> an expression in the ring generated by the integers, x, sin xn, and sin(x sin xn)

We can always write AML trees for expressions generated by the integers, x, sin xn, and sin(x sin xn), right?

So we should be able to write EML trees for any two such expressions, A and B. If they're equal everywhere, then A - B = 0 everywhere. A - B is also in the aforementioned ring.

If there was a decision procedure always to determine if EML trees represent the same function, then that contradicts Miklós Laczkovich's extension, right?

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DoctorOetker 2 days ago | parent [-]

no Miklós Laczkovich's extension as described on wikipedia only says that both of the following questions are proven undecidable:

1) is there some value x such that some function F(x)=A(x)-B(x)=0?

2) is there some value x such that F(x)>0?

while you asked:

> I'm pretty sure it's not decidable if two EML trees describe the same function.

that would be

3) is for every x F(x)=A(x)-B(x)==0?

which Miklós Laczkovich's extension does not provide.

And you ignore the fact that Miklós Laczkovich's extension applies to real numbers and functions...

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vintermann a day ago | parent [-]

If it's undecidable whether it's 0 at even ONE point, clearly you can't prove that it's 0 everywhere.

Likewise, if it's not decidable for real-valued functions, clearly it's not decidable for complex valued functions.

	▲	DoctorOetker a day ago \| parent [-]
		thats not how this works decidability does not distribute over pointwise question asking on sets, or if you believe it does, show us the proof. Telling if an EML(x,y),1 constructed expression is identically 0 is in the gray zone, as far as I can tell, it has neither been proven decidable nor been proven undecidable. Nevertheless regardless of decidability the authors clearly show the multipoint sampling/testing is a decent filter, and the shorter resulting expressions have been proven correct in the results for the construction at least.