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(* Title: HoTT/ex/Synthesis.thy
Author: Josh Chen
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 routine
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 compute
prefer 4 apply compute
prefer 3 apply compute
apply (rule Nat_routine Nat_elim | compute | assumption)+
done
text "
The above proof finds a candidate, namely \<open>\<^bold>\<lambda>n. ind\<^sub>\<nat> (\<lambda>a b. a) 0 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) 0 n"
lemma pred_type: "pred: \<nat> \<rightarrow> \<nat>" unfolding pred_def by routine
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 (routine lems: pred_type)
have *: "pred`0 \<equiv> 0" unfolding pred_def
proof compute
show "\<And>n. n: \<nat> \<Longrightarrow> ind\<^sub>\<nat> (\<lambda>a b. a) 0 n: \<nat>" by routine
show "ind\<^sub>\<nat> (\<lambda>a b. a) 0 0 \<equiv> 0"
proof compute
show "\<nat>: U(O)" ..
qed routine
qed rule
then show "refl(0): (pred`0) =\<^sub>\<nat> 0" by (subst *) routine
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) 0 n)`succ(n)) =\<^sub>\<nat> n"
proof compute
show "\<And>n. n: \<nat> \<Longrightarrow> ind\<^sub>\<nat> (\<lambda>a b. a) 0 n: \<nat>" by routine
show "\<And>n. n: \<nat> \<Longrightarrow> refl(n): ind\<^sub>\<nat> (\<lambda>a b. a) 0 (succ n) =\<^sub>\<nat> n"
proof compute
show "\<nat>: U(O)" ..
qed routine
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 (routine lems: pred_welltyped pred_type pred_props)
end
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