Why does the (( compound command assign to a variable in this case?

$ test=false
$ echo $test
false
$ ((!test))
((test=false))
$ echo $?
1
$ echo $test
0

Why is test now 0? I dont understand where or why (( is assigning to test or why it prints ((test=false)). This only happens when writing the commands manually on the terminal; when running those same commands as a script test stays on "false".

Firstly, test and false are also the names of builtins which aren't relevant here, so for the sake of clarity, I'm going to change them to foo and bar

$ foo=bar

$ echo $foo
bar

!foo is history expansion, and the matching command is foo=bar.

$ echo !foo
echo foo=bar
foo=bar

$ ((!foo))
((foo=bar))

$ echo $?
1

• The extra line below the command there is a preview of the command that's being run. I'm not sure what the technical term for that is.
• History expansion is turned off in scripts.

In an arithmetic context, the string bar is treated as a variable name, but that variable isn't defined, so it's evaluated as 0. Hence foo=0.

$ echo $foo
0