3 ms·
It is undecidable, see https://en.wikipedia.org/wiki/Richardson%27s_theorem https://en.wikipedia.org/wiki/Richardson%27s_theorem
by tyilo 4y ago
It is undecidable, see https://en.wikipedia.org/wiki/Richardson%27s_theorem https://en.wikipedia.org/wiki/Richardson%27s_theorem
- fdej 4y agoNo, this is wrong. Richardson's theorem is about functions, not constants. Equality of constants constructed from exponentials and logarithms is decidable (assuming Schanuel's conjecture) by another theorem (and algorithm!) of Richardson.