(* Title: HOL/ex/Fundefs.thy
Author: Alexander Krauss, TU Muenchen
*)
section \<open>Examples of function definitions\<close>
theory Fundefs
imports Main "~~/src/HOL/Library/Monad_Syntax"
begin
subsection \<open>Very basic\<close>
fun fib :: "nat \<Rightarrow> nat"
where
"fib 0 = 1"
| "fib (Suc 0) = 1"
| "fib (Suc (Suc n)) = fib n + fib (Suc n)"
text \<open>partial simp and induction rules:\<close>
thm fib.psimps
thm fib.pinduct
text \<open>There is also a cases rule to distinguish cases along the definition\<close>
thm fib.cases
text \<open>total simp and induction rules:\<close>
thm fib.simps
thm fib.induct
text \<open>elimination rules\<close>
thm fib.elims
subsection \<open>Currying\<close>
fun add
where
"add 0 y = y"
| "add (Suc x) y = Suc (add x y)"
thm add.simps
thm add.induct -- \<open>Note the curried induction predicate\<close>
subsection \<open>Nested recursion\<close>
function nz
where
"nz 0 = 0"
| "nz (Suc x) = nz (nz x)"
by pat_completeness auto
lemma nz_is_zero: -- \<open>A lemma we need to prove termination\<close>
assumes trm: "nz_dom x"
shows "nz x = 0"
using trm
by induct (auto simp: nz.psimps)
termination nz
by (relation "less_than") (auto simp:nz_is_zero)
thm nz.simps
thm nz.induct
text \<open>Here comes McCarthy's 91-function\<close>
function f91 :: "nat => nat"
where
"f91 n = (if 100 < n then n - 10 else f91 (f91 (n + 11)))"
by pat_completeness auto
(* Prove a lemma before attempting a termination proof *)
lemma f91_estimate:
assumes trm: "f91_dom n"
shows "n < f91 n + 11"
using trm by induct (auto simp: f91.psimps)
termination
proof
let ?R = "measure (%x. 101 - x)"
show "wf ?R" ..
fix n::nat assume "~ 100 < n" (* Inner call *)
thus "(n + 11, n) : ?R" by simp
assume inner_trm: "f91_dom (n + 11)" (* Outer call *)
with f91_estimate have "n + 11 < f91 (n + 11) + 11" .
with \<open>~ 100 < n\<close> show "(f91 (n + 11), n) : ?R" by simp
qed
text\<open>Now trivial (even though it does not belong here):\<close>
lemma "f91 n = (if 100 < n then n - 10 else 91)"
by (induct n rule:f91.induct) auto
subsection \<open>More general patterns\<close>
subsubsection \<open>Overlapping patterns\<close>
text \<open>Currently, patterns must always be compatible with each other, since
no automatic splitting takes place. But the following definition of
gcd is ok, although patterns overlap:\<close>
fun gcd2 :: "nat \<Rightarrow> nat \<Rightarrow> nat"
where
"gcd2 x 0 = x"
| "gcd2 0 y = y"
| "gcd2 (Suc x) (Suc y) = (if x < y then gcd2 (Suc x) (y - x)
else gcd2 (x - y) (Suc y))"
thm gcd2.simps
thm gcd2.induct
subsubsection \<open>Guards\<close>
text \<open>We can reformulate the above example using guarded patterns\<close>
function gcd3 :: "nat \<Rightarrow> nat \<Rightarrow> nat"
where
"gcd3 x 0 = x"
| "gcd3 0 y = y"
| "x < y \<Longrightarrow> gcd3 (Suc x) (Suc y) = gcd3 (Suc x) (y - x)"
| "\<not> x < y \<Longrightarrow> gcd3 (Suc x) (Suc y) = gcd3 (x - y) (Suc y)"
apply (case_tac x, case_tac a, auto)
apply (case_tac ba, auto)
done
termination by lexicographic_order
thm gcd3.simps
thm gcd3.induct
text \<open>General patterns allow even strange definitions:\<close>
function ev :: "nat \<Rightarrow> bool"
where
"ev (2 * n) = True"
| "ev (2 * n + 1) = False"
proof - -- \<open>completeness is more difficult here \dots\<close>
fix P :: bool
and x :: nat
assume c1: "\<And>n. x = 2 * n \<Longrightarrow> P"
and c2: "\<And>n. x = 2 * n + 1 \<Longrightarrow> P"
have divmod: "x = 2 * (x div 2) + (x mod 2)" by auto
show "P"
proof cases
assume "x mod 2 = 0"
with divmod have "x = 2 * (x div 2)" by simp
with c1 show "P" .
next
assume "x mod 2 \<noteq> 0"
hence "x mod 2 = 1" by simp
with divmod have "x = 2 * (x div 2) + 1" by simp
with c2 show "P" .
qed
qed presburger+ -- \<open>solve compatibility with presburger\<close>
termination by lexicographic_order
thm ev.simps
thm ev.induct
thm ev.cases
subsection \<open>Mutual Recursion\<close>
fun evn od :: "nat \<Rightarrow> bool"
where
"evn 0 = True"
| "od 0 = False"
| "evn (Suc n) = od n"
| "od (Suc n) = evn n"
thm evn.simps
thm od.simps
thm evn_od.induct
thm evn_od.termination
thm evn.elims
thm od.elims
subsection \<open>Definitions in local contexts\<close>
locale my_monoid =
fixes opr :: "'a \<Rightarrow> 'a \<Rightarrow> 'a"
and un :: "'a"
assumes assoc: "opr (opr x y) z = opr x (opr y z)"
and lunit: "opr un x = x"
and runit: "opr x un = x"
begin
fun foldR :: "'a list \<Rightarrow> 'a"
where
"foldR [] = un"
| "foldR (x#xs) = opr x (foldR xs)"
fun foldL :: "'a list \<Rightarrow> 'a"
where
"foldL [] = un"
| "foldL [x] = x"
| "foldL (x#y#ys) = foldL (opr x y # ys)"
thm foldL.simps
lemma foldR_foldL: "foldR xs = foldL xs"
by (induct xs rule: foldL.induct) (auto simp:lunit runit assoc)
thm foldR_foldL
end
thm my_monoid.foldL.simps
thm my_monoid.foldR_foldL
subsection \<open>@{text fun_cases}\<close>
subsubsection \<open>Predecessor\<close>
fun pred :: "nat \<Rightarrow> nat" where
"pred 0 = 0" |
"pred (Suc n) = n"
thm pred.elims
lemma assumes "pred x = y"
obtains "x = 0" "y = 0" | "n" where "x = Suc n" "y = n"
by (fact pred.elims[OF assms])
text \<open>If the predecessor of a number is 0, that number must be 0 or 1.\<close>
fun_cases pred0E[elim]: "pred n = 0"
lemma "pred n = 0 \<Longrightarrow> n = 0 \<or> n = Suc 0"
by (erule pred0E) metis+
text \<open>Other expressions on the right-hand side also work, but whether the
generated rule is useful depends on how well the simplifier can
simplify it. This example works well:\<close>
fun_cases pred42E[elim]: "pred n = 42"
lemma "pred n = 42 \<Longrightarrow> n = 43"
by (erule pred42E)
subsubsection \<open>List to option\<close>
fun list_to_option :: "'a list \<Rightarrow> 'a option" where
"list_to_option [x] = Some x" |
"list_to_option _ = None"
fun_cases list_to_option_NoneE: "list_to_option xs = None"
and list_to_option_SomeE: "list_to_option xs = Some x"
lemma "list_to_option xs = Some y \<Longrightarrow> xs = [y]"
by (erule list_to_option_SomeE)
subsubsection \<open>Boolean Functions\<close>
fun xor :: "bool \<Rightarrow> bool \<Rightarrow> bool" where
"xor False False = False" |
"xor True True = False" |
"xor _ _ = True"
thm xor.elims
text \<open>@{text fun_cases} does not only recognise function equations, but also works with
functions that return a boolean, e.g.:\<close>
fun_cases xor_TrueE: "xor a b" and xor_FalseE: "\<not>xor a b"
print_theorems
subsubsection \<open>Many parameters\<close>
fun sum4 :: "nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> nat" where
"sum4 a b c d = a + b + c + d"
fun_cases sum40E: "sum4 a b c d = 0"
lemma "sum4 a b c d = 0 \<Longrightarrow> a = 0"
by (erule sum40E)
subsection \<open>Partial Function Definitions\<close>
text \<open>Partial functions in the option monad:\<close>
partial_function (option)
collatz :: "nat \<Rightarrow> nat list option"
where
"collatz n =
(if n \<le> 1 then Some [n]
else if even n
then do { ns \<leftarrow> collatz (n div 2); Some (n # ns) }
else do { ns \<leftarrow> collatz (3 * n + 1); Some (n # ns)})"
declare collatz.simps[code]
value "collatz 23"
text \<open>Tail-recursive functions:\<close>
partial_function (tailrec) fixpoint :: "('a \<Rightarrow> 'a) \<Rightarrow> 'a \<Rightarrow> 'a"
where
"fixpoint f x = (if f x = x then x else fixpoint f (f x))"
subsection \<open>Regression tests\<close>
text \<open>The following examples mainly serve as tests for the
function package\<close>
fun listlen :: "'a list \<Rightarrow> nat"
where
"listlen [] = 0"
| "listlen (x#xs) = Suc (listlen xs)"
(* Context recursion *)
fun f :: "nat \<Rightarrow> nat"
where
zero: "f 0 = 0"
| succ: "f (Suc n) = (if f n = 0 then 0 else f n)"
(* A combination of context and nested recursion *)
function h :: "nat \<Rightarrow> nat"
where
"h 0 = 0"
| "h (Suc n) = (if h n = 0 then h (h n) else h n)"
by pat_completeness auto
(* Context, but no recursive call: *)
fun i :: "nat \<Rightarrow> nat"
where
"i 0 = 0"
| "i (Suc n) = (if n = 0 then 0 else i n)"
(* Tupled nested recursion *)
fun fa :: "nat \<Rightarrow> nat \<Rightarrow> nat"
where
"fa 0 y = 0"
| "fa (Suc n) y = (if fa n y = 0 then 0 else fa n y)"
(* Let *)
fun j :: "nat \<Rightarrow> nat"
where
"j 0 = 0"
| "j (Suc n) = (let u = n in Suc (j u))"
(* There were some problems with fresh names\<dots> *)
function k :: "nat \<Rightarrow> nat"
where
"k x = (let a = x; b = x in k x)"
by pat_completeness auto
function f2 :: "(nat \<times> nat) \<Rightarrow> (nat \<times> nat)"
where
"f2 p = (let (x,y) = p in f2 (y,x))"
by pat_completeness auto
(* abbreviations *)
fun f3 :: "'a set \<Rightarrow> bool"
where
"f3 x = finite x"
(* Simple Higher-Order Recursion *)
datatype 'a tree =
Leaf 'a
| Branch "'a tree list"
fun treemap :: "('a \<Rightarrow> 'a) \<Rightarrow> 'a tree \<Rightarrow> 'a tree"
where
"treemap fn (Leaf n) = (Leaf (fn n))"
| "treemap fn (Branch l) = (Branch (map (treemap fn) l))"
fun tinc :: "nat tree \<Rightarrow> nat tree"
where
"tinc (Leaf n) = Leaf (Suc n)"
| "tinc (Branch l) = Branch (map tinc l)"
fun testcase :: "'a tree \<Rightarrow> 'a list"
where
"testcase (Leaf a) = [a]"
| "testcase (Branch x) =
(let xs = concat (map testcase x);
ys = concat (map testcase x) in
xs @ ys)"
(* Pattern matching on records *)
record point =
Xcoord :: int
Ycoord :: int
function swp :: "point \<Rightarrow> point"
where
"swp \<lparr> Xcoord = x, Ycoord = y \<rparr> = \<lparr> Xcoord = y, Ycoord = x \<rparr>"
proof -
fix P x
assume "\<And>xa y. x = \<lparr>Xcoord = xa, Ycoord = y\<rparr> \<Longrightarrow> P"
thus "P"
by (cases x)
qed auto
termination by rule auto
(* The diagonal function *)
fun diag :: "bool \<Rightarrow> bool \<Rightarrow> bool \<Rightarrow> nat"
where
"diag x True False = 1"
| "diag False y True = 2"
| "diag True False z = 3"
| "diag True True True = 4"
| "diag False False False = 5"
(* Many equations (quadratic blowup) *)
datatype DT =
A | B | C | D | E | F | G | H | I | J | K | L | M | N | P
| Q | R | S | T | U | V
fun big :: "DT \<Rightarrow> nat"
where
"big A = 0"
| "big B = 0"
| "big C = 0"
| "big D = 0"
| "big E = 0"
| "big F = 0"
| "big G = 0"
| "big H = 0"
| "big I = 0"
| "big J = 0"
| "big K = 0"
| "big L = 0"
| "big M = 0"
| "big N = 0"
| "big P = 0"
| "big Q = 0"
| "big R = 0"
| "big S = 0"
| "big T = 0"
| "big U = 0"
| "big V = 0"
(* automatic pattern splitting *)
fun
f4 :: "nat \<Rightarrow> nat \<Rightarrow> bool"
where
"f4 0 0 = True"
| "f4 _ _ = False"
(* polymorphic partial_function *)
partial_function (option) f5 :: "'a list \<Rightarrow> 'a option"
where
"f5 x = f5 x"
end