(*  Title:  HoTT/Proj.thy
    Author: Josh Chen
    Date:   Jun 2018

Projection functions \<open>fst\<close> and \<open>snd\<close> for the dependent sum type.
*)

theory Proj
  imports
    HoTT_Methods
    Prod
    Sum
begin

consts
  fst :: "[Term, 'a] \<Rightarrow> Term"  ("(1fst[/_,/ _])")
  snd :: "[Term, 'a] \<Rightarrow> Term"  ("(1snd[/_,/ _])")


section \<open>Overloaded syntax for dependent and nondependent pairs\<close>

overloading
  fst_dep \<equiv> fst
  fst_nondep \<equiv> fst
begin
  definition fst_dep :: "[Term, Typefam] \<Rightarrow> Term" where
    "fst_dep A B \<equiv> \<^bold>\<lambda>p: (\<Sum>x:A. B x). indSum[A,B] (\<lambda>_. A) (\<lambda>x y. x) p"
  
  definition fst_nondep :: "[Term, Term] \<Rightarrow> Term" where
    "fst_nondep A B \<equiv> \<^bold>\<lambda>p: A \<times> B. indSum[A, \<lambda>_. B] (\<lambda>_. A) (\<lambda>x y. x) p"
end

overloading
  snd_dep \<equiv> snd
  snd_nondep \<equiv> snd
begin
  definition snd_dep :: "[Term, Typefam] \<Rightarrow> Term" where
    "snd_dep A B \<equiv> \<^bold>\<lambda>p: (\<Sum>x:A. B x). indSum[A,B] (\<lambda>q. B (fst[A,B]`q)) (\<lambda>x y. y) p"
  
  definition snd_nondep :: "[Term, Term] \<Rightarrow> Term" where
    "snd_nondep A B \<equiv> \<^bold>\<lambda>p: A \<times> B. indSum[A, \<lambda>_. B] (\<lambda>_. B) (\<lambda>x y. y) p"
end


section \<open>Properties\<close>

text "Typing judgments and computation rules for the dependent and non-dependent projection functions."

lemma fst_dep_type:
  assumes "p : \<Sum>x:A. B x"
  shows "fst[A,B]`p : A"

proof \<comment> \<open>The standard reasoner knows to backchain with the product elimination rule here...\<close>
  \<comment> \<open>Also write about this proof: compare the effect of letting the standard reasoner do simplifications, as opposed to using the minus switch and writing everything out explicitly from first principles.\<close>

  have *: "\<Sum>x:A. B x : U" using assms ..
  
  show "fst[A,B]: (\<Sum>x:A. B x) \<rightarrow> A"

  proof (unfold fst_dep_def, standard) \<comment> \<open>...and with the product introduction rule here...\<close>
    show "\<And>p. p : \<Sum>x:A. B x \<Longrightarrow> indSum[A,B] (\<lambda>_. A) (\<lambda>x y. x) p : A"
    
    proof \<comment> \<open>...and with sum elimination here.\<close>
      show "A : U" using * ..
    qed

  qed (rule *)

qed (rule assms)


lemma fst_dep_comp:  (* Potential for automation *)
  assumes "B: A \<rightarrow> U" and "a : A" and "b : B a"
  shows "fst[A,B]`(a,b) \<equiv> a"

proof -
  \<comment> "Write about this proof: unfolding, how we set up the introduction rules (explain \<open>..\<close>), do a trace of the proof, explain the meaning of keywords, etc."

  have *: "A : U" using assms(2) .. (* I keep thinking this should not have to be done explicitly, but rather automated. *)

  have "fst[A,B]`(a,b) \<equiv> indSum[A,B] (\<lambda>_. A) (\<lambda>x y. x) (a,b)"
  proof (unfold fst_dep_def, standard)
    show "\<And>p. p : \<Sum>x:A. B x \<Longrightarrow> indSum[A,B] (\<lambda>_. A) (\<lambda>x y. x) p : A" using * ..
    show "(a,b) : \<Sum>x:A. B x" using assms ..
  qed
  
  also have "indSum[A,B] (\<lambda>_. A) (\<lambda>x y. x) (a,b) \<equiv> a"
    by (rule Sum_comp) (rule *, assumption, (rule assms)+)
  
  finally show "fst[A,B]`(a,b) \<equiv> a" .
qed


text "In proving results about the second dependent projection function we often use the following two lemmas."

lemma snd_dep_welltyped:
  assumes "p : \<Sum>x:A. B x"
  shows "B (fst[A,B]`p) : U"

proof -
  have "\<Sum>x:A. B x : U" using assms ..
  then have *: "B: A \<rightarrow> U" ..

  have "fst[A,B]`p : A" using assms by (rule fst_dep_type)
  then show "B (fst[A,B]`p) : U" by (rule *)
qed


lemma snd_dep_const_type:
  assumes "B: A \<rightarrow> U" and "x : A" and "y : B x"
  shows "y : B (fst[A,B]`(x,y))"

proof -
  have "fst[A,B]`(x,y) \<equiv> x" using assms by (rule fst_dep_comp)
  then show "y : B (fst[A,B]`(x,y))" using assms by simp
qed


lemma snd_dep_type:
  assumes "p : \<Sum>x:A. B x"
  shows "snd[A,B]`p : B (fst[A,B]`p)"

proof
  have *: "\<Sum>x:A. B x : U" using assms ..

  show "snd[A, B] : \<Prod>p:(\<Sum>x:A. B x). B (fst[A,B]`p)"

  proof (unfold snd_dep_def, standard)
    show "\<And>p. p : \<Sum>x:A. B x \<Longrightarrow> indSum[A,B] (\<lambda>q. B (fst[A,B]`q)) (\<lambda>x y. y) p : B (fst[A,B]`p)"

    proof (standard, elim snd_dep_welltyped)
      fix x y assume 1: "x : A" and 2: "y : B x"
      show "y : B (fst[A,B]`(x,y))"
      proof -
        have "B: A \<rightarrow> U" using * ..
        then show "y : B (fst[A,B]`(x,y))" using 1 2 by (rule snd_dep_const_type)
      qed
    qed

  qed (rule *)

qed (rule assms)


lemma snd_dep_comp:
  assumes "B: A \<rightarrow> U" and "a : A" and "b : B a"
  shows "snd[A,B]`(a,b) \<equiv> b"

proof -
  have "snd[A,B]`(a,b) \<equiv> indSum[A, B] (\<lambda>q. B (fst[A,B]`q)) (\<lambda>x y. y) (a,b)"
  proof (unfold snd_dep_def, standard)
    show "(a,b) : \<Sum>x:A. B x" using assms ..

    fix p assume *: "p : \<Sum>x:A. B x"
    show "indSum[A,B] (\<lambda>q. B (fst[A,B]`q)) (\<lambda>x y. y) p : B (fst[A,B]`p)"
    proof (standard, elim snd_dep_welltyped)
      show "\<And>x y. \<lbrakk>x : A; y : B x\<rbrakk> \<Longrightarrow> y : B (fst[A,B]`(x,y))" using assms
        by (elim snd_dep_const_type)
    qed (rule *)
  qed

  also have "indSum[A, B] (\<lambda>q. B (fst[A,B]`q)) (\<lambda>x y. y) (a,b) \<equiv> b"
  proof (standard, elim snd_dep_welltyped)
    show "\<And>x y. \<lbrakk>x : A; y : B x\<rbrakk> \<Longrightarrow> y : B (fst[A,B]`(x,y))" using assms
      by (elim snd_dep_const_type)
  qed (rule assms)+

  finally show "snd[A,B]`(a,b) \<equiv> b" .
qed


text "For non-dependent projection functions:"

print_statement fst_dep_type
print_statement fst_dep_type[where ?p = p and ?A = A and ?B = "\<lambda>_. B"]


lemma fst_nondep_type: "p : A \<times> B \<Longrightarrow> fst[A,B]`p : A"
by (rule fst_dep_type[where ?B = "\<lambda>_. B"])
  

lemma fst_nondep_comp:
  assumes "a : A" and "b : B"
  shows "fst[A,B]`(a,b) \<equiv> a"
proof (unfold fst_nondep_def)
  have "A : U" using assms(1) ..
  then show "fst[A,B]`(a,b) \<equiv> a" unfolding fst_nondep_def 
qed

lemma snd_nondep_comp: "\<lbrakk>a : A; b : B\<rbrakk> \<Longrightarrow> snd[A,B]`(a,b) \<equiv> b"
proof -
  assume "a : A" and "b : B"
  then have "(a, b) : A \<times> B" ..
  then show "snd[A,B]`(a,b) \<equiv> b" unfolding snd_nondep_def by simp
qed