more explicit HOL-Proofs sessions, including former ex/Hilbert_Classical.thy which works in parallel mode without the antiquotation option "margin" (which is still critical);
(* Title: HOL/Proofs/Extraction/Pigeonhole.thy
Author: Stefan Berghofer, TU Muenchen
*)
header {* The pigeonhole principle *}
theory Pigeonhole
imports Util Efficient_Nat
begin
text {*
We formalize two proofs of the pigeonhole principle, which lead
to extracted programs of quite different complexity. The original
formalization of these proofs in {\sc Nuprl} is due to
Aleksey Nogin \cite{Nogin-ENTCS-2000}.
This proof yields a polynomial program.
*}
theorem pigeonhole:
"\<And>f. (\<And>i. i \<le> Suc n \<Longrightarrow> f i \<le> n) \<Longrightarrow> \<exists>i j. i \<le> Suc n \<and> j < i \<and> f i = f j"
proof (induct n)
case 0
hence "Suc 0 \<le> Suc 0 \<and> 0 < Suc 0 \<and> f (Suc 0) = f 0" by simp
thus ?case by iprover
next
case (Suc n)
{
fix k
have
"k \<le> Suc (Suc n) \<Longrightarrow>
(\<And>i j. Suc k \<le> i \<Longrightarrow> i \<le> Suc (Suc n) \<Longrightarrow> j < i \<Longrightarrow> f i \<noteq> f j) \<Longrightarrow>
(\<exists>i j. i \<le> k \<and> j < i \<and> f i = f j)"
proof (induct k)
case 0
let ?f = "\<lambda>i. if f i = Suc n then f (Suc (Suc n)) else f i"
have "\<not> (\<exists>i j. i \<le> Suc n \<and> j < i \<and> ?f i = ?f j)"
proof
assume "\<exists>i j. i \<le> Suc n \<and> j < i \<and> ?f i = ?f j"
then obtain i j where i: "i \<le> Suc n" and j: "j < i"
and f: "?f i = ?f j" by iprover
from j have i_nz: "Suc 0 \<le> i" by simp
from i have iSSn: "i \<le> Suc (Suc n)" by simp
have S0SSn: "Suc 0 \<le> Suc (Suc n)" by simp
show False
proof cases
assume fi: "f i = Suc n"
show False
proof cases
assume fj: "f j = Suc n"
from i_nz and iSSn and j have "f i \<noteq> f j" by (rule 0)
moreover from fi have "f i = f j"
by (simp add: fj [symmetric])
ultimately show ?thesis ..
next
from i and j have "j < Suc (Suc n)" by simp
with S0SSn and le_refl have "f (Suc (Suc n)) \<noteq> f j"
by (rule 0)
moreover assume "f j \<noteq> Suc n"
with fi and f have "f (Suc (Suc n)) = f j" by simp
ultimately show False ..
qed
next
assume fi: "f i \<noteq> Suc n"
show False
proof cases
from i have "i < Suc (Suc n)" by simp
with S0SSn and le_refl have "f (Suc (Suc n)) \<noteq> f i"
by (rule 0)
moreover assume "f j = Suc n"
with fi and f have "f (Suc (Suc n)) = f i" by simp
ultimately show False ..
next
from i_nz and iSSn and j
have "f i \<noteq> f j" by (rule 0)
moreover assume "f j \<noteq> Suc n"
with fi and f have "f i = f j" by simp
ultimately show False ..
qed
qed
qed
moreover have "\<And>i. i \<le> Suc n \<Longrightarrow> ?f i \<le> n"
proof -
fix i assume "i \<le> Suc n"
hence i: "i < Suc (Suc n)" by simp
have "f (Suc (Suc n)) \<noteq> f i"
by (rule 0) (simp_all add: i)
moreover have "f (Suc (Suc n)) \<le> Suc n"
by (rule Suc) simp
moreover from i have "i \<le> Suc (Suc n)" by simp
hence "f i \<le> Suc n" by (rule Suc)
ultimately show "?thesis i"
by simp
qed
hence "\<exists>i j. i \<le> Suc n \<and> j < i \<and> ?f i = ?f j"
by (rule Suc)
ultimately show ?case ..
next
case (Suc k)
from search [OF nat_eq_dec] show ?case
proof
assume "\<exists>j<Suc k. f (Suc k) = f j"
thus ?case by (iprover intro: le_refl)
next
assume nex: "\<not> (\<exists>j<Suc k. f (Suc k) = f j)"
have "\<exists>i j. i \<le> k \<and> j < i \<and> f i = f j"
proof (rule Suc)
from Suc show "k \<le> Suc (Suc n)" by simp
fix i j assume k: "Suc k \<le> i" and i: "i \<le> Suc (Suc n)"
and j: "j < i"
show "f i \<noteq> f j"
proof cases
assume eq: "i = Suc k"
show ?thesis
proof
assume "f i = f j"
hence "f (Suc k) = f j" by (simp add: eq)
with nex and j and eq show False by iprover
qed
next
assume "i \<noteq> Suc k"
with k have "Suc (Suc k) \<le> i" by simp
thus ?thesis using i and j by (rule Suc)
qed
qed
thus ?thesis by (iprover intro: le_SucI)
qed
qed
}
note r = this
show ?case by (rule r) simp_all
qed
text {*
The following proof, although quite elegant from a mathematical point of view,
leads to an exponential program:
*}
theorem pigeonhole_slow:
"\<And>f. (\<And>i. i \<le> Suc n \<Longrightarrow> f i \<le> n) \<Longrightarrow> \<exists>i j. i \<le> Suc n \<and> j < i \<and> f i = f j"
proof (induct n)
case 0
have "Suc 0 \<le> Suc 0" ..
moreover have "0 < Suc 0" ..
moreover from 0 have "f (Suc 0) = f 0" by simp
ultimately show ?case by iprover
next
case (Suc n)
from search [OF nat_eq_dec] show ?case
proof
assume "\<exists>j < Suc (Suc n). f (Suc (Suc n)) = f j"
thus ?case by (iprover intro: le_refl)
next
assume "\<not> (\<exists>j < Suc (Suc n). f (Suc (Suc n)) = f j)"
hence nex: "\<forall>j < Suc (Suc n). f (Suc (Suc n)) \<noteq> f j" by iprover
let ?f = "\<lambda>i. if f i = Suc n then f (Suc (Suc n)) else f i"
have "\<And>i. i \<le> Suc n \<Longrightarrow> ?f i \<le> n"
proof -
fix i assume i: "i \<le> Suc n"
show "?thesis i"
proof (cases "f i = Suc n")
case True
from i and nex have "f (Suc (Suc n)) \<noteq> f i" by simp
with True have "f (Suc (Suc n)) \<noteq> Suc n" by simp
moreover from Suc have "f (Suc (Suc n)) \<le> Suc n" by simp
ultimately have "f (Suc (Suc n)) \<le> n" by simp
with True show ?thesis by simp
next
case False
from Suc and i have "f i \<le> Suc n" by simp
with False show ?thesis by simp
qed
qed
hence "\<exists>i j. i \<le> Suc n \<and> j < i \<and> ?f i = ?f j" by (rule Suc)
then obtain i j where i: "i \<le> Suc n" and ji: "j < i" and f: "?f i = ?f j"
by iprover
have "f i = f j"
proof (cases "f i = Suc n")
case True
show ?thesis
proof (cases "f j = Suc n")
assume "f j = Suc n"
with True show ?thesis by simp
next
assume "f j \<noteq> Suc n"
moreover from i ji nex have "f (Suc (Suc n)) \<noteq> f j" by simp
ultimately show ?thesis using True f by simp
qed
next
case False
show ?thesis
proof (cases "f j = Suc n")
assume "f j = Suc n"
moreover from i nex have "f (Suc (Suc n)) \<noteq> f i" by simp
ultimately show ?thesis using False f by simp
next
assume "f j \<noteq> Suc n"
with False f show ?thesis by simp
qed
qed
moreover from i have "i \<le> Suc (Suc n)" by simp
ultimately show ?thesis using ji by iprover
qed
qed
extract pigeonhole pigeonhole_slow
text {*
The programs extracted from the above proofs look as follows:
@{thm [display] pigeonhole_def}
@{thm [display] pigeonhole_slow_def}
The program for searching for an element in an array is
@{thm [display,eta_contract=false] search_def}
The correctness statement for @{term "pigeonhole"} is
@{thm [display] pigeonhole_correctness [no_vars]}
In order to analyze the speed of the above programs,
we generate ML code from them.
*}
instantiation nat :: default
begin
definition "default = (0::nat)"
instance ..
end
instantiation prod :: (default, default) default
begin
definition "default = (default, default)"
instance ..
end
definition
"test n u = pigeonhole n (\<lambda>m. m - 1)"
definition
"test' n u = pigeonhole_slow n (\<lambda>m. m - 1)"
definition
"test'' u = pigeonhole 8 (op ! [0, 1, 2, 3, 4, 5, 6, 3, 7, 8])"
ML "timeit (@{code test} 10)"
ML "timeit (@{code test'} 10)"
ML "timeit (@{code test} 20)"
ML "timeit (@{code test'} 20)"
ML "timeit (@{code test} 25)"
ML "timeit (@{code test'} 25)"
ML "timeit (@{code test} 500)"
ML "timeit @{code test''}"
consts_code
"default :: nat" ("{* 0::nat *}")
"default :: nat \<times> nat" ("{* (0::nat, 0::nat) *}")
code_module PH
contains
test = test
test' = test'
test'' = test''
ML "timeit (PH.test 10)"
ML "timeit (PH.test' 10)"
ML "timeit (PH.test 20)"
ML "timeit (PH.test' 20)"
ML "timeit (PH.test 25)"
ML "timeit (PH.test' 25)"
ML "timeit (PH.test 500)"
ML "timeit PH.test''"
end