That works. Note that I needed the single quotes. Without that, Zsh and Bash will interpret the $ symbol as a shell variable. Or I can do it like this:
The dollar sign marks a shell variable, so you need to escape it. Right now it’s being read as inform6 -r =1, because $TRANSCRIPT_FORMAT is not an existing variable.
The -- syntax above is definitely good to remember (particularly if you even want to call Inform6 from other places, like a Makefile or a script). The amount of quoting you need to get a literal-dollar-sign can be frustrating at times, and the --define skips all that ugliness.
that part i, thankfully, did figure out. i had no idea about the rest, though. is this handling of shell variables a specific quirk of macos? or is this standard behavior and linux does the same?