( 3 >= 3 && !(true || true) )
according to precedence, brackets, not, and, or, evaluates in the following order…
Since they enclose the entire expression they are moot and can be removed.
3 >= 3 && !(true || true)
means evaluating TRUE OR TRUE which we know yields,
3 >= 3 && !true
which leaves us with,
3 >= 3 && false
Owing to short-circuiting, we can immediately conclude the expression is false, regardless the truthiness of the first operand. Technically, the
>= operator does have precedence over AND so the operand would be evaluated first, yielding
true; however, this is moot.
So we have an outcome of
false Now it comes to you to set
tricky to that value and follow through with the remaining instruction.