Simplest way to calculate the intersect area of two rectangles

I have a problem where I have TWO NON-rotated rectangles (given as two point tuples {x1 x2 y1 y2}) and I like to calculate their intersect area. I have seen more general answers to this question, e.g. more rectangles or even rotated ones, and I was wondering whether there is a much simpler solution as I only have two non-rotated rectangles.

What I imagine should be achievable is an algorithm that only uses addition, subtraction and multiplication, possibly abs() as well. What certainly should not be used are min/max, equal, greater/smaller and so on, which would make the question obsolete.

Thank you!

EDIT 2: okay, it's become too easy using min/max or abs(). Can somebody show or disprove the case only using add/sub/mul?

EDIT: let's relax it a little bit, only conditional expressions (e.g. if, case) are prohibited!

PS: I have been thinking about it for a half hour, without success, maybe I am now too old for this :)

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You seem to hope to get an answer using only the field operations ($+$, $-$, $\times$, and $/$). But can you get a formula for the distance between two points on the real line using only those tools? Failing to do so should persuade you that your hope is forlorn. –  Lubin Jan 16 '12 at 20:06
I thought because the area of a square would be calculated as (a-b)*(c-d), a negative area would mean that the area is rotating in the other direction (if you use vector). But I think the question is ill-posed anyways. Sorry for wasting time.... –  chaiy Jan 17 '12 at 13:20

You can also reduce min to max here (or the opposite), i.e. $min\{a,b\} = -max\{-a,-b\}$.