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19 votes
Accepted

Is eta-equality provable in Coq?

Yes, this is provable—indeed, it is provable by reflexivity. This is because Coq has eta-equality of functions built into its notion of definitional equality. ...
sarahzrf's user avatar
  • 306
10 votes

Is eta-equality provable in Coq?

As for your follow-up question: Coq implements η-conversion, but not η-reduction/expansion, because η really is treated at the level of the conversion test and not at that of reduction (which is a sub-...
Meven Lennon-Bertrand's user avatar
5 votes
Accepted

State of art for observational equality, on the injectivity requirement

Is the injectivity of 𝑇𝑇𝑜𝑏𝑠/𝐶𝐶𝑜𝑏𝑠 also anti-classical? It isn't. The setoid model in the TTobs paper supports LEM (meaning that every setoid is empty or nonempty). This note explains that ...
András Kovács's user avatar
5 votes

Has extensionality ever caused any problems in a mathematical proof?

Kind of fun to note that extensionality is inconsistent with the internal Church Thesis which states that every function $\mathbb{N}\rightarrow\mathbb{N}$ is realized by some Turing machine. In turn ...
cody's user avatar
  • 394
4 votes
Accepted

Has extensionality ever caused any problems in a mathematical proof?

Function extensionality is “safe“ in the sense that it is valid in traditional constructive mathematics, for example as practiced by Erret Bishop, as well as in any topos. It is also valid in homotopy ...
Andrej Bauer's user avatar
  • 9,812
3 votes
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Why did this proof succeed without function extensionality?

To expand on @NaïmFavier's answer: The key here is the difference between definitional and propositional equality. Most times, when people talk about "function extensionality" they mean ...
Joey Eremondi's user avatar
2 votes
Accepted

Provable `fun-ext`and "rewriting under binders" in intensional MLTT

The new rule seems to be equivalent to funext plus the definitional equality that says that funext of constant refl is refl. As ...
András Kovács's user avatar
1 vote

Why did this proof succeed without function extensionality?

You don't need function extensionality to prove that, only the $\eta$-rule for functions: then both sides are equal to $\lambda x y.\ f x y$. I don't know if that's how Coq does it.
Naïm Favier's user avatar

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