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Unsure of case (2) in my proof:

Question: Assume Trans$(x)$, show Trans$(S(x))$

Definition: Trans$(x)\leftrightarrow_{df}\forall p\in x(p\subseteq x)$

Question symbollically: Show [$y\in x\rightarrow y\subseteq x$]$\rightarrow$ [$y\in x\cup\{x\}\rightarrow y\subseteq x\cup\{x\}$]

We have two cases for $y\in x\cup\{x\}$, either $y\in x$ or $y\in \{x\}$

(1) If $y\in x$, then immediately from definition of Trans$(x)$, $y\subseteq x\subseteq x\cup\{x\}$

(2) If $y\in\{x\}$, then we have $x\in\{x\}$ since the only element of $\{x\}$ is $x$ therefore $x=y$, therefore in this case $y=x\subseteq x\cup\{x\}$

Unsure if I can say $y=x$ in part (2) of this proof and if this is sufficient to show case (2).

Does $y\in\{x\}$ imply $y\subseteq x\cup\{x\}$? How?

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Yes, it's fine to say that $x=y$ in the second case.

If $y\in\{x\}$ then $y=x$, because $\{x\}$ is a shorthand for the unique set $U$ satisfying $\forall z(z\in U\leftrightarrow z=x)$.

And therefore $y=x\subseteq x\cup\{x\}$.

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