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(* Title: HoTT/Sum.thy
Author: Josh Chen
Date: Jun 2018
Dependent sum type.
*)
theory Sum
imports HoTT_Base
begin
axiomatization
Sum :: "[Term, Typefam] \<Rightarrow> Term" and
pair :: "[Term, Term] \<Rightarrow> Term" ("(1'(_,/ _'))") and
indSum :: "[Term, Typefam, Typefam, [Term, Term] \<Rightarrow> Term, Term] \<Rightarrow> Term" ("(1indSum[_,/ _])")
section \<open>Syntax\<close>
syntax
"_SUM" :: "[idt, Term, Term] \<Rightarrow> Term" ("(3\<Sum>_:_./ _)" 20)
"_SUM_ASCII" :: "[idt, Term, Term] \<Rightarrow> Term" ("(3SUM _:_./ _)" 20)
translations
"\<Sum>x:A. B" \<rightleftharpoons> "CONST Sum A (\<lambda>x. B)"
"SUM x:A. B" \<rightharpoonup> "CONST Sum A (\<lambda>x. B)"
section \<open>Type rules\<close>
axiomatization where
Sum_form: "\<And>A B. \<lbrakk>A : U; B: A \<rightarrow> U\<rbrakk> \<Longrightarrow> \<Sum>x:A. B x : U"
and
Sum_form_cond1: "\<And>A B. (\<Sum>x:A. B x : U) \<Longrightarrow> A : U"
and
Sum_form_cond2: "\<And>A B. (\<Sum>x:A. B x : U) \<Longrightarrow> B: A \<rightarrow> U"
and
Sum_intro: "\<And>A B a b. \<lbrakk>B: A \<rightarrow> U; a : A; b : B a\<rbrakk> \<Longrightarrow> (a,b) : \<Sum>x:A. B x"
and
Sum_elim: "\<And>A B C f p. \<lbrakk>
C: \<Sum>x:A. B x \<rightarrow> U;
\<And>x y. \<lbrakk>x : A; y : B x\<rbrakk> \<Longrightarrow> f x y : C (x,y);
p : \<Sum>x:A. B x
\<rbrakk> \<Longrightarrow> indSum[A,B] C f p : C p"
and
Sum_comp: "\<And>A B C f a b. \<lbrakk>
C: \<Sum>x:A. B x \<rightarrow> U;
\<And>x y. \<lbrakk>x : A; y : B x\<rbrakk> \<Longrightarrow> f x y : C (x,y);
a : A;
b : B a
\<rbrakk> \<Longrightarrow> indSum[A,B] C f (a,b) \<equiv> f a b"
lemmas Sum_rules [intro] = Sum_form Sum_intro Sum_elim Sum_comp
lemmas Sum_form_conds [elim, wellform] = Sum_form_cond1 Sum_form_cond2
lemmas Sum_comps [comp] = Sum_comp
\<comment> \<open>Nondependent pair\<close>
abbreviation Pair :: "[Term, Term] \<Rightarrow> Term" (infixr "\<times>" 50)
where "A \<times> B \<equiv> \<Sum>_:A. B"
text "The nondependent pair needs its own separate introduction rule."
lemma Pair_intro [intro]: "\<And>A B a b. \<lbrakk>a : A; b : B\<rbrakk> \<Longrightarrow> (a,b) : A \<times> B"
proof
fix b B assume "b : B"
then show "B : U" ..
qed
end
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