Show Trans$(x)\rightarrow$ Trans$(S(x))$ 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?

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