$ perl -MO=Deparse,-p -le"print - --$i"
BEGIN { $/ = "\n"; $\ = "\n"; }
print((-(--$i)));
-e syntax OK
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print(-( ++$i - --$i ))
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While the operand evaluation order for the ops you've used in your example isn't defined, it doesn't matter (unless there's an overflow) for that example. It's just a clarity issue.

$i appears only once, the behaviour is defined

Except it ($Counter) appeared twice.

duh, you're referring to the fact that he's assigning to the variable he just modified. I was still asleep.

While such constructs should be avoided over clarity concerns, it's wrong to say it's undefined. The precedence and evaluation order for that statement is actually clearly defined in perlop.

Quoting perlop:

Note that just as in C, Perl doesn’t define when the variable is
incremented or decremented. You just know it will be done sometime
before or after the value is returned. This also means that modifying 
+a
variable twice in the same statement will lead to undefined behaviour.
Avoid statements like:

    $i = $i ++;
    print ++ $i + $i ++;

Perl will not guarantee what the result of the above statements is.
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To me, that seems that perlop quite explicitly says the behaviour is undefined.

Perhaps you have an other reading of perlop which defines the order?