In general, one solution is to just keep the original inputs around as necessary. So instead of x -> x mod 4, take the function x -> (x, x mod 4); the inverse i

In general, one solution is to just keep the original inputs around as necessary. So instead of x -> x mod 4, take the function x -> (x, x mod 4); the inverse is just (x, y) -> x. The actual % operator in this language is defined as (x, y) -> (y, x / y, x % y), acting as a divmod. The + and - operators are defined as (x, y) -> (x + y, y) and (x, y) -> (x - y, y). And so on, making sure enough inputs are left on the stack to reconstruct the others.