coq

I noticed that Coq synthesizes different induction principles on equality for Prop and Type. Does anybody have an explanation for that? Equality is defined as Inductive eq (A : Type) (x : A) : A -> Prop := eq_refl : x = x And the associated induction principle has the following type: eq_ind : forall (A : Type) (x : A) (P : A -> Prop), P x -> forall y : A, x = y -> P y Now let's define a Type pendant of eq: Inductive eqT {A:Type}(x:A):A->Type:= eqT_refl: eqT x x. The automatically generated induction principle is eqT_ind : forall (A : Type) (x : A) (P : forall a : A, eqT x a -> Prop), P x (eqT

I would like to be able to define the same Coq notations for different inductive definitions, and distinguish the notations based on the types of their arguments. Here is a minimal example: Inductive type : Type := | TBool : type. Inductive term1 : Type := | tvar1 : term1. Inductive term2 : Type := | tvar2 : term2. Definition context := nat -> (option type). Reserved Notation "G '⊢' t '::' T" (at level 40, t at level 59). Inductive typing1 : context -> term1 -> type -> Prop := | T_Var1 : forall G T, G ⊢ tvar1 :: T where "G '⊢' t1 '::' T" := (typing1 G t1 T) with typing2 : context -> term2 ->

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

I have defined a Sygma-Type that looks like: { R : nat -> nat -> bool | Reflexive R } I have two elements r1 r2 : { R : nat -> nat -> bool | Reflexive R } and I am to prove r1 = r2 . How can I do that? If you want to show such an equality, you need to (1) show that the underlying functions are equal (i.e., the R component of your sigma type), and (2) show that the corresponding proofs are equal. There are two problems, however. The first one is that equality of functions is too weak in Coq. According to common mathematical practice, we expect two functions to be equal if they yield equal

Given a valid Coq proof using the ; tactical, is there a general formula for converting it to a valid equivalent proof with . substituted for ; ? Many Coq proofs use the ; or tactic sequencing tactical. As a beginner, I want to watch the individual steps execute, so I want to substitute . for ; , but to my surprise I find that this may break the proof. Documentation on ; is sparse, and I haven't found an explicit discussion of . anywhere. I did see a paper that says informal meaning of t1; t2 is apply t2 to every subgoal produced by the execution of t1 in the current proof context, and I

They seem to serve similar purposes. The one difference I've noticed so far is that while Program Fixpoint will accept a compound measure like {measure (length l1 + length l2) } , Function seems to reject this and will only allow {measure length l1} . Is Program Fixpoint strictly more powerful than Function , or are they better suited for different use cases? This may not be a complete list, but it is what I have found so far: As you already mentioned, Program Fixpoint allows the measure to look at more than one argument. Function creates a foo_equation lemma that can be used to rewrite calls

I was trying to prove the following simple theorem from an online course that excluded middle is irrefutable, but got stuck pretty much at step 1: Theorem excluded_middle_irrefutable: forall (P:Prop), ~~(P \/ ~ P). Proof. intros P. unfold not. intros H. Now I get: 1 subgoals P : Prop H : P \/ (P -> False) -> False ______________________________________(1/1) False If I apply H , then the goal would be P \/ ~P , which is excluded middle and can't be proven constructively. But other than apply , I don't know what can be done about the hypothesis P \/ (P -> False) -> False : implication -> is