Automatically create a formula from the output of Reduce function

Question

Suppose an output of Reduce[] function looks like this:

x2 > 0 && ((0 < x1 < x2 && y2 > 0 && 0 < y1 < y2 && z1 > 0 && z2 > z1 && t1 > c1 && 0 < t2 < c2) || (x1 > x2 && y2 > 0 && y1 > y2 && z1 > 0 && z2 > z1 && t1 > 0 && t2 > c2))

I.e., Boolean formulae with inequalities of two types:

1) $x>a$, to which we associate $\left(\frac{1}{x-a}\right)$,

2) $a<x<b$, to which we associate $\left(\frac{1}{x-a}-\frac{1}{x-b}\right)$.

I would like to automatically create from the above output the following formula, which preserves all Boolean logic and for which && means multiplied by and || means add the corresponding piece:

$\frac{1}{x_2}\Big(\big((\frac{1}{x_1}-\frac{1}{x_1-x_2})\frac{1}{y_2}(\frac{1}{y_1}-\frac{1}{y_1-y_2})\frac{1}{z_1}\frac{1}{z_2-z_1}\frac{1}{t_1-c_1}(\frac{1}{t_2}-\frac{1}{t_2-c_2})+\frac{1}{x_1-x_2}\frac{1}{y_2}\frac{1}{y_1-y_2}\frac{1}{z_1}\frac{1}{z_2-z_1}\frac{1}{t_1}\frac{1}{t_2-c_2}\big)\Big)$

Is there an easy way to do it, given that the number of parameters $x,y,z,t,\dots$, as well as the number of nested parentheses and boolean symbols ( smthng && ( smthng || smthng ) && ( smthng && smthng && smthng ))) can be anything?

UPDATE: Also, sometimes the output contains an equality like this one:

x2==x1-a

buried somewhere inside of the parentheses:

x2 > 0 && ((x2==x1-a || 0 < x1 < x2 && y2 > 0 && 0 < y1 < y2 && z1 > 0 && z2 > z1 && t1 > c1 && 0 < t2 < c2) || (x1 > x2 && y2 > 0 && y1 > y2 && z1 > 0 && z2 > z1 && t1 > 0 && t2 > c2))

which I would like to completely ignore and not to include in the final formula. What replacement rule should I use then?

Answer 1

Use some replacement rules. We make a function,

toBooleanFunction[expr_] := Simplify@expr //.
  {a_ > b_ :> 1/(a - b), a_ < x_ < b_ :> 1/(x - a) - 1/(x - b), a_ < b_ :> -1/(a - b), a_ > x_ > b_ :> -1/(x - a) + 1/(x - b), _Equal :> (## &[])} //. 
  {And -> Times, Or -> Plus}

Then:

Reduce[{x1 - x2 > a, a > 0, x1 > 0, x2 > 0, x3 == 0}, {x1, x2, x3}]
% // toBooleanFunction
(* -(1/(a (-a + x1) x2 (a - x1 + x2))) *)