(** {1 Defunctionalization} This is inspired from student exercises proposed by {h Olivier Danvy} at the {h JFLA 2014 conference} *) (** {2 Simple Arithmetic Expressions} *) module Expr use export int.Int (** Grammar of expressions {h
} *) type expr = Cte int | Sub expr expr type prog = expr (** Examples: {hn : int e : expression e ::= n | e - e p : program p ::= e
} *) let constant p0 : prog = Cte 0 let constant p1 : prog = Sub (Cte 10) (Cte 6) let constant p2 : prog = Sub (Sub (Cte 10) (Cte 6)) (Sub (Cte 7) (Cte 2)) let constant p3 : prog = Sub (Sub (Cte 7) (Cte 2)) (Sub (Cte 10) (Cte 6)) let constant p4 : prog = Sub (Cte 10) (Sub (Cte 2) (Cte 3)) end (** {2 Direct Semantics} *) module DirectSem use Expr (** Values: {hp0 = 0 p1 = 10 - 6 p2 = (10 - 6) - (7 - 2) p3 = (7 - 2) - (10 - 6) p4 = 10 - (2 - 3)
} Expressible Values: {hv : value v ::= n
} Interpretation: {hve : expressible_value ve ::= v
------
n => n
e1 => n1 e2 => n2 n1 - n2 = n3
----------------------------------
e1 - e2 => n3
}
A program e is interpreted into a value n if judgement
{h } holds. *) (** {4 Exercise 0.0} Program the interpreter above and test it on the examples. {he => n
} *) (* Note: Why3 definitions introduced by "function" belong to the logic part of Why3 language *) let rec function eval_0 (e:expr) : int = match e with | Cte n -> n | Sub e1 e2 -> eval_0 e1 - eval_0 e2 end let function interpret_0 (p:prog) : int = eval_0 p (** Tests, can be replayed using {heval_0 : expression -> expressible_value interpret_0 : program -> expressible_value
} (Why3 version at least 0.82 required) *) let test () = interpret_0 p0, interpret_0 p1, interpret_0 p2, interpret_0 p3, interpret_0 p4 constant v3 : int = eval_0 p3 goal eval_p3 : v3 = 1 end (** {2 CPS: Continuation Passing Style} *) module CPS use Expr (** {4 Exercise 0.1} CPS-transform (call by value, left to right) the function `eval_0`, and call it from the main interpreter with the identity function as initial continuation {hwhy3 defunctionalization.mlw --exec DirectSem.test
eval_1 : expression -> (expressible_value -> 'a) -> 'a
interpret_1 : program -> expressible_value
}
*)
use DirectSem
let rec eval_1 (e:expr) (k: int->'a) : 'a
variant { e }
ensures { result = k (eval_0 e) }
= match e with
| Cte n -> k n
| Sub e1 e2 ->
eval_1 e1 (fun v1 -> eval_1 e2 (fun v2 -> k (v1 - v2)))
end
let interpret_1 (p : prog) : int
ensures { result = eval_0 p }
= eval_1 p (fun n -> n)
end
(** {2 Defunctionalization} *)
module Defunctionalization
use Expr
use DirectSem
(** {4 Exercise 0.2}
De-functionalize the continuation of `eval_1`.
{h
cont ::= ...
continue_2 : cont -> value -> value
eval_2 : expression -> cont -> value
interpret_2 : program -> value
}
The data type `cont` represents the grammar of contexts.
The two mutually recursive functions `eval_2` and `continue_2`
implement an abstract machine, that is a state transition system.
*)
type cont = A1 expr cont | A2 int cont | I
(** One would want to write in Why:
{h
function eval_cont (c:cont) (v:int) : int =
match c with
| A1 e2 k ->
let v2 = eval_0 e2 in
eval_cont (A2 v k) v2
| A2 v1 k -> eval_cont k (v1 - v)
| I -> v
end
}
But since the recursion is not structural, Why3 kernel rejects it
(definitions in the logic part of the language must be total)
We replace that with a relational definition, an inductive one.
*)
inductive eval_cont cont int int =
| a1 :
forall e2:expr, k:cont, v:int, r:int.
eval_cont (A2 v k) (eval_0 e2) r -> eval_cont (A1 e2 k) v r
| a2 :
forall v1:int, k:cont, v:int, r:int.
eval_cont k (v1 - v) r -> eval_cont (A2 v1 k) v r
| a3 :
forall v:int. eval_cont I v v
(** Some functions to serve as measures for the termination proof *)
function size_e (e:expr) : int =
match e with
| Cte _ -> 1
| Sub e1 e2 -> 3 + size_e e1 + size_e e2
end
lemma size_e_pos: forall e: expr. size_e e >= 1
function size_c (c:cont) : int =
match c with
| I -> 0
| A1 e2 k -> 2 + size_e e2 + size_c k
| A2 _ k -> 1 + size_c k
end
lemma size_c_pos: forall c: cont. size_c c >= 0
(** WhyML programs (declared with `let` instead of `function`),
mutually recursive, resulting from de-functionalization *)
let rec continue_2 (c:cont) (v:int) : int
variant { size_c c }
ensures { eval_cont c v result }
=
match c with
| A1 e2 k -> eval_2 e2 (A2 v k)
| A2 v1 k -> continue_2 k (v1 - v)
| I -> v
end
with eval_2 (e:expr) (c:cont) : int
variant { size_c c + size_e e }
ensures { eval_cont c (eval_0 e) result }
=
match e with
| Cte n -> continue_2 c n
| Sub e1 e2 -> eval_2 e1 (A1 e2 c)
end
(** The interpreter. The post-condition specifies that this program
computes the same thing as `eval_0` *)
let interpret_2 (p:prog) : int
ensures { result = eval_0 p }
= eval_2 p I
let test () =
interpret_2 p0,
interpret_2 p1,
interpret_2 p2,
interpret_2 p3,
interpret_2 p4
end
(** {2 Defunctionalization with an algebraic variant} *)
module Defunctionalization2
use Expr
use DirectSem
(** {4 Exercise 0.2}
De-functionalize the continuation of `eval_1`.
{h
cont ::= ...
continue_2 : cont -> value -> value
eval_2 : expression -> cont -> value
interpret_2 : program -> value
}
The data type `cont` represents the grammar of contexts.
The two mutually recursive functions `eval_2` and `continue_2`
implement an abstract machine, that is a state transition system.
*)
type cont = A1 expr cont | A2 int cont | I
(** One would want to write in Why:
{h
function eval_cont (c:cont) (v:int) : int =
match c with
| A1 e2 k ->
let v2 = eval_0 e2 in
eval_cont (A2 v k) v2
| A2 v1 k -> eval_cont k (v1 - v)
| I -> v
end
}
But since the recursion is not structural, Why3 kernel rejects it
(definitions in the logic part of the language must be total)
We replace that with a relational definition, an inductive one.
*)
inductive eval_cont cont int int =
| a1 :
forall e2:expr, k:cont, v:int, r:int.
eval_cont (A2 v k) (eval_0 e2) r -> eval_cont (A1 e2 k) v r
| a2 :
forall v1:int, k:cont, v:int, r:int.
eval_cont k (v1 - v) r -> eval_cont (A2 v1 k) v r
| a3 :
forall v:int. eval_cont I v v
(** Peano naturals to serve as the measure for the termination proof *)
type nat = S nat | O
function size_e (e:expr) (acc:nat) : nat =
match e with
| Cte _ -> S acc
| Sub e1 e2 -> S (size_e e1 (S (size_e e2 (S acc))))
end
function size_c (c:cont) (acc:nat) : nat =
match c with
| I -> acc
| A1 e2 k -> S (size_e e2 (S (size_c k acc)))
| A2 _ k -> S (size_c k acc)
end
(** WhyML programs (declared with `let` instead of `function`),
mutually recursive, resulting from de-functionalization *)
let rec continue_2 (c:cont) (v:int) : int
variant { size_c c O }
ensures { eval_cont c v result }
=
match c with
| A1 e2 k -> eval_2 e2 (A2 v k)
| A2 v1 k -> continue_2 k (v1 - v)
| I -> v
end
with eval_2 (e:expr) (c:cont) : int
variant { size_e e (size_c c O) }
ensures { eval_cont c (eval_0 e) result }
=
match e with
| Cte n -> continue_2 c n
| Sub e1 e2 -> eval_2 e1 (A1 e2 c)
end
(** The interpreter. The post-condition specifies that this program
computes the same thing as `eval_0` *)
let interpret_2 (p:prog) : int
ensures { result = eval_0 p }
= eval_2 p I
let test () =
interpret_2 p0,
interpret_2 p1,
interpret_2 p2,
interpret_2 p3,
interpret_2 p4
end
(** {2 Semantics with errors} *)
module SemWithError
use Expr
(** Errors:
{h } Expressible values: {hs : error
} *) type value = Vnum int | Underflow (* in (Vnum n), n should always be >= 0 *) (** Interpretation: {hve : expressible_value ve ::= v | s
------
n => n
e1 => s
------------
e1 - e2 => s
e1 => n1 e2 => s
------------------
e1 - e2 => s
e1 => n1 e2 => n2 n1 < n2
-----------------------------
e1 - e2 => "underflow"
e1 => n1 e2 => n2 n1 >= n2 n1 - n2 = n3
---------------------------------------------
e1 - e2 => n3
}
We interpret the program `e` into value `n` if the judgement
{h } holds, and into error `s` if the judgement {he => n
} holds. {4 Exercise 1.0} Program the interpreter above and test it on the examples. {he => s
} *) function eval_0 (e:expr) : value = match e with | Cte n -> if n >= 0 then Vnum n else Underflow | Sub e1 e2 -> match eval_0 e1 with | Underflow -> Underflow | Vnum v1 -> match eval_0 e2 with | Underflow -> Underflow | Vnum v2 -> if v1 >= v2 then Vnum (v1 - v2) else Underflow end end end function interpret_0 (p:prog) : value = eval_0 p (** {4 Exercise 1.1} CPS-transform (call by value, from left to right) the function `eval_0`, call it from the main interpreter with the identity function as initial continuation. {heval_0 : expr -> expressible_value interpret_0 : program -> expressible_value
eval_1 : expr -> (expressible_value -> 'a) -> 'a
interpret_1 : program -> expressible_value
}
*)
function eval_1 (e:expr) (k:value -> 'a) : 'a =
match e with
| Cte n -> k (if n >= 0 then Vnum n else Underflow)
| Sub e1 e2 ->
eval_1 e1 (fun v1 ->
match v1 with
| Underflow -> k Underflow
| Vnum v1 ->
eval_1 e2 (fun v2 ->
match v2 with
| Underflow -> k Underflow
| Vnum v2 -> k (if v1 >= v2 then Vnum (v1 - v2) else Underflow)
end)
end)
end
function interpret_1 (p : prog) : value = eval_1 p (fun n -> n)
lemma cps_correct_expr:
forall e: expr. forall k:value -> 'a. eval_1 e k = k (eval_0 e)
lemma cps_correct: forall p. interpret_1 p = interpret_0 p
(** {4 Exercise 1.2}
Divide the continuation
{h
expressible_value -> 'a
}
in two:
{h
(value -> 'a) * (error -> 'a)
}
and adapt `eval_1` and `interpret_1` as
{h
eval_2 : expr -> (value -> 'a) -> (error -> 'a) -> 'a
interpret_2 : program -> expressible_value
}
*)
(*
function eval_2 (e:expr) (k:int -> 'a) (kerr: unit -> 'a) : 'a =
match e with
| Cte n -> if n >= 0 then k n else kerr ()
| Sub e1 e2 ->
eval_2 e1 (fun v1 ->
eval_2 e2 (fun v2 ->
if v1 >= v2 then k (v1 - v2) else kerr ())
kerr) kerr
end
*)
function eval_2 (e:expr) (k:int -> 'a) (kerr: unit -> 'a) : 'a =
match e with
| Cte n -> if n >= 0 then k n else kerr ()
| Sub e1 e2 ->
eval_2 e1 (eval_2a e2 k kerr) kerr
end
with eval_2a (e2:expr) (k:int -> 'a) (kerr : unit -> 'a) : int -> 'a =
(fun v1 -> eval_2 e2 (eval_2b v1 k kerr) kerr)
with eval_2b (v1:int) (k:int -> 'a) (kerr : unit -> 'a) : int -> 'a =
(fun v2 -> if v1 >= v2 then k (v1 - v2) else kerr ())
function interpret_2 (p : prog) : value =
eval_2 p (fun n -> Vnum n) (fun _ -> Underflow)
lemma cps2_correct_expr_aux:
forall k: int -> 'a, e1 e2, kerr: unit -> 'a.
eval_2 (Sub e1 e2) k kerr = eval_2 e1 (eval_2a e2 k kerr) kerr
lemma cps2_correct_expr:
forall e, kerr: unit->'a, k:int -> 'a. eval_2 e k kerr =
match eval_0 e with Vnum n -> k n | Underflow -> kerr () end
lemma cps2_correct: forall p. interpret_2 p = interpret_0 p
(** {4 Exercise 1.3}
Specialize the codomain of the continuations and of the evaluation function
so that it is not polymorphic anymore but is `expressible_value`, and
then short-circuit the second continuation to stop in case of error
{h
eval_3 : expr -> (value -> expressible_value) -> expressible_value
interpret_3 : program -> expressible_value
}
NB: Now there is only one continuation and it is applied only in
absence of error.
*)
function eval_3 (e:expr) (k:int -> value) : value =
match e with
| Cte n -> if n >= 0 then k n else Underflow
| Sub e1 e2 ->
eval_3 e1 (eval_3a e2 k)
end
with eval_3a (e2:expr) (k:int -> value) : int -> value =
(fun v1 -> eval_3 e2 (eval_3b v1 k))
with eval_3b (v1:int) (k:int -> value) : int -> value =
(fun v2 -> if v1 >= v2 then k (v1 - v2) else Underflow)
function interpret_3 (p : prog) : value = eval_3 p (fun n -> Vnum n)
let rec lemma cps3_correct_expr (e:expr) : unit
variant { e }
ensures { forall k. eval_3 e k =
match eval_0 e with Vnum n -> k n | Underflow -> Underflow end }
= match e with
| Cte _ -> ()
| Sub e1 e2 ->
cps3_correct_expr e1;
cps3_correct_expr e2;
assert {forall k. eval_3 e k = eval_3 e1 (eval_3a e2 k) }
end
lemma cps3_correct: forall p. interpret_3 p = interpret_0 p
(** {4 Exercise 1.4}
De-functionalize the continuation of `eval_3`.
{h
cont ::= ...
continue_4 : cont -> value -> expressible_value
eval_4 : expr -> cont -> expressible_value
interprete_4 : program -> expressible_value
}
The data type `cont` represents the grammar of contexts.
(NB. has it changed w.r.t to previous exercise?)
The two mutually recursive functions `eval_4` and `continue_4`
implement an abstract machine, that is a transition system that
stops immediately in case of error, or and the end of computation.
*)
type cont = I | A expr cont | B int cont
(**
One would want to write in Why:
{h } But since the recursion is not structural, Why3 kernel rejects it (definitions in the logic part of the language must be total) We replace that with a relational definition, an inductive one. *) inductive eval_cont cont value value = | underflow : forall k:cont. eval_cont k Underflow Underflow | a1 : forall e2:expr, k:cont, v:int, r:value. eval_cont (B v k) (eval_0 e2) r -> eval_cont (A e2 k) (Vnum v) r | a2 : forall v1:int, k:cont, v:int, r:value. eval_cont k (if v1 >= v then Vnum (v1 - v) else Underflow) r -> eval_cont (B v1 k) (Vnum v) r | a3 : forall v:int. eval_cont I (Vnum v) (Vnum v) (** Some functions to serve as measures for the termination proof *) function size_e (e:expr) : int = match e with | Cte _ -> 1 | Sub e1 e2 -> 3 + size_e e1 + size_e e2 end lemma size_e_pos: forall e: expr. size_e e >= 1 use Defunctionalization as D function size_c (c:cont) : int = match c with | I -> 0 | A e2 k -> 2 + D.size_e e2 + size_c k | B _ k -> 1 + size_c k end lemma size_c_pos: forall c: cont. size_c c >= 0 let rec continue_4 (c:cont) (v:int) : value requires { v >= 0 } variant { size_c c } ensures { eval_cont c (Vnum v) result } = match c with | A e2 k -> eval_4 e2 (B v k) | B v1 k -> if v1 >= v then continue_4 k (v1 - v) else Underflow | I -> Vnum v end with eval_4 (e:expr) (c:cont) : value variant { size_c c + D.size_e e } ensures { eval_cont c (eval_0 e) result } = match e with | Cte n -> if n >= 0 then continue_4 c n else Underflow | Sub e1 e2 -> eval_4 e1 (A e2 c) end (** The interpreter. The post-condition specifies that this program computes the same thing as `eval_0` *) let interpret_4 (p:prog) : value ensures { result = eval_0 p } = eval_4 p I let test () = interpret_4 p0, interpret_4 p1, interpret_4 p2, interpret_4 p3, interpret_4 p4 end (** {2 Reduction Semantics} *) module ReductionSemantics (** A small step semantics, defined iteratively with reduction contexts *) use Expr use DirectSem (** Expressions: {hfunction eval_cont (c:cont) (v:value) : value = match v with | Underflow -> Underflow | Vnum v -> match c with | A e2 k -> eval_cont (B (Vnum v) k) (eval_0 e2) | B v1 k -> eval_cont k (if v1 >= v then Vnum (v1 - v) else Underflow) | I -> Vnum v (* v >= 0 by construction *) end
} Values: {hn : int e : expression e ::= n | e - e p : program p ::= e
} Potential Redexes: {hv : value v ::= n
} Reduction Rule: {hrp ::= n1 - n2
} (in the case of relative integers Z, all potential redexes are true redexes; but for natural numbers, not all of them are true ones: {hn1 - n2 -> n3
} a potential redex that is not a true one is stuck.) Contraction Function: {hn1 - n2 -> n3 if n1 >= n2
} and if there are only non-negative numbers: {hcontract : redex_potentiel -> expression + stuck contract (n1 - n2) = n3 si n3 = n1 - n2
} *) predicate is_a_redex (e:expr) = match e with | Sub (Cte _) (Cte _) -> true | _ -> false end let contract (e:expr) : expr requires { is_a_redex e } ensures { eval_0 result = eval_0 e } = match e with | Sub (Cte v1) (Cte v2) -> Cte (v1 - v2) | _ -> absurd end (** Reduction Contexts: {hcontract (n1 - n2) = n3 if n1 >= n2 and n3 = n1 - n2 contract (n1 - n2) = stuck if n1 < n2
} *) type context = Empty | Left context expr | Right int context (** Recomposition: {hC : cont C ::= [] | [C e] | [v C]
recompose : cont * expression -> expression
recompose ([], e) = e
recompose ([C e2], e1) = recompose (C, e1 - e2)
recompose ([n1 C], e2) = recompose (C, n1 - e2)
}
*)
let rec function recompose (c:context) (e:expr) : expr =
match c with
| Empty -> e
| Left c e2 -> recompose c (Sub e e2)
| Right n1 c -> recompose c (Sub (Cte n1) e)
end
let rec lemma recompose_values (c:context) (e1 e2:expr) : unit
requires { eval_0 e1 = eval_0 e2 }
variant { c }
ensures { eval_0 (recompose c e1) = eval_0 (recompose c e2) }
= match c with
| Empty -> ()
| Left c e -> recompose_values c (Sub e1 e) (Sub e2 e)
| Right n c -> recompose_values c (Sub (Cte n) e1) (Sub (Cte n) e2)
end
(* not needed anymore
let rec lemma recompose_inversion (c:context) (e:expr)
requires {
match c with Empty -> false | _ -> true end \/
match e with Cte _ -> false | Sub _ _ -> true end }
variant { c }
ensures { match recompose c e with
Cte _ -> false | Sub _ _ -> true end }
= match c with
| Empty -> ()
| Left c e2 -> recompose_inversion c (Sub e e2)
| Right n1 c -> recompose_inversion c (Sub (Cte n1) e)
end
*)
(**
Decomposition:
{h } Decomposition function: {hdec_or_val = (C, rp) | v
decompose_term : expression * cont -> dec_or_val
decompose_term (n, C) = decompose_cont (C, n)
decompose_term (e1 - e2, C) = decompose_term (e1, [C e2])
decompose_cont : cont * value -> dec_or_val
decompose_cont ([], n) = n
decompose_cont ([C e], n) = decompose_term (e, [n c])
decompose_term ([n1 C], n2) = (C, n1 - n2)
decompose : expression -> dec_or_val
decompose e = decompose_term (e, [])
}
*)
exception Stuck
predicate is_a_value (e:expr) =
match e with
| Cte _ -> true
| _ -> false
end
predicate is_empty_context (c:context) =
match c with
| Empty -> true
| _ -> false
end
use Defunctionalization as D (* for size_e *)
function size_e (e:expr) : int = D.size_e e
function size_c (c:context) : int =
match c with
| Empty -> 0
| Left c e -> 2 + size_c c + size_e e
| Right _ c -> 1 + size_c c
end
lemma size_c_pos: forall c: context. size_c c >= 0
let rec decompose_term (e:expr) (c:context) : (context, expr)
variant { size_e e + size_c c }
ensures { let c1,e1 = result in
recompose c1 e1 = recompose c e /\
is_a_redex e1 }
raises { Stuck -> is_empty_context c /\ is_a_value e }
(* raises { Stuck -> is_a_value (recompose c e) } *)
=
match e with
| Cte n -> decompose_cont c n
| Sub e1 e2 -> decompose_term e1 (Left c e2)
end
with decompose_cont (c:context) (n:int) : (context, expr)
variant { size_c c }
ensures { let c1,e1 = result in
recompose c1 e1 = recompose c (Cte n) /\
is_a_redex e1 }
raises { Stuck -> is_empty_context c }
(* raises { Stuck -> is_a_value (recompose c (Cte n)) } *)
=
match c with
| Empty -> raise Stuck
| Left c e -> decompose_term e (Right n c)
| Right n1 c -> c, Sub (Cte n1) (Cte n)
end
let decompose (e:expr) : (context, expr)
ensures { let c1,e1 = result in recompose c1 e1 = e /\
is_a_redex e1 }
raises { Stuck -> is_a_value e }
=
decompose_term e Empty
(**
One reduction step:
{h } *) (** {4 Exercise 2.0} Implement the reduction semantics above and test it *) let reduce (e:expr) : expr ensures { eval_0 result = eval_0 e } raises { Stuck -> is_a_value e } = let c,r = decompose e in recompose c (contract r) (** Evaluation based on iterated reduction: {hreduce : expression -> expression + stuck if decompose e = v then reduce e = stuck if decompose e = (C, rp) and contract rp = stuck then reduce e = stuck if decompose e = (C, rp) and contract rp = c then reduce e = recompose (C, c)
} *) let rec itere (e:expr) : int diverges (* termination could be proved but that's not the point *) ensures { eval_0 e = result } = match reduce e with | e' -> itere e' | exception Stuck -> match e with | Cte n -> n | _ -> absurd end end (** {4 Exercise 2.1} Optimize the step recomposition / decomposition into a single function `refocus`. *) let refocus c e ensures { let c1,e1 = result in recompose c1 e1 = recompose c e /\ is_a_redex e1 } raises { Stuck -> is_empty_context c /\ is_a_value e } = decompose_term e c let rec itere_opt (c:context) (e:expr) : int diverges ensures { result = eval_0 (recompose c e) } = match refocus c e with | c, r -> itere_opt c (contract r) | exception Stuck -> assert { is_empty_context c }; match e with | Cte n -> n | _ -> absurd end end let rec normalize (e:expr) diverges ensures { result = eval_0 e } = itere_opt Empty e (** {4 Exercise 2.2} Derive an abstract machine *) (** {hitere : red_ou_val -> value + erreur itere v = v if contract rp = stuck then itere (C, rp) = stuck if contract rp = c then itere (C, rp) = itere (decompose (recompose (C, c)))
} *) let rec eval_1 c e variant { 2 * (size_c c + size_e e) } ensures { result = eval_0 (recompose c e) } = match e with | Cte n -> eval_2 c n | Sub e1 e2 -> eval_1 (Left c e2) e1 end with eval_2 c n variant { 1 + 2 * size_c c } ensures { result = eval_0 (recompose c (Cte n)) } = match c with | Empty -> n | Left c e -> eval_1 (Right n c) e | Right n1 c -> eval_1 c (Cte (n1 - n)) end let interpret p ensures { result = eval_0 p } = eval_1 Empty p let test () = interpret p0, interpret p1, interpret p2, interpret p3, interpret p4 end module RWithError use Expr use SemWithError (** {4 Exercise 2.3} An abstract machine for the case with errors *) (** {h(c,Cte n)_1 -> (c,n)_2 (c,Sub e1 e2)_1 -> (Left c e2,e1)_1 (Empty,n)_2 -> stop with result = n (Left c e,n)_2 -> (Right n c,e)_1 (Right n1 c,n)_2 -> (c,Cte (n1 - n))_1
} *) type context = Empty | Left context expr | Right int context use Defunctionalization as D (* for size_e *) function size_e (e:expr) : int = D.size_e e function size_c (c:context) : int = match c with | Empty -> 0 | Left c e -> 2 + size_c c + size_e e | Right _ c -> 1 + size_c c end lemma size_c_pos: forall c: context. size_c c >= 0 function recompose (c:context) (e:expr) : expr = match c with | Empty -> e | Left c e2 -> recompose c (Sub e e2) | Right n1 c -> recompose c (Sub (Cte n1) e) end let rec lemma recompose_values (c:context) (e1 e2:expr) : unit requires { eval_0 e1 = eval_0 e2 } variant { c } ensures { eval_0 (recompose c e1) = eval_0 (recompose c e2) } = match c with | Empty -> () | Left c e -> recompose_values c (Sub e1 e) (Sub e2 e) | Right n c -> recompose_values c (Sub (Cte n) e1) (Sub (Cte n) e2) end let rec lemma recompose_underflow (c:context) (e:expr) : unit requires { eval_0 e = Underflow } variant { c } ensures { eval_0 (recompose c e) = Underflow } = match c with | Empty -> () | Left c e1 -> recompose_underflow c (Sub e e1) | Right n c -> recompose_underflow c (Sub (Cte n) e) end let rec eval_1 c e variant { 2 * (size_c c + size_e e) } ensures { result = eval_0 (recompose c e) } = match e with | Cte n -> if n >= 0 then eval_2 c n else Underflow | Sub e1 e2 -> eval_1 (Left c e2) e1 end with eval_2 c n variant { 1 + 2 * size_c c } requires { n >= 0 } ensures { result = eval_0 (recompose c (Cte n)) } = match c with | Empty -> Vnum n | Left c e -> eval_1 (Right n c) e | Right n1 c -> if n1 >= n then eval_1 c (Cte (n1 - n)) else Underflow end let interpret p ensures { result = eval_0 p } = eval_1 Empty p let test () = interpret p0, interpret p1, interpret p2, interpret p3, interpret p4 end(c,Cte n)_1 -> stop on Underflow if n < 0 (c,Cte n)_1 -> (c,n)_2 if n >= 0 (c,Sub e1 e2)_1 -> (Left c e2,e1)_1 (Empty,n)_2 -> stop on Vnum n (Left c e,n)_2 -> (Right n c,e)_1 (Right n1 c,n)_2 -> stop on Underflow if n1 < n (Right n1 c,n)_2 -> (c,Cte (n1 - n))_1 if n1 >= n