Coq：Prop与类型中的设置（n）

Question

I want to consider the following three (related?) Coq definitions.

Inductive nat1: Prop :=
  | z1 : nat1
  | s1 : nat1 -> nat1.

Inductive nat2 : Set := 
  | z2 : nat2
  | s2 : nat2 -> nat2.

Inductive nat3 : Type :=
  | z3 : nat3
  | s3 : nat3 -> nat3.

All three types give induction principles to prove a proposition holds.

nat1_ind
     : forall P : Prop, P -> (nat1 -> P -> P) -> nat1 -> P

nat2_ind
     : forall P : nat2 -> Prop,
       P z2 -> (forall n : nat2, P n -> P (s2 n)) -> forall n : nat2, P n

nat3_ind
     : forall P : nat3 -> Prop,
       P z3 -> (forall n : nat3, P n -> P (s3 n)) -> forall n : nat3, P n

The set and type versions also contain induction principles for definitions over set and type (rec and rect respectively). This is the extent of my knowledge about the difference between Prop and Set; Prop has a weaker induction.

I have also read that Prop is impredicative while Set is predicative, but this seems like a property rather than a defining quality.

While some practical (moral?) differences between Set and Prop are clear, the exact, defining differences between Set and Prop, as well as where they fit into the universe of types is unclear (running check on Prop and Set gives Type (* (Set) + 1*)), and I'm not exactly sure how to interpret this...

Answer 1

Type : Type is inconsistent.

Impredicative Set with excluded middle implies proof irrelevance, so impredicative Set with proof relevance, eg true <> false , refutes excluded middle, which intuitionism isn't supposed to do.

Therefore we leave impredicativity in Prop and the rest of the type hierarchy gives us predicativity.

By the way,

forall P : nat1 -> Prop, P z1 -> (forall n : nat1, P n -> P (s1 n)) -> forall n : nat1, P n

is provable. Don't ask me what's the benefit of Coq only automatically proving that other weaker induction principle...

Also, have you read this chapter of CPDT ?

Answer 2

Just read about this in an hour. This is because Coq will assume equality of two proof object of a same Prop . This is an axiom and is called proof irrelevance.

https://coq.inria.fr/library/Coq.Logic.ProofIrrelevance.html

It just thinks a predicate over Prop (Here P ) doesn't really need to have some proof passed as its argument (or hypothesis) and removed it.

Consider this. Because of every nat1 are the same, whenever we try to proof some property P , we can just abstract over some nat1 , while use the axiom to rewrite it to required ones. Thus Coq generated the "simplified" version of induction principle.

To generate the "full" version, you can use

Scheme nat1_ind_full := Induction for nat1 Sort Prop.

ref. Different induction principles for Prop and Type