What is the easiest way to generate code for a sorting algorithm that sorts its argument in reverse order, while building on top of the existing List.sort?
I came up with two solutions that are shown below in my answer. But both of them are not really satisfactory.
Any other ideas how this could be done?
I came up with two possible solutions. But both have (severe) drawbacks. (I would have liked to obtain the result almost automatically.)
Introduce a Haskell-style newtype. E.g., if we wanted to sort lists of nats, something like
datatype 'a new = New (old : 'a)
instantiation new :: (linorder) linorder
begin
definition "less_eq_new x y ⟷ old x ≥ old y"
definition "less_new x y ⟷ old x > old y"
instance by (default, case_tac [!] x) (auto simp: less_eq_new_def less_new_def)
end
At this point
value [code] "sort_key New [0::nat, 1, 0, 0, 1, 2]"
yields the desired reverse sorting. While this is comparatively easy, it is not as automatic as I would like the solution to be and in addition has a small runtime overhead (since Isabelle doesn't have Haskell's newtype).
Via a locale for the dual of a linear order. First we more or less copy the existing code for insertion sort (but instead of relying on a type class, we make the parameter that represents the comparison explicit).
fun insort_by_key :: "('b ⇒ 'b ⇒ bool) ⇒ ('a ⇒ 'b) ⇒ 'a ⇒ 'a list ⇒ 'a list"
where
"insort_by_key P f x [] = [x]"
| "insort_by_key P f x (y # ys) =
(if P (f x) (f y) then x # y # ys else y # insort_by_key P f x ys)"
definition "revsort_key f xs = foldr (insort_by_key (op ≥) f) xs []"
at this point we have code for revsort_key.
value [code] "revsort_key id [0::nat, 1, 0, 0, 1, 2]"
but we also want all the nice results that have already been proved in the linorder locale (that derives from the linorder class). To this end, we introduce the dual of a linear order and use a "mixin" (not sure if I'm using the correct naming here) to replace all occurrences of linorder.sort_key (which does not allow for code generation) by our new "code constant" revsort_key.
interpretation dual_linorder!: linorder "op ≥ :: 'a::linorder ⇒ 'a ⇒ bool" "op >"
where
"linorder.sort_key (op ≥ :: 'a ⇒ 'a ⇒ bool) f xs = revsort_key f xs"
proof -
show "class.linorder (op ≥ :: 'a ⇒ 'a ⇒ bool) (op >)" by (rule dual_linorder)
then interpret rev_order: linorder "op ≥ :: 'a ⇒ 'a ⇒ bool" "op >" .
have "rev_order.insort_key f = insort_by_key (op ≥) f"
by (intro ext) (induct_tac xa; simp)
then show "rev_order.sort_key f xs = revsort_key f xs"
by (simp add: rev_order.sort_key_def revsort_key_def)
qed
While with this solution we do not have any runtime penalty, it is far too verbose for my taste and is not easily adaptable to changes in the standard code setup (e.g., if we wanted to use the mergesort implementation from the Archive of Formal Proofs for all of our sorting operations).
Related
In Isabelle HOL, I have a predicate on two numbers like this:
definition f :: "nat ⇒ nat ⇒ bool"
where
...
I can prove that this predicate is morally a function:
lemma f_function:
fixes x :: nat
shows "∃! y . f x y""
...
Intuitively, this should be enough for me to construct a function f' :: nat ⇒ nat that is provably equivalent to f', i.e.:
lemma f'_correct:
"f x y = (f' x = y)"
But how do I do that?
definition f' :: "nat ⇒ nat"
where
"f' x ≡ ?"
What do I put in for the question mark?
The typical approach is to use the definite description operator THE:
definition f' :: "nat ⇒ nat" where "f' x = (THE y. f x y)"
If you have already proven that this y is unique, you can then use e.g. the theorem theI' to show that f x (f' x) holds and theI_unique to show that if f x y holds, then y = f' x.
For more information about THE, SOME, etc. see the following:
Isabelle/HOL: What does the THE construct denote?
Proving intuitive statements about THE in Isabelle
Here's a small theorem in mathematics:
Suppose u is not an element of A, and v is not an element of B, and f is an injective function from A to B. Let A' = A union {u} and B' = B union {v}, and define g: A' -> B' by g(x) = f(x) if x is in A, and g(u) = v. Then g is injective as well.
If I were writing OCaml-like code, I'd represent A and B as types, and f as an A->B function, something like
module type Q =
sig
type 'a
type 'b
val f: 'a -> 'b
end
and then define a functor
module Extend (M : Q) : Q =
struct
type a = OrdinaryA of M.a | ExoticA
type b = OrdinaryB of M.b | ExoticB
let f x = match x with
OrdinaryA t -> OrdinaryB ( M.f t)
| Exotic A -> ExoticB
end;;
and my theorem would be that if Q.f is injective, then so is (Extend Q).f, where I'm hoping I've gotten the syntax more or less correct.
I'd like to do the same thing in Isabelle/Isar. Normally, that'd mean writing something like
definition injective :: "('a ⇒ 'b) ⇒ bool"
where "injective f ⟷ ( ∀ P Q. (f(P) = f(Q)) ⟷ (P = Q))"
proposition: "injective f ⟹ injective (Q(f))"
and Q is ... something. I don't know how to make, in Isabelle a single operation analogous to the functor Q in OCaml that creates two new datatypes and a function between them. The proof of injectivity seems as if it'd be fairly straightforward --- merely a four-case split. But I'd like help defining the new function that I've called Q f, given the function f.
Here's a solution. I tried to make a "definition" for the function Q, but could not do so; instead, creating a constant Q (built in strong analogy to map) let me state and prove the theorem:
theory Extensions
imports Main
begin
text ‹We show that if we have f: 'a → 'b that's injective, and we extend
both the domain and codomain types by a new element, and extend f in the
obvious way, then the resulting function is still injective.›
definition injective :: "('a ⇒ 'b) ⇒ bool"
where "injective f ⟷ ( ∀ P Q. (f(P) = f(Q)) ⟷ (P = Q))"
datatype 'a extension = Ordinary 'a | Exotic
fun Q :: "('a ⇒ 'b) ⇒ (('a extension) ⇒ ('b extension))" where
"Q f (Ordinary u) = Ordinary (f u)" |
"Q f (Exotic) = Exotic"
lemma "⟦injective f⟧ ⟹ injective (Q f)"
by (smt Q.elims extension.distinct(1) extension.inject injective_def)
end
I'm working on a theory that requires usage of rings, so I imported the following theories: https://www.isa-afp.org/browser_info/devel/AFP/Group-Ring-Module/
Right now, I have defined a set X of a certain type and I'd like to define operations on it to make it a ring, as in the locale "Ring" of the imported theory.
How do I define a ring with carrier X and have it recognized as an instance of the locale "Ring"?
The locale "Ring" is declared by extending "aGroup", which in turn is declared by extending "Group", which is in the theory "Algebra2.thy":
record 'a Group = "'a carrier" +
top :: "['a, 'a ] ⇒ 'a" (infixl "⋅ı" 70)
iop :: "'a ⇒ 'a" ("ρı _" [81] 80)
one :: "'a" ("𝟭ı")
locale Group =
fixes G (structure)
assumes top_closed: "top G ∈ carrier G → carrier G → carrier G"
and tassoc : "⟦a ∈ carrier G; b ∈ carrier G; c ∈ carrier G⟧ ⟹
(a ⋅ b) ⋅ c = a ⋅ (b ⋅ c)"
and iop_closed:"iop G ∈ carrier G → carrier G"
and l_i :"a ∈ carrier G ⟹ (ρ a) ⋅ a = 𝟭"
and unit_closed: "𝟭 ∈ carrier G"
and l_unit:"a ∈ carrier G ⟹ 𝟭 ⋅ a = a"
Another possible problem I antecipate: if I'm not mistaken, the carrier must be of type 'a set, but my set X is of type ('a set \times 'a) set set. Is there a workaround?
EDIT: In order to better formulate the sequential question in the comments, here are some pieces of what I did. All that follows is within the context of a locale presheaf, that fixes (among other things):
T :: 'a set set and
objectsmap :: "'a set ⇒ ('a, 'm) Ring_scheme" and
restrictionsmap:: "('a set ×'a set) ⇒ ('a ⇒ 'a)"
I then introduced the following:
definition prestalk :: "'a ⇒('a set × 'a) set" where
"prestalk x = { (U,s). (U ∈ T) ∧ x ∈U ∧ (s ∈ carrier (objectsmap U))}"
definition stalkrel :: "'a ⇒ ( ('a set × 'a) × ('a set × 'a) ) set" where
"stalkrel x = {( (U,s), (V,t) ). (U,s) ∈ prestalk x ∧ (V,t) ∈ prestalk x ∧ (∃W. W ⊆ U∩V ∧ x∈W ∧
restrictionsmap (V,W) t = restrictionsmap (U,W)) s} "
I then proved that for each x, stalkrel x is an equivalence relation, and defined:
definition germ:: "'a ⇒ 'a set ⇒ 'a ⇒ ('a set × 'a) set" where
"germ x U s = {(V,t). ((U,s),(V,t)) ∈ stalkrel x}"
definition stalk:: "'a ⇒( ('a set × 'a) set) set" where
"stalk x = {w. (∃ U s. w = germ x U s ∧ (U,s) ∈ prestalk x) }"
I'm trying to show that for each x this stalk x is a ring, and the ring operation is "built" out of the ring operations of rings objectsmap (U∩V) , i.e, I'd like germ x U s + germ x V t to be germ x (U∩V) (restrictionsmap (U, (U∩V)) s + restrictionsmap (V, (U∩V)) t), where this last sum is the sum of ring objectsmap (U∩V).
A multiplicative Group in the AFP entry mentioned is a record with four fields: a set carrier for the carrier, the binary group operation top, the inverse operation iop and the neutral element one. Similarly, a Ring is a record which extends an additive group (record aGroup with fields carrier, pop, mop, zero) with the binary multiplicative operation tp and the multiplicative unit un. If you want to define an instance of a group or record, you must define something of the appropriate record type. For example,
definition my_ring :: "<el> Ring" where
"my_ring =
(|carrier = <c>,
pop = <plus>,
mop = <minus>,
zero = <0>,
tp = <times>,
un = <unit>|)"
where you have to replace all the <...> by the types and terms for your ring. That is, <el> is the type of the ring elements, <c> is the carrier set, etc. Note that you can specialise the type of ring elements as needed.
In order to prove that my_ring is indeed a ring, you must show that it satisfies the assumptions of the corresponding locale Ring:
lemma "Ring my_ring"
proof unfold_locales
...
qed
If you want to use the theorems that have been proven abstractly for arbitrary rings, you may want to interpret the locale using interpretation.
Up until several days ago, I always defined a type, and then proved theorems directly about the type. Now I'm trying to use type classes.
Problem
The problem is that I can't instantiate cNAT for my type myD below, and it appears it's because simp has no effect on the abstract function cNAT, which I've made concrete with my primrec function cNAT_myD. I can only guess what's happening because of the automation that happens after instance proof.
Questions
Q1: Below, at the statement instantiation myD :: (type) cNAT, can you tell me how to finish the proof, and why I can prove the following theorem, but not the type class proof, which requires injective?
theorem dNAT_1_to_1: "(dNAT n = dNAT m) ==> n = m"
assumes injective: "(cNAT n = cNAT m) ==> n = m"
Q2: This is not as important, but at the bottom is this statement:
instantiation myD :: (type) cNAT2
It involves another way I was trying to instantiate cNAT. Can you tell me why I get Failed to refine any pending goal at shows? I put some comments in the source to explain some of what I did to set it up. I used this slightly modified formula for the requirement injective:
assumes injective: "!!n m. (cNAT2 n = cNAT2 m) --> n = m"
Specifics
My contrived datatype is this, which may be useful to me someday: (Update: Well, for another example maybe. A good mental exercise is for me to try and figure out how I can actually get something inside a 'a myD list, other than []. With BNF, something like datatype_new 'a myD = myS "'a myD fset" gives me the warning that there's an unused type variable on the right-hand side)
datatype 'a myD = myL "'a myD list"
The type class is this, which requires an injective function from nat to 'a:
class cNAT =
fixes cNAT :: "nat => 'a"
assumes injective: "(cNAT n = cNAT m) ==> n = m"
dNAT: this non-type class version of cNAT works
fun get_myL :: "'a myD => 'a myD list" where
"get_myL (myL L) = L"
primrec dNAT :: "nat => 'a myD" where
"dNAT 0 = myL []"
|"dNAT (Suc n) = myL (myL [] # get_myL(dNAT n))"
fun myD2nat :: "'a myD => nat" where
"myD2nat (myL []) = 0"
|"myD2nat (myL (x # xs)) = Suc(myD2nat (myL xs))"
theorem left_inverse_1 [simp]:
"myD2nat(dNAT n) = n"
apply(induct n, auto)
by(metis get_myL.cases get_myL.simps)
theorem dNAT_1_to_1:
"(dNAT n = dNAT m) ==> n = m"
apply(induct n)
apply(simp) (*
The simp method expanded dNAT.*)
apply(metis left_inverse_1 myD2nat.simps(1))
by (metis left_inverse_1)
cNAT: type class version that I can't instantiate
instantiation myD :: (type) cNAT
begin
primrec cNAT_myD :: "nat => 'a myD" where
"cNAT_myD 0 = myL []"
|"cNAT_myD (Suc n) = myL (myL [] # get_myL(cNAT_myD n))"
instance
proof
fix n m :: nat
show "cNAT n = cNAT m ==> n = m"
apply(induct n)
apply(simp) (*
The simp method won't expand cNAT to cNAT_myD's definition.*)
by(metis injective)+ (*
Metis proved it without unfolding cNAT_myD. It's useless. Goals always remain,
and the type variables in the output panel are all weird.*)
oops
end
cNAT2: Failed to refine any pending goal at show
(*I define a variation of `injective` in which the `assumes` definition, the
goal, and the `show` statement are exactly the same, and that strange `fails
to refine any pending goal shows up.*)
class cNAT2 =
fixes cNAT2 :: "nat => 'a"
assumes injective: "!!n m. (cNAT2 n = cNAT2 m) --> n = m"
instantiation myD :: (type) cNAT2
begin
primrec cNAT2_myD :: "nat => 'a myD" where
"cNAT2_myD 0 = myL []"
|"cNAT2_myD (Suc n) = myL (myL [] # get_myL(cNAT2_myD n))"
instance
proof (*
goal: !!n m. cNAT2 n = cNAT2 m --> n = m.*)
show
"!!n m. cNAT2 n = cNAT2 m --> n = m"
(*Failed to refine any pending goal
Local statement fails to refine any pending goal
Failed attempt to solve goal by exported rule:
cNAT2 (n::nat) = cNAT2 (m::nat) --> n = m *)
Your function cNAT is polymorphic in its result type, but the type variable does not appear among the parameters. This often causes type inference to compute a type which is more general than you want. In your case for cNAT, Isabelle infers for the two occurrences of cNAT in the show statement the type nat => 'b for some 'b of sort cNAT, but their type in the goal is nat => 'a myD. You can see this in jEdit by Ctrl-hovering over the cNAT occurrences to inspect the types. In ProofGeneral, you can enable printing of types with using [[show_consts]].
Therefore, you have to explicitly constrain types in the show statement as follows:
fix n m
assume "(cNAT n :: 'a myD) = cNAT m"
then show "n = m"
Note that it is usually not a good idea to use Isabelle's meta-connectives !! and ==> inside a show statement, you better rephrase them using fix/assume/show.
I want to create a definition which has a tuple as its argument.
definition my_def :: "('a × 'a) ⇒ bool" where
"my_def (a, b) ⟷ a = b"
However, this is not accepted. The error message is
Bad arguments on lhs: "(a, b)"
The error(s) above occurred in definition:
"my_def (a, b) ≡ a = b"
Using fun instead of definition works but this is not what I want. The following notation also works but is somewhat ugly:
definition my_def :: "('a × 'a) ⇒ bool" where
"my_def t ⟷ fst t = snd t"
What is the best way to use tuples as arguments in a definition?
Using fun is probably the least painful way to do this, the definition package doesn't recognise patterns on the left hand side.
Another option is to use tuple patterns for lambda expressions:
my_def ≡ %(a,b). a = b