Steps to reproduce
Enable strict line breaks and enter the following
> [!check]- Proof
> For the identity property, it is clear that $f(a \circ e) = f(a) = f(a) • f(e)$ for any $a \in G$, hence $f(e) = e$.
> For the latter property, notice that for any $a \in G$ it follows $f(a \circ a^{-1}) = f(a) • f(a^{-1}) = f(e) = e$,
> so $f(a^{-1}) = f(a)^{-1}$.
Did you follow the troubleshooting guide? Y
Expected result
In reading mode, the callout should display no differently to if there were no line breaks within the callout.
Actual result
Environment
SYSTEM INFO:
Obsidian version: v1.6.3
Installer version: v1.5.8
Operating system: Darwin Kernel Version 23.5.0: Wed May 1 20:09:52 PDT 2024; root:xnu-10063.121.3~5/RELEASE_X86_64 23.5.0
Login status: not logged in
Insider build toggle: off
Live preview: on
Base theme: adapt to system
Community theme: none
Snippets enabled: 0
Restricted mode: on
RECOMMENDATIONS:
none
Additional information
As far as I can tell, this only stopped working properly in a recent update.