Equality inverse and composition done. Cleaned up methods and method example theory.

1 files changed, 152 insertions, 40 deletions

diff --git a/EqualProps.thy b/EqualProps.thy
index d645fb6..3d99456 100644
--- a/EqualProps.thy
+++ b/EqualProps.thy
@@ -1,6 +1,6 @@
 (*  Title:  HoTT/EqualProps.thy
     Author: Josh Chen
-    Date:   Jun 2018
+    Date:   Aug 2018
 
 Properties of equality.
 *)
@@ -18,22 +18,28 @@ 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
-proof (rule Equal_elim[where ?x=x and ?y=y])  \<comment> \<open>Path induction\<close>
-  show "\<And>x y. \<lbrakk>x: A; y: A\<rbrakk> \<Longrightarrow> y =\<^sub>A x: U(i)" using assms(1) ..
-  show "\<And>x. x: A \<Longrightarrow> refl x: x =\<^sub>A x" ..
-qed (fact assms)+
+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)"
+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" ..
-  show "\<And>x. x: A \<Longrightarrow> x =\<^sub>A x: U(i)" using assms(1) ..
-qed (fact assms)
+qed (simple lem: assms)
 
 
 section \<open>Transitivity / Path composition\<close>
@@ -45,67 +51,173 @@ text "
 axiomatization rcompose :: Term where
   rcompose_def: "rcompose \<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 rcompose_type:
   assumes "A: U(i)"
   shows "rcompose: \<Prod>x:A. \<Prod>y:A. x =\<^sub>A y \<rightarrow> (\<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z)"
 unfolding rcompose_def
 proof
-  show "\<And>x. x: A \<Longrightarrow>
-    \<^bold>\<lambda>y p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z p. ind\<^sub>= refl p) p: \<Prod>y:A. x =\<^sub>A y \<rightarrow> (\<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z)"
+  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
-    show "\<And>x y. \<lbrakk>x: A ; y: A\<rbrakk> \<Longrightarrow>
-       \<^bold>\<lambda>p. ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z p. ind\<^sub>= refl p) p: x =\<^sub>A y \<rightarrow> (\<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z)"
+    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 x y p assume asm: "x: A" "y: A" "p: x =\<^sub>A y"
-      show "ind\<^sub>= (\<lambda>_. \<^bold>\<lambda>z p. ind\<^sub>= refl p) p: \<Prod>z:A. y =[A] z \<rightarrow> x =[A] z"
+      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>x y. \<lbrakk>x: A; y: A\<rbrakk> \<Longrightarrow> \<Prod>z:A. y =\<^sub>A z \<rightarrow> x =\<^sub>A z: U(i)"
+        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>x y z. \<lbrakk>x: A; y: A; z: A\<rbrakk> \<Longrightarrow> y =\<^sub>A z \<rightarrow> x =\<^sub>A z: U(i)"
-            by (rule Prod_form Equal_form assms | assumption)+
-        qed (rule assms)
-
-        show "\<And>x. x: A \<Longrightarrow> \<^bold>\<lambda>z p. ind\<^sub>= refl p: \<Prod>z:A. x =\<^sub>A z \<rightarrow> x =\<^sub>A z"
-        proof
-          show "\<And>x z. \<lbrakk>x: A; z: A\<rbrakk> \<Longrightarrow> \<^bold>\<lambda>p. ind\<^sub>= refl p: x =\<^sub>A z \<rightarrow> x =\<^sub>A z"
+          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 x z p assume asm: "x: A" "z: A" "p: x =\<^sub>A z"
-            show "ind\<^sub>= refl p: x =[A] z"
-            proof (rule Equal_elim[where ?x=x and ?y=z])
-              show "\<And>x y. \<lbrakk>x: A; y: A\<rbrakk> \<Longrightarrow> x =\<^sub>A y: U(i)" by standard (rule assms)
-              show "\<And>x. x: A \<Longrightarrow> refl x: x =\<^sub>A x" ..
-            qed (fact asm)+ }
-            show "\<And>x z. \<lbrakk>x: A; z: A\<rbrakk> \<Longrightarrow> x =\<^sub>A z: U(i)" by standard (rule assms)
-          qed
+            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 (rule asm)+ }
-      show "\<And>x y. \<lbrakk>x: A; y: A\<rbrakk> \<Longrightarrow> x =\<^sub>A y: U(i)" by standard (rule assms)
-    qed
+      qed (simple lem: assms 1 2 3)
+    qed (simple lem: assms 1 2)
   qed (rule assms)
-qed (fact 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 "rcompose`x`y`p`z`q: x =\<^sub>A z"
-  by standard+ (rule rcompose_type assms)+
+  by (simple lem: assms rcompose_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 rcompose_comp:
+  assumes "A: U(i)" and "a: A"
+  shows "rcompose`a`a`refl(a)`a`refl(a) \<equiv> refl(a)"
+unfolding rcompose_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 compose :: "[Term, Term] \<Rightarrow> Term"  (infixl "\<bullet>" 60) where
-  compose_comp: "\<lbrakk>
+  compose_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> rcompose`x`y`p`z`q"
 
 
+lemma compose_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 compose_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 rcompose_type)
+
 lemma compose_comp:
-  assumes "A : U(i)" and "a : A" shows "compose[A,a,a,a]`refl(a)`refl(a) \<equiv> refl(a)"
-  unfolding rcompose_def
-  by (simplify lems: assms)
+  assumes "A : U(i)" and "a : A" shows "refl(a) \<bullet> refl(a) \<equiv> refl(a)"
+by (subst compose_def) (simple lem: assms rcompose_comp)
 
 
-lemmas EqualProps_rules [intro] = inv_type inv_comp compose_type compose_comp
+lemmas EqualProps_rules [intro] = inv_type compose_type
 lemmas EqualProps_comps [comp] = inv_comp compose_comp