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section \<open>Ackermann's Function and the PR Functions\<close>
text \<open>
This proof has been adopted from a development by Nora Szasz \cite{szasz93}.
\medskip
\<close>
theory Primrec imports Main begin
subsection\<open>Ackermann's Function\<close>
fun ack :: "[nat,nat] \<Rightarrow> nat" where
"ack 0 n = Suc n"
| "ack (Suc m) 0 = ack m 1"
| "ack (Suc m) (Suc n) = ack m (ack (Suc m) n)"
text \<open>PROPERTY A 4\<close>
lemma less_ack2 [iff]: "j < ack i j"
by (induct i j rule: ack.induct) simp_all
text \<open>PROPERTY A 5-, the single-step lemma\<close>
lemma ack_less_ack_Suc2 [iff]: "ack i j < ack i (Suc j)"
by (induct i j rule: ack.induct) simp_all
text \<open>PROPERTY A 5, monotonicity for \<open><\<close>\<close>
lemma ack_less_mono2: "j < k \<Longrightarrow> ack i j < ack i k"
by (simp add: lift_Suc_mono_less)
text \<open>PROPERTY A 5', monotonicity for \<open>\<le>\<close>\<close>
lemma ack_le_mono2: "j \<le> k \<Longrightarrow> ack i j \<le> ack i k"
by (simp add: ack_less_mono2 less_mono_imp_le_mono)
text \<open>PROPERTY A 6\<close>
lemma ack2_le_ack1 [iff]: "ack i (Suc j) \<le> ack (Suc i) j"
proof (induct j)
case 0 show ?case by simp
next
case (Suc j) show ?case
by (metis Suc ack.simps(3) ack_le_mono2 le_trans less_ack2 less_eq_Suc_le)
qed
text \<open>PROPERTY A 7-, the single-step lemma\<close>
lemma ack_less_ack_Suc1 [iff]: "ack i j < ack (Suc i) j"
by (blast intro: ack_less_mono2 less_le_trans)
text \<open>PROPERTY A 4'? Extra lemma needed for \<^term>\<open>CONSTANT\<close> case, constant functions\<close>
lemma less_ack1 [iff]: "i < ack i j"
proof (induct i)
case 0
then show ?case
by simp
next
case (Suc i)
then show ?case
using less_trans_Suc by blast
qed
text \<open>PROPERTY A 8\<close>
lemma ack_1 [simp]: "ack (Suc 0) j = j + 2"
by (induct j) simp_all
text \<open>PROPERTY A 9. The unary \<open>1\<close> and \<open>2\<close> in \<^term>\<open>ack\<close> is essential for the rewriting.\<close>
lemma ack_2 [simp]: "ack (Suc (Suc 0)) j = 2 * j + 3"
by (induct j) simp_all
text \<open>Added in 2022 just for fun\<close>
lemma ack_3: "ack (Suc (Suc (Suc 0))) j = 2 ^ (j+3) - 3"
proof (induct j)
case 0
then show ?case by simp
next
case (Suc j)
with less_le_trans show ?case
by (fastforce simp add: power_add algebra_simps)
qed
text \<open>PROPERTY A 7, monotonicity for \<open><\<close> [not clear why
@{thm [source] ack_1} is now needed first!]\<close>
lemma ack_less_mono1_aux: "ack i k < ack (Suc (i +i')) k"
proof (induct i k rule: ack.induct)
case (1 n) show ?case
using less_le_trans by auto
next
case (2 m) thus ?case by simp
next
case (3 m n) thus ?case
using ack_less_mono2 less_trans by fastforce
qed
lemma ack_less_mono1: "i < j \<Longrightarrow> ack i k < ack j k"
using ack_less_mono1_aux less_iff_Suc_add by auto
text \<open>PROPERTY A 7', monotonicity for \<open>\<le>\<close>\<close>
lemma ack_le_mono1: "i \<le> j \<Longrightarrow> ack i k \<le> ack j k"
using ack_less_mono1 le_eq_less_or_eq by auto
text \<open>PROPERTY A 10\<close>
lemma ack_nest_bound: "ack i1 (ack i2 j) < ack (2 + (i1 + i2)) j"
proof -
have "ack i1 (ack i2 j) < ack (i1 + i2) (ack (Suc (i1 + i2)) j)"
by (meson ack_le_mono1 ack_less_mono1 ack_less_mono2 le_add1 le_trans less_add_Suc2 not_less)
also have "... = ack (Suc (i1 + i2)) (Suc j)"
by simp
also have "... \<le> ack (2 + (i1 + i2)) j"
using ack2_le_ack1 add_2_eq_Suc by presburger
finally show ?thesis .
qed
text \<open>PROPERTY A 11\<close>
lemma ack_add_bound: "ack i1 j + ack i2 j < ack (4 + (i1 + i2)) j"
proof -
have "ack i1 j \<le> ack (i1 + i2) j" "ack i2 j \<le> ack (i1 + i2) j"
by (simp_all add: ack_le_mono1)
then have "ack i1 j + ack i2 j < ack (Suc (Suc 0)) (ack (i1 + i2) j)"
by simp
also have "... < ack (4 + (i1 + i2)) j"
by (metis ack_nest_bound add.assoc numeral_2_eq_2 numeral_Bit0)
finally show ?thesis .
qed
text \<open>PROPERTY A 12. Article uses existential quantifier but the ALF proof
used \<open>k + 4\<close>. Quantified version must be nested \<open>\<exists>k'. \<forall>i j. ...\<close>\<close>
lemma ack_add_bound2:
assumes "i < ack k j" shows "i + j < ack (4 + k) j"
proof -
have "i + j < ack k j + ack 0 j"
using assms by auto
also have "... < ack (4 + k) j"
by (metis ack_add_bound add.right_neutral)
finally show ?thesis .
qed
subsection\<open>Primitive Recursive Functions\<close>
primrec hd0 :: "nat list \<Rightarrow> nat" where
"hd0 [] = 0"
| "hd0 (m # ms) = m"
text \<open>Inductive definition of the set of primitive recursive functions of type \<^typ>\<open>nat list \<Rightarrow> nat\<close>.\<close>
definition SC :: "nat list \<Rightarrow> nat"
where "SC l = Suc (hd0 l)"
definition CONSTANT :: "nat \<Rightarrow> nat list \<Rightarrow> nat"
where "CONSTANT k l = k"
definition PROJ :: "nat \<Rightarrow> nat list \<Rightarrow> nat"
where "PROJ i l = hd0 (drop i l)"
definition COMP :: "[nat list \<Rightarrow> nat, (nat list \<Rightarrow> nat) list, nat list] \<Rightarrow> nat"
where "COMP g fs l = g (map (\<lambda>f. f l) fs)"
fun PREC :: "[nat list \<Rightarrow> nat, nat list \<Rightarrow> nat, nat list] \<Rightarrow> nat"
where
"PREC f g [] = 0"
| "PREC f g (x # l) = rec_nat (f l) (\<lambda>y r. g (r # y # l)) x"
\<comment> \<open>Note that \<^term>\<open>g\<close> is applied first to \<^term>\<open>PREC f g y\<close> and then to \<^term>\<open>y\<close>!\<close>
inductive PRIMREC :: "(nat list \<Rightarrow> nat) \<Rightarrow> bool" where
SC: "PRIMREC SC"
| CONSTANT: "PRIMREC (CONSTANT k)"
| PROJ: "PRIMREC (PROJ i)"
| COMP: "PRIMREC g \<Longrightarrow> \<forall>f \<in> set fs. PRIMREC f \<Longrightarrow> PRIMREC (COMP g fs)"
| PREC: "PRIMREC f \<Longrightarrow> PRIMREC g \<Longrightarrow> PRIMREC (PREC f g)"
text \<open>Useful special cases of evaluation\<close>
lemma SC [simp]: "SC (x # l) = Suc x"
by (simp add: SC_def)
lemma PROJ_0 [simp]: "PROJ 0 (x # l) = x"
by (simp add: PROJ_def)
lemma COMP_1 [simp]: "COMP g [f] l = g [f l]"
by (simp add: COMP_def)
lemma PREC_0: "PREC f g (0 # l) = f l"
by simp
lemma PREC_Suc [simp]: "PREC f g (Suc x # l) = g (PREC f g (x # l) # x # l)"
by auto
subsection \<open>Main Result: Ackermann's Function is not Primitive Recursive\<close>
lemma SC_case: "SC l < ack 1 (sum_list l)"
unfolding SC_def
by (induct l) (simp_all add: le_add1 le_imp_less_Suc)
lemma CONSTANT_case: "CONSTANT k l < ack k (sum_list l)"
by (simp add: CONSTANT_def)
lemma PROJ_case: "PROJ i l < ack 0 (sum_list l)"
unfolding PROJ_def
proof (induct l arbitrary: i)
case Nil
then show ?case
by simp
next
case (Cons a l)
then show ?case
by (metis ack.simps(1) add.commute drop_Cons' hd0.simps(2) leD leI lessI not_less_eq sum_list.Cons trans_le_add2)
qed
text \<open>\<^term>\<open>COMP\<close> case\<close>
lemma COMP_map_aux: "\<forall>f \<in> set fs. PRIMREC f \<and> (\<exists>kf. \<forall>l. f l < ack kf (sum_list l))
\<Longrightarrow> \<exists>k. \<forall>l. sum_list (map (\<lambda>f. f l) fs) < ack k (sum_list l)"
proof (induct fs)
case Nil
then show ?case
by auto
next
case (Cons a fs)
then show ?case
by simp (blast intro: add_less_mono ack_add_bound less_trans)
qed
lemma COMP_case:
assumes 1: "\<forall>l. g l < ack kg (sum_list l)"
and 2: "\<forall>f \<in> set fs. PRIMREC f \<and> (\<exists>kf. \<forall>l. f l < ack kf (sum_list l))"
shows "\<exists>k. \<forall>l. COMP g fs l < ack k (sum_list l)"
unfolding COMP_def
using 1 COMP_map_aux [OF 2] by (meson ack_less_mono2 ack_nest_bound less_trans)
text \<open>\<^term>\<open>PREC\<close> case\<close>
lemma PREC_case_aux:
assumes f: "\<And>l. f l + sum_list l < ack kf (sum_list l)"
and g: "\<And>l. g l + sum_list l < ack kg (sum_list l)"
shows "PREC f g l + sum_list l < ack (Suc (kf + kg)) (sum_list l)"
proof (cases l)
case Nil
then show ?thesis
by (simp add: Suc_lessD)
next
case (Cons m l)
have "rec_nat (f l) (\<lambda>y r. g (r # y # l)) m + (m + sum_list l) < ack (Suc (kf + kg)) (m + sum_list l)"
proof (induct m)
case 0
then show ?case
using ack_less_mono1_aux f less_trans by fastforce
next
case (Suc m)
let ?r = "rec_nat (f l) (\<lambda>y r. g (r # y # l)) m"
have "\<not> g (?r # m # l) + sum_list (?r # m # l) < g (?r # m # l) + (m + sum_list l)"
by force
then have "g (?r # m # l) + (m + sum_list l) < ack kg (sum_list (?r # m # l))"
by (meson assms(2) leI less_le_trans)
moreover
have "... < ack (kf + kg) (ack (Suc (kf + kg)) (m + sum_list l))"
using Suc.hyps by simp (meson ack_le_mono1 ack_less_mono2 le_add2 le_less_trans)
ultimately show ?case
by auto
qed
then show ?thesis
by (simp add: local.Cons)
qed
proposition PREC_case:
"\<lbrakk>\<And>l. f l < ack kf (sum_list l); \<And>l. g l < ack kg (sum_list l)\<rbrakk>
\<Longrightarrow> \<exists>k. \<forall>l. PREC f g l < ack k (sum_list l)"
by (metis le_less_trans [OF le_add1 PREC_case_aux] ack_add_bound2)
lemma ack_bounds_PRIMREC: "PRIMREC f \<Longrightarrow> \<exists>k. \<forall>l. f l < ack k (sum_list l)"
by (erule PRIMREC.induct) (blast intro: SC_case CONSTANT_case PROJ_case COMP_case PREC_case)+
theorem ack_not_PRIMREC:
"\<not> PRIMREC (\<lambda>l. case l of [] \<Rightarrow> 0 | x # l' \<Rightarrow> ack x x)"
proof
assume *: "PRIMREC (\<lambda>l. case l of [] \<Rightarrow> 0 | x # l' \<Rightarrow> ack x x)"
then obtain m where m: "\<And>l. (case l of [] \<Rightarrow> 0 | x # l' \<Rightarrow> ack x x) < ack m (sum_list l)"
using ack_bounds_PRIMREC by metis
show False
using m [of "[m]"] by simp
qed
end
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