That expression,

```
( 3 >= 3 && !(true || true) )
```

according to precedence, brackets, not, and, or, evaluates in the following order…

```
outer brackets
```

Since they enclose the entire expression they are moot and can be removed.

```
3 >= 3 && !(true || true)
inner brackets
```

means evaluating TRUE OR TRUE which we know yields, `true`

.

```
3 >= 3 && !true
!true
```

which leaves us with,

```
3 >= 3 && false
and
```

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.