Bash Boolean testing

Your understanding of booleans in shell context is incorrect.

var1=true
var2=false

Both the above variables are true since those are non-empty strings.

You could instead make use of arithmetic context:

$ a=1
$ b=0
$ ((a==1 && b==0)) && echo y
y
$ ((a==0 && b==0)) && echo y
$
$ ((a && !(b))) && echo y;       # This seems to be analogous to what you were attempting
y

The shell does not have Boolean variables, per se. However, there are commands named true and false whose exit statuses are 0 and 1, respectively, and so can be used similarly to Boolean values.

var1=true
var2=false
if $var1 && ! $var2; then var2="something"; fi

The difference is that instead of testing if var1 is set to a true value, you expand it to the name of a command, which runs and succeeds. Likewise, var2 is expanded to a command name which runs and fails, but because it is prefixed with ! the exit status is inverted to indicate success.

(Note that unlike most programming languages, an exit status of 0 indicates success because while most commands have 1 way to succeed, there are many different ways they could fail, so different non-zero values can be assigned different meanings.)

true and false are evaluated as strings ;)

[[ $var ]] is an equivalent of [[ -n $var ]] that check if $var is empty or not.

Then, no need to quote your variables inside [[. See this reminder.

Finally, here is an explication of the difference between && inside brackets and outside.

The closest you can come seems to be use functions instead of variables because you can use their return status in conditionals.

$  var1() { return 0; }
$  var2() { return 1; }  # !0 = failure ~ false

and we can test this way

$  var1 && echo "it's true" || echo "it's false"
it's true
$  var2 && echo "it's true" || echo "it's false"
it's false

or this way

$  if var1; then echo "it's true"; else echo "it's false"; fi
it's true
$  if var2; then echo "it's true"; else echo "it's false"; fi
it's false

Hope this helps.