that result does not apply for EML: EML doesn't have the | . | absolute value function, a prerequisite for Richardson's theorem.
Yes it does; you can build the absolute value as sqrt(x²), and sqrt(x) and x² are both constructible using eml.
If I understand the page correctly, the extension by Miklós Laczkovich should be enough to show that it's undecidable.
Yes it does; you can build the absolute value as sqrt(x²), and sqrt(x) and x² are both constructible using eml.