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(* Title: HoTT/EqualProps.thy
Author: Josh Chen
Properties of equality
*)
theory EqualProps
imports
HoTT_Methods
Equal
Prod
begin
section \<open>Symmetry / Path inverse\<close>
definition inv :: "Term \<Rightarrow> Term" ("_\<inverse>" [1000] 1000) where "p\<inverse> \<equiv> 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]) (routine lems: 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 compute
show "\<And>x. x: A \<Longrightarrow> refl x: x =\<^sub>A x" ..
qed (routine lems: 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>.
"
definition rpathcomp :: Term where "rpathcomp \<equiv> \<^bold>\<lambda>_ _ p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>_ 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 rpathcomp_type:
assumes "A: U(i)"
shows "rpathcomp: \<Prod>x:A. \<Prod>y:A. x =\<^sub>A y \<rightarrow> (\<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z)"
unfolding rpathcomp_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]) (routine lems: assms asm)
qed (routine lems: assms)
qed (rule assms)
qed (routine lems: assms 1 2 3)
qed (routine lems: 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 "rpathcomp`x`y`p`z`q: x =\<^sub>A z"
by (routine lems: assms rpathcomp_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 rpathcomp_comp:
assumes "A: U(i)" and "a: A"
shows "rpathcomp`a`a`refl(a)`a`refl(a) \<equiv> refl(a)"
unfolding rpathcomp_def
proof compute
{ 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]) (routine lems: assms asm)
qed (routine lems: assms)
qed (rule assms)
qed (routine lems: assms 1 2 3)
qed (routine lems: 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 compute
{ 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 =\<^sub>A z \<rightarrow> u =\<^sub>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]) (routine lems: assms 3 4)
qed (routine lems: assms 3 4)
qed fact
qed (routine lems: assms 1 2)
qed (routine lems: 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 compute
{ fix p assume 1: "p: a =\<^sub>A a"
show "ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q) p: \<Prod>z:A. a =\<^sub>A z \<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 z \<rightarrow> u =\<^sub>A z"
proof
fix z assume 3: "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]) (routine lems: assms 2 3)
qed (routine lems: assms 2 3)
qed fact
qed (routine lems: assms 1) }
show "(ind\<^sub>=(\<lambda>_. \<^bold>\<lambda>z q. ind\<^sub>= refl q)(refl(a)))`a`refl(a) \<equiv> refl(a)"
proof compute
{ fix u assume 1: "u: A"
show "\<^bold>\<lambda>z q. ind\<^sub>= refl q: \<Prod>z:A. u =\<^sub>A z \<rightarrow> u =\<^sub>A z"
proof
fix z assume 2: "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]) (routine lems: assms 1 2)
qed (routine lems: assms 1 2)
qed fact }
show "(\<^bold>\<lambda>z q. ind\<^sub>= refl q)`a`refl(a) \<equiv> refl(a)"
proof compute
{ 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]) (routine lems: assms 1)
qed (routine lems: assms 1) }
show "(\<^bold>\<lambda>q. ind\<^sub>= refl q)`refl(a) \<equiv> refl(a)"
proof compute
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]) (routine lems: assms)
show "ind\<^sub>= refl (refl(a)) \<equiv> refl(a)"
proof compute
show "\<And>x. x: A \<Longrightarrow> refl(x): x =\<^sub>A x" ..
qed (routine lems: assms)
qed (routine lems: assms)
qed fact
qed (routine lems: assms)
qed (routine lems: assms)
qed fact
qed fact
text "The raw object lambda term is cumbersome to use, so we define a simpler constant instead."
axiomatization pathcomp :: "[Term, Term] \<Rightarrow> Term" (infixl "\<bullet>" 60) where
pathcomp_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> rpathcomp`x`y`p`z`q"
lemma pathcomp_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 pathcomp_def)
show "A: U(i)" "x: A" "y: A" "z: A" "p: x =\<^sub>A y" "q: y =\<^sub>A z" by fact+
qed (routine lems: assms rpathcomp_type)
lemma pathcomp_comp:
assumes "A : U(i)" and "a : A" shows "refl(a) \<bullet> refl(a) \<equiv> refl(a)"
by (subst pathcomp_def) (routine lems: assms rpathcomp_comp)
lemmas EqualProps_type [intro] = inv_type pathcomp_type
lemmas EqualProps_comp [comp] = inv_comp pathcomp_comp
end
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