Can we assign cardinals in Coq to Prop, Set and each Type_i ? I only see the definition of finite cardinals in Coq's library, so maybe we need that of big cardinals to begin with.
According to the proof-irrelevance semantics, for example exposed here, Set and the Type_i form an increasing sequence of inaccessible cardinals. Can this be proven in Coq ?
Prop seems more complex because of impredicativity. Proof-irrelevance means we identify all proofs of the same P : Prop, and interpret Prop itself as the pair {false, true}. So the cardinal of Prop would be 2. However, for any two proofs p1 p2 : P, Coq does not accept eq_refl p1 as a proof of p1 = p2. So Coq does not completely identify p1 and p2. And on the other hand, impredicativity means that for any A : Type and P : Prop, A -> P is of type Prop. That makes a lot more inhabitants than in Set.
If this is too hard, can Coq prove that Prop and Set are uncountable ? By Cantor's theorem, Coq easily proves that there is not surjection nat -> (nat -> Prop). This does not seem very far from proving there is not surjection nat -> Prop. But then we need the filter Prop -> (nat -> Prop), that isolates which Prop's have a free nat variable. Can we define this filter in Coq, since we cannot pattern-match on Prop ?