$ if   false; then :; else (if [ $? -eq 0 ]; then echo msg; fi); fi
  $ if ! false; then (if [ $? -eq 0 ]; then echo msg; fi); else :; fi
  msg

You should do

if ARG=false

Then use $ARG not $? in the if statement.

I can't think of an example of the if statement being confusing that isn't relying on $?

  error=0
  if ! some_command; then error=$?; bar; fi
  echo "blah"
  return $error

Status=0

Somecommand || status=$?

Then do logic based on status

false the command returns 1, ! the command turns a 0 return code into a 1 and anything other than a zero into a 0, if is vaguely backwards from most languages in that it deals in return codes a rc of 0 is true and anything else is false

in the first case false has a rc of 1 this will execute the else list then the test command 1 is equal to 0(rc = 1 false list) so "msg" does not print

in the second case false(rc 1), !(rc 0), if(true list) now you check if rc is 0, it is and "msg" is printed.

I don't know what kind of shell scripts the designers of this write, but the number of times I've wanted or found it helpful for ! to modify $? is has been exactly zero.

I am almost certain that >99% of people who write shell scripts do not find this intuitive.

I don't think that is true, consider:

  # if  false ; then  echo "true $?" ; else echo "false $?" ; fi
  false 1
  # if ! false ; then  echo "true $?" ; else echo "false $?" ; fi
  true 0
  # if ! ! false ; then  echo "true $?" ; else echo "false $?" ; fi
  false 1

One of the first things I learned when starting to use $? was to assign it to a variable asap if I needed to use it more than once.

Of course

  cond = false
  if cond; then x; else (if cond; then echo msg); fi

  cond = true
  if cond; then (if cond; then echo msg); else x; fi