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(*
Title: EqualProps.thy
Author: Joshua Chen
Date: 2018
Properties of equality
*)
theory EqualProps
imports HoTT_Methods Equal Prod
begin
section \<open>Symmetry of equality/Path inverse\<close>
definition inv :: "t \<Rightarrow> t" ("_\<inverse>" [1000] 1000) where "p\<inverse> \<equiv> ind\<^sub>= (\<lambda>x. refl x) p"
lemma inv_type: "\<lbrakk>A: U i; x: A; y: A; p: x =\<^sub>A y\<rbrakk> \<Longrightarrow> p\<inverse>: y =\<^sub>A x"
unfolding inv_def by (elim Equal_elim) routine
lemma inv_comp: "\<lbrakk>A: U i; a: A\<rbrakk> \<Longrightarrow> (refl a)\<inverse> \<equiv> refl a"
unfolding inv_def by compute routine
declare
inv_type [intro]
inv_comp [comp]
section \<open>Transitivity of equality/Path composition\<close>
text \<open>
Composition function, of type @{term "x =\<^sub>A y \<rightarrow> (y =\<^sub>A z) \<rightarrow> (x =\<^sub>A z)"} polymorphic over @{term A}, @{term x}, @{term y}, and @{term z}.
\<close>
definition pathcomp :: t where "pathcomp \<equiv> \<^bold>\<lambda>p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>q. ind\<^sub>= (\<lambda>x. (refl x)) q) p"
syntax "_pathcomp" :: "[t, t] \<Rightarrow> t" (infixl "\<bullet>" 120)
translations "p \<bullet> q" \<rightleftharpoons> "CONST pathcomp`p`q"
lemma pathcomp_type:
assumes "A: U i" "x: A" "y: A" "z: A"
shows "pathcomp: x =\<^sub>A y \<rightarrow> (y =\<^sub>A z) \<rightarrow> (x =\<^sub>A z)"
unfolding pathcomp_def by rule (elim Equal_elim, routine add: assms)
corollary pathcomp_trans:
assumes "A: U i" and "x: A" "y: A" "z: A" and "p: x =\<^sub>A y" "q: y =\<^sub>A z"
shows "p \<bullet> q: x =\<^sub>A z"
by (routine add: assms pathcomp_type)
lemma pathcomp_comp:
assumes "A: U i" and "a: A"
shows "(refl a) \<bullet> (refl a) \<equiv> refl a"
unfolding pathcomp_def proof compute
show "(ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>q. ind\<^sub>= (\<lambda>x. refl x) q) (refl a))`(refl a) \<equiv> refl a"
proof compute
show "\<^bold>\<lambda>q. (ind\<^sub>= (\<lambda>x. refl x) q): a =\<^sub>A a \<rightarrow> a =\<^sub>A a"
by (routine add: assms)
show "(\<^bold>\<lambda>q. (ind\<^sub>= (\<lambda>x. refl x) q))`(refl a) \<equiv> refl a"
proof compute
show "\<And>q. q: a =\<^sub>A a \<Longrightarrow> ind\<^sub>= (\<lambda>x. refl x) q: a =\<^sub>A a" by (routine add: assms)
qed (derive lems: assms)
qed (routine add: assms)
show "\<And>p. p: a =\<^sub>A a \<Longrightarrow> ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>q. ind\<^sub>= (\<lambda>x. refl x) q) p: a =\<^sub>A a \<rightarrow> a =\<^sub>A a"
by (routine add: assms)
qed (routine add: assms)
declare
pathcomp_type [intro]
pathcomp_trans [intro]
pathcomp_comp [comp]
section \<open>Higher groupoid structure of types\<close>
schematic_goal pathcomp_right_id:
assumes "A: U(i)" "x: A" "y: A" "p: x =\<^sub>A y"
shows "?a: p \<bullet> (refl y) =[x =\<^sub>A y] p"
proof (rule Equal_elim[where ?x=x and ?y=y and ?p=p]) \<comment> \<open>@{method elim} does not seem to work with schematic goals.\<close>
show "\<And>x. x: A \<Longrightarrow> refl(refl x): (refl x) \<bullet> (refl x) =[x =\<^sub>A x] (refl x)"
by (derive lems: assms)
qed (routine add: assms)
schematic_goal pathcomp_left_id:
assumes "A: U(i)" "x: A" "y: A" "p: x =\<^sub>A y"
shows "?a: (refl x) \<bullet> p =[x =\<^sub>A y] p"
proof (rule Equal_elim[where ?x=x and ?y=y and ?p=p])
show "\<And>x. x: A \<Longrightarrow> refl(refl x): (refl x) \<bullet> (refl x) =[x =\<^sub>A x] (refl x)"
by (derive lems: assms)
qed (routine add: assms)
schematic_goal pathcomp_left_inv:
assumes "A: U(i)" "x: A" "y: A" "p: x =\<^sub>A y"
shows "?a: (p\<inverse> \<bullet> p) =[y =\<^sub>A y] refl(y)"
proof (rule Equal_elim[where ?x=x and ?y=y and ?p=p])
show "\<And>x. x: A \<Longrightarrow> refl(refl x): (refl x)\<inverse> \<bullet> (refl x) =[x =\<^sub>A x] (refl x)"
by (derive lems: assms)
qed (routine add: assms)
schematic_goal pathcomp_right_inv:
assumes "A: U(i)" "x: A" "y: A" "p: x =\<^sub>A y"
shows "?a: (p \<bullet> p\<inverse>) =[x =\<^sub>A x] refl(x)"
proof (rule Equal_elim[where ?x=x and ?y=y and ?p=p])
show "\<And>x. x: A \<Longrightarrow> refl(refl x): (refl x) \<bullet> (refl x)\<inverse> =[x =\<^sub>A x] (refl x)"
by (derive lems: assms)
qed (routine add: assms)
schematic_goal inv_involutive:
assumes "A: U(i)" "x: A" "y: A" "p: x =\<^sub>A y"
shows "?a: p\<inverse>\<inverse> =[x =\<^sub>A y] p"
proof (rule Equal_elim[where ?x=x and ?y=y and ?p=p])
show "\<And>x. x: A \<Longrightarrow> refl(refl x): (refl x)\<inverse>\<inverse> =[x =\<^sub>A x] (refl x)"
by (derive lems: assms)
qed (routine add: assms)
text \<open>All of the propositions above have the same proof term, which we abbreviate here.\<close>
abbreviation \<iota> :: "t \<Rightarrow> t" where "\<iota> p \<equiv> ind\<^sub>= (\<lambda>x. refl (refl x)) p"
text \<open>The next proof has a triply-nested path induction.\<close>
lemma pathcomp_assoc:
assumes "A: U i" "x: A" "y: A" "z: A" "w: A"
shows "\<^bold>\<lambda>p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>q. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>r. \<iota> r) q) p:
\<Prod>p: x =\<^sub>A y. \<Prod>q: y =\<^sub>A z. \<Prod>r: z =\<^sub>A w. p \<bullet> (q \<bullet> r) =[x =\<^sub>A w] (p \<bullet> q) \<bullet> r"
proof
show "\<And>p. p: x =\<^sub>A y \<Longrightarrow> ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>q. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>r. \<iota> r) q) p:
\<Prod>q: y =\<^sub>A z. \<Prod>r: z =\<^sub>A w. p \<bullet> (q \<bullet> r) =[x =\<^sub>A w] p \<bullet> q \<bullet> r"
proof (elim Equal_elim)
fix x assume 1: "x: A"
show "\<^bold>\<lambda>q. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>r. \<iota> r) q:
\<Prod>q: x =\<^sub>A z. \<Prod>r: z =\<^sub>A w. refl x \<bullet> (q \<bullet> r) =[x =\<^sub>A w] refl x \<bullet> q \<bullet> r"
proof
show "\<And>q. q: x =\<^sub>A z \<Longrightarrow> ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>r. \<iota> r) q:
\<Prod>r: z =\<^sub>A w. refl x \<bullet> (q \<bullet> r) =[x =\<^sub>A w] refl x \<bullet> q \<bullet> r"
proof (elim Equal_elim)
fix x assume *: "x: A"
show "\<^bold>\<lambda>r. \<iota> r: \<Prod>r: x =\<^sub>A w. refl x \<bullet> (refl x \<bullet> r) =[x =\<^sub>A w] refl x \<bullet> refl x \<bullet> r"
proof
show "\<And>r. r: x =[A] w \<Longrightarrow> \<iota> r: refl x \<bullet> (refl x \<bullet> r) =[x =[A] w] refl x \<bullet> refl x \<bullet> r"
by (elim Equal_elim, derive lems: * assms)
qed (routine add: * assms)
qed (routine add: 1 assms)
qed (routine add: 1 assms)
text \<open>
In the following part @{method derive} fails to obtain the correct subgoals, so we have to prove the statement manually.
\<close>
fix y p assume 2: "y: A" "p: x =\<^sub>A y"
show "\<Prod>q: y =\<^sub>A z. \<Prod>r: z =\<^sub>A w. p \<bullet> (q \<bullet> r) =[x =\<^sub>A w] p \<bullet> q \<bullet> r: U i"
proof (routine add: assms)
fix q assume 3: "q: y =\<^sub>A z"
show "\<Prod>r: z =\<^sub>A w. p \<bullet> (q \<bullet> r) =[x =\<^sub>A w] p \<bullet> q \<bullet> r: U i"
proof (routine add: assms)
show "\<And>r. r: z =[A] w \<Longrightarrow> p \<bullet> (q \<bullet> r): x =[A] w" and "\<And>r. r: z =[A] w \<Longrightarrow> p \<bullet> q \<bullet> r: x =[A] w"
by (routine add: 1 2 3 assms)
qed (routine add: 1 assms)
qed fact+
qed fact+
qed (routine add: assms)
section \<open>Transport\<close>
definition transport :: "t \<Rightarrow> t" where "transport p \<equiv> ind\<^sub>= (\<lambda>_. (\<^bold>\<lambda>x. x)) p"
text \<open>Note that @{term transport} is a polymorphic function in our formulation.\<close>
lemma transport_type: "\<lbrakk>p: x =\<^sub>A y; x: A; y: A; A: U i; P: A \<longrightarrow> U i\<rbrakk> \<Longrightarrow> transport p: P x \<rightarrow> P y"
unfolding transport_def by (elim Equal_elim) routine
lemma transport_comp: "\<lbrakk>A: U i; x: A\<rbrakk> \<Longrightarrow> transport (refl x) \<equiv> id"
unfolding transport_def by derive
end
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