used the fact that $x\notin s$.
We have that if $s\neq t$ then $s\in t$ or $t\in s$.
Now, here I am stuck, I should arrive at a contradiction.

REPLY [3 votes]: Suppose $s\neq t$ and $t\cup\{t\}=s\cup\{s\}$.
Then $t\in t\cup\{t\}=s\cup\{s\}$. So $t\in s$ since $t\neq s$.
Now do the same thing but switch the roles of $s$ and $t$ (or use what you've already proved).