The compatibility of function extensionality with impredicative Set in Coq is a topic where the consistency status is not definitively settled. According to the Coq FAQ:

"Function extensionality is consistent with predicative Set (the default in Coq). It is not known to be inconsistent with impredicative Set (when the option ImpredicativeSet is enabled), but it is also not known to be c...

Short answer: Yes, as far as is known, function extensionality is compatible with making Set impredicative in Coq. There is no known inconsistency between the two, and there are semantic models in which both hold.

A bit more detail:

• Function extensionality (FunExt) is strictly weaker than univalence. While impredicative Set is known to be incompatible with univalence, there is no such obstructi...

In fact nobody knows of a counter‐model, but equally no one has given a proof that adding

Axiom fun_ext : ∀ {A B} (f g : A→B), (∀ x, f x = g x) → f = g

to Coq’s impredicative Set is consistent. All of the published consistency proofs of Coq+funext (in particular the PER–/parametricity‐style models) crucially use the fact that Set (and the universe hierarchy) is predicative. As so...

<think> We are given a question about the compatibility of impredicative Set and function extensionality in Coq. The Coq FAQ states that function extensionality is consistent with predicative Set, but the user is asking about impredicative Set.

First, recall:

• Predicative Set: In Coq, by default, Set is predicative, meaning that when defining a type in Set, we cannot quantify over all ty...

Compatibility of Impredicative Set and Function Extensionality in Coq

The Coq FAQ indeed states that function extensionality is consistent with predicative Set, but doesn't explicitly address its consistency with impredicative Set.

Based on my understanding of type theory:

Current Knowledge

Function extensionality is believed to be consistent with impredicative Set. The consistency of fun...

Short answer
Yes. Assuming the ordinary set-theoretic strength that is usually used to justify Coq, making Set impredicative (-impredicative-set) does not conflict with adding the axiom

FunExt : ∀ {A B} (f g : A → B), (∀ x, f x = g x) → f = g.

so the two features are compatible and known to be consistent relative to ZFC (in practice one works in ZFC plus the existence of one inaccessible ...

The consistency of function extensionality with impredicative Set in Coq is indeed a subtle question. Let me clarify the situation:

The Short Answer

Function extensionality is consistent with impredicative Set. This has been known for quite some time, and there are no known inconsistencies between these two axioms.

The Details

What is Function Extensionality?

Function extensi...

This is an excellent and subtle question about the logical foundations of Coq. The short answer is:

Yes, function extensionality is believed to be consistent with impredicative Set.

The phrasing in the FAQ is likely a matter of precision and historical context rather than an implication of known inconsistency.

Here's a more detailed breakdown.

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1. Understanding the Concepts

Firs...

Yes. Adding the usual axiom of (dependent) functional extensionality to Coq is believed to be sound whether Set is left predicative (the factory settings) or made impredicative with the flag -impredicative-set.

Why we are confident:

1. What the axiom says
   Functional extensionality is the sentence

   ∀ (A:Type) (B:A→Type) (f g : ∀x, B x),
        (∀x, f x = g...