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(* Title: HoTT/EqualProps.thy
Author: Josh Chen
Date: Aug 2018
Properties of equality.
*)
theory EqualProps
imports
HoTT_Methods
Equal
Prod
begin
section \<open>Symmetry / Path inverse\<close>
axiomatization inv :: "Term \<Rightarrow> Term" ("_\<inverse>" [1000] 1000)
where inv_def: "inv \<equiv> \<lambda>p. ind\<^sub>= (\<lambda>x. refl(x)) p"
text "
In the proof below we begin by using path induction on \<open>p\<close> with the application of \<open>rule Equal_elim\<close>, telling Isabelle the specific substitutions to use.
The proof is finished with a standard application of the relevant type rules.
"
lemma inv_type:
assumes "A : U(i)" and "x : A" and "y : A" and "p: x =\<^sub>A y" shows "p\<inverse>: y =\<^sub>A x"
unfolding inv_def
by (rule Equal_elim[where ?x=x and ?y=y]) (simple lem: assms)
\<comment> \<open>The type doesn't depend on \<open>p\<close> so we don't need to specify \<open>?p\<close> in the \<open>where\<close> clause above.\<close>
text "
The next proof requires explicitly telling Isabelle what to substitute on the RHS of the typing judgment after the initial application of the type rules.
(If viewing this inside Isabelle, place the cursor after the \<open>proof\<close> statement and observe the second subgoal.)
"
lemma inv_comp:
assumes "A : U(i)" and "a : A" shows "(refl a)\<inverse> \<equiv> refl(a)"
unfolding inv_def
proof
show "\<And>x. x: A \<Longrightarrow> refl x: x =\<^sub>A x" ..
qed (simple lem: assms)
section \<open>Transitivity / Path composition\<close>
text "
Raw composition function, of type \<open>\<Prod>x:A. \<Prod>y:A. x =\<^sub>A y \<rightarrow> (\<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z)\<close> polymorphic over the type \<open>A\<close>.
"
axiomatization reqcompose :: Term where
reqcompose_def: "reqcompose \<equiv> \<^bold>\<lambda>x y p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= (\<lambda>x. refl(x)) q) p"
text "
More complicated proofs---the nested path inductions require more explicit step-by-step rule applications:
"
lemma reqcompose_type:
assumes "A: U(i)"
shows "reqcompose: \<Prod>x:A. \<Prod>y:A. x =\<^sub>A y \<rightarrow> (\<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z)"
unfolding reqcompose_def
proof
fix x assume 1: "x: A"
show "\<^bold>\<lambda>y p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q) p: \<Prod>y:A. x =\<^sub>A y \<rightarrow> (\<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z)"
proof
fix y assume 2: "y: A"
show "\<^bold>\<lambda>p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q) p: x =\<^sub>A y \<rightarrow> (\<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z)"
proof
fix p assume 3: "p: x =\<^sub>A y"
show "ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q) p: \<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z"
proof (rule Equal_elim[where ?x=x and ?y=y])
show "\<And>u. u: A \<Longrightarrow> \<^bold>\<lambda>z q. ind\<^sub>= refl q: \<Prod>z:A. u =\<^sub>A z \<rightarrow> u =\<^sub>A z"
proof
show "\<And>u z. \<lbrakk>u: A; z: A\<rbrakk> \<Longrightarrow> \<^bold>\<lambda>q. ind\<^sub>= refl q: u =\<^sub>A z \<rightarrow> u =\<^sub>A z"
proof
fix u z q assume asm: "u: A" "z: A" "q: u =\<^sub>A z"
show "ind\<^sub>= refl q: u =\<^sub>A z"
by (rule Equal_elim[where ?x=u and ?y=z]) (simple lem: assms asm)
qed (simple lem: assms)
qed (rule assms)
qed (simple lem: assms 1 2 3)
qed (simple lem: assms 1 2)
qed (rule assms)
qed fact
corollary
assumes "A: U(i)" "x: A" "y: A" "z: A" "p: x =\<^sub>A y" "q: y =\<^sub>A z"
shows "reqcompose`x`y`p`z`q: x =\<^sub>A z"
by (simple lem: assms reqcompose_type)
text "
The following proof is very long, chiefly because for every application of \<open>`\<close> we have to show the wellformedness of the type family appearing in the equality computation rule.
"
lemma reqcompose_comp:
assumes "A: U(i)" and "a: A"
shows "reqcompose`a`a`refl(a)`a`refl(a) \<equiv> refl(a)"
unfolding reqcompose_def
proof (subst comp)
{ fix x assume 1: "x: A"
show "\<^bold>\<lambda>y p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q) p: \<Prod>y:A. x =\<^sub>A y \<rightarrow> (\<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z)"
proof
fix y assume 2: "y: A"
show "\<^bold>\<lambda>p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q) p: x =\<^sub>A y \<rightarrow> (\<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z)"
proof
fix p assume 3: "p: x =\<^sub>A y"
show "ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q) p: \<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z"
proof (rule Equal_elim[where ?x=x and ?y=y])
show "\<And>u. u: A \<Longrightarrow> \<^bold>\<lambda>z q. ind\<^sub>= refl q: \<Prod>z:A. u =\<^sub>A z \<rightarrow> u =\<^sub>A z"
proof
show "\<And>u z. \<lbrakk>u: A; z: A\<rbrakk> \<Longrightarrow> \<^bold>\<lambda>q. ind\<^sub>= refl q: u =\<^sub>A z \<rightarrow> u =\<^sub>A z"
proof
fix u z assume asm: "u: A" "z: A"
show "\<And>q. q: u =\<^sub>A z \<Longrightarrow> ind\<^sub>= refl q: u =\<^sub>A z"
by (rule Equal_elim[where ?x=u and ?y=z]) (simple lem: assms asm)
qed (simple lem: assms)
qed (rule assms)
qed (simple lem: assms 1 2 3)
qed (simple lem: assms 1 2)
qed (rule assms) }
show "(\<^bold>\<lambda>y p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q) p)`a`refl(a)`a`refl(a) \<equiv> refl(a)"
proof (subst comp)
{ fix y assume 1: "y: A"
show "\<^bold>\<lambda>p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q) p: a =\<^sub>A y \<rightarrow> (\<Prod>z:A. y =\<^sub>A z \<rightarrow> a =\<^sub>A z)"
proof
fix p assume 2: "p: a =\<^sub>A y"
show "ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q) p: \<Prod>z:A. y =\<^sub>A z \<rightarrow> a =\<^sub>A z"
proof (rule Equal_elim[where ?x=a and ?y=y])
fix u assume 3: "u: A"
show "\<^bold>\<lambda>z q. ind\<^sub>= refl q: \<Prod>z:A. u =[A] z \<rightarrow> u =[A] z"
proof
fix z assume 4: "z: A"
show "\<^bold>\<lambda>q. ind\<^sub>= refl q: u =\<^sub>A z \<rightarrow> u =\<^sub>A z"
proof
show "\<And>q. q: u =\<^sub>A z \<Longrightarrow> ind\<^sub>= refl q: u =\<^sub>A z"
by (rule Equal_elim[where ?x=u and ?y=z]) (simple lem: assms 3 4)
qed (simple lem: assms 3 4)
qed fact
qed (simple lem: assms 1 2)
qed (simple lem: assms 1) }
show "(\<^bold>\<lambda>p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z. \<^bold>\<lambda>q. ind\<^sub>= refl q) p)`refl(a)`a`refl(a) \<equiv> refl(a)"
proof (subst comp)
{ fix p assume 1: "p: a =\<^sub>A a"
show "ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z. \<^bold>\<lambda>q. ind\<^sub>= refl q) p: \<Prod>z:A. a =\<^sub>A a \<rightarrow> a =\<^sub>A z"
proof (rule Equal_elim[where ?x=a and ?y=a])
fix u assume 2: "u: A"
show "\<^bold>\<lambda>z q. ind\<^sub>= refl q: \<Prod>z:A. u =\<^sub>A u\<rightarrow> u =\<^sub>A z"
proof
fix z assume 3: "z: A"
show "\<^bold>\<lambda>q. ind\<^sub>= refl q: u =\<^sub>A u \<rightarrow> u =\<^sub>A z"
proof
show "\<And>q. q: u =\<^sub>A u \<Longrightarrow> ind\<^sub>= refl q: u =\<^sub>A z"
by (rule Equal_elim[where ?x=u and ?y=z]) (simple lem: assms 2 3)
qed (simple lem: assms 2)
qed fact
qed (simple lem: assms 1) }
show "(ind\<^sub>=(\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q)(refl(a)))`a`refl(a) \<equiv> refl(a)"
proof (subst comp)
{ fix u assume 1: "u: A"
show "\<^bold>\<lambda>z q. ind\<^sub>= refl q: \<Prod>z:A. u =\<^sub>A u\<rightarrow> u =\<^sub>A z"
proof
fix z assume 2: "z: A"
show "\<^bold>\<lambda>q. ind\<^sub>= refl q: u =\<^sub>A u \<rightarrow> u =\<^sub>A z"
proof
show "\<And>q. q: u =\<^sub>A u \<Longrightarrow> ind\<^sub>= refl q: u =\<^sub>A z"
by (rule Equal_elim[where ?x=u and ?y=z]) (simple lem: assms 1 2)
qed (simple lem: assms 1)
qed fact }
show "(\<^bold>\<lambda>z q. ind\<^sub>= refl q)`a`refl(a) \<equiv> refl(a)"
proof (subst comp)
{ fix a assume 1: "a: A"
show "\<^bold>\<lambda>q. ind\<^sub>= refl q: a =\<^sub>A a \<rightarrow> a =\<^sub>A a"
proof
show "\<And>q. q: a =\<^sub>A a \<Longrightarrow> ind\<^sub>= refl q: a =\<^sub>A a"
by (rule Equal_elim[where ?x=a and ?y=a]) (simple lem: assms 1)
qed (simple lem: assms 1) }
show "(\<^bold>\<lambda>q. ind\<^sub>= refl q)`refl(a) \<equiv> refl(a)"
proof (subst comp)
show "\<And>p. p: a =\<^sub>A a \<Longrightarrow> ind\<^sub>= refl p: a =\<^sub>A a"
by (rule Equal_elim[where ?x=a and ?y=a]) (simple lem: assms)
show "ind\<^sub>= refl (refl(a)) \<equiv> refl(a)"
proof
show "\<And>x. x: A \<Longrightarrow> refl(x): x =\<^sub>A x" ..
qed (simple lem: assms)
qed (simple lem: assms)
qed fact
qed (simple lem: assms)
qed (simple lem: assms)
qed fact
qed fact
text "The raw object lambda term is cumbersome to use, so we define a simpler constant instead."
axiomatization eqcompose :: "[Term, Term] \<Rightarrow> Term" (infixl "\<bullet>" 60) where
eqcompose_def: "\<lbrakk>
A: U(i);
x: A; y: A; z: A;
p: x =\<^sub>A y; q: y =\<^sub>A z
\<rbrakk> \<Longrightarrow> p \<bullet> q \<equiv> reqcompose`x`y`p`z`q"
lemma eqcompose_type:
assumes "A: U(i)" "x: A" "y: A" "z: A" "p: x =\<^sub>A y" "q: y =\<^sub>A z"
shows "p \<bullet> q: x =\<^sub>A z"
proof (subst eqcompose_def)
show "A: U(i)" "x: A" "y: A" "z: A" "p: x =\<^sub>A y" "q: y =\<^sub>A z" by fact+
qed (simple lem: assms reqcompose_type)
lemma eqcompose_comp:
assumes "A : U(i)" and "a : A" shows "refl(a) \<bullet> refl(a) \<equiv> refl(a)"
by (subst eqcompose_def) (simple lem: assms reqcompose_comp)
lemmas EqualProps_rules [intro] = inv_type eqcompose_type
lemmas EqualProps_comps [comp] = inv_comp eqcompose_comp
end
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