From e6473c383b479610cee4c0119e5811a12a938cf9 Mon Sep 17 00:00:00 2001
From: Josh Chen
Date: Tue, 14 Aug 2018 17:43:03 +0200
Subject: Well-formation rules are back in the methods; new theory synthesizing
 the natural number predecessor function.

---
 ex/Synthesis.thy | 81 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 81 insertions(+)
 create mode 100644 ex/Synthesis.thy

(limited to 'ex')

diff --git a/ex/Synthesis.thy b/ex/Synthesis.thy
new file mode 100644
index 0000000..60655e5
--- /dev/null
+++ b/ex/Synthesis.thy
@@ -0,0 +1,81 @@
+(*  Title:  HoTT/ex/Synthesis.thy
+    Author: Josh Chen
+    Date:   Aug 2018
+
+Examples of synthesis.
+*)
+
+
+theory Synthesis
+  imports "../HoTT"
+begin
+
+
+section \<open>Synthesis of the predecessor function\<close>
+
+text "
+  In this example we construct, with the help of Isabelle, a predecessor function for the natural numbers.
+
+  This is also done in \<open>CTT.thy\<close>; there the work is easier as the equality type is extensional, and also the methods are set up a little more nicely.
+"
+
+text "
+  First we show that the property we want is well-defined.
+"
+
+lemma pred_welltyped: "\<Sum>pred:\<nat>\<rightarrow>\<nat> . ((pred`0) =\<^sub>\<nat> 0) \<times> (\<Prod>n:\<nat>. (pred`(succ n)) =\<^sub>\<nat> n): U(O)"
+by simple
+
+text "
+  Now we look for an inhabitant of this type.
+  Observe that we're looking for a lambda term \<open>pred\<close> satisfying \<open>(pred`0) =\<^sub>\<nat> 0\<close> and \<open>\<Prod>n:\<nat>. (pred`(succ n)) =\<^sub>\<nat> n\<close>.
+  What if we require **definitional** equality instead of just propositional equality?
+"
+
+schematic_goal "?p`0 \<equiv> 0" and "\<And>n. n: \<nat> \<Longrightarrow> (?p`(succ n)) \<equiv> n"
+apply (subst comp, rule Nat_rules)
+prefer 3 apply (subst comp, rule Nat_rules)
+prefer 3 apply (rule Nat_rules)
+prefer 8 apply (rule Nat_rules | assumption)+
+done
+
+text "
+  The above proof finds a candidate, namely \<open>\<^bold>\<lambda>n. ind\<^sub>\<nat> (\<lambda>a b. a) n n\<close>.
+  We prove this has the required type and properties.
+"
+
+definition pred :: Term where "pred \<equiv> \<^bold>\<lambda>n. ind\<^sub>\<nat> (\<lambda>a b. a) n n"
+
+lemma pred_type: "pred: \<nat> \<rightarrow> \<nat>" unfolding pred_def by simple
+
+lemma pred_props: "<refl(0), \<^bold>\<lambda>n. refl(n)>: ((pred`0) =\<^sub>\<nat> 0) \<times> (\<Prod>n:\<nat>. (pred`(succ n)) =\<^sub>\<nat> n)"
+proof (simple lem: pred_type)
+  have *: "pred`0 \<equiv> 0" unfolding pred_def
+  proof (subst comp)
+    show "\<And>n. n: \<nat> \<Longrightarrow> ind\<^sub>\<nat> (\<lambda>a b. a) n n: \<nat>" by simple
+    show "ind\<^sub>\<nat> (\<lambda>a b. a) 0 0 \<equiv> 0"
+    proof (rule Nat_comps)
+      show "\<nat>: U(O)" ..
+    qed simple
+  qed rule
+  then show "refl(0): (pred`0) =\<^sub>\<nat> 0" by (subst *) simple
+
+  show "\<^bold>\<lambda>n. refl(n): \<Prod>n:\<nat>. (pred`(succ(n))) =\<^sub>\<nat> n"
+  unfolding pred_def proof
+    show "\<And>n. n: \<nat> \<Longrightarrow> refl(n): ((\<^bold>\<lambda>n. ind\<^sub>\<nat> (\<lambda>a b. a) n n)`succ(n)) =\<^sub>\<nat> n"
+    proof (subst comp)
+      show "\<And>n. n: \<nat> \<Longrightarrow> ind\<^sub>\<nat> (\<lambda>a b. a) n n: \<nat>" by simple
+      show "\<And>n. n: \<nat> \<Longrightarrow> refl(n): ind\<^sub>\<nat> (\<lambda>a b. a) (succ n) (succ n) =\<^sub>\<nat> n"
+      proof (subst comp)
+        show "\<nat>: U(O)" ..
+      qed simple
+    qed rule
+  qed rule
+qed
+
+theorem
+  "<pred, <refl(0), \<^bold>\<lambda>n. refl(n)>>: \<Sum>pred:\<nat>\<rightarrow>\<nat> . ((pred`0) =\<^sub>\<nat> 0) \<times> (\<Prod>n:\<nat>. (pred`(succ n)) =\<^sub>\<nat> n)"
+by (simple lem: pred_welltyped pred_type pred_props)
+
+
+end
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