(** * Hoare: Hoare Logic, Part I *)
Require Export Imp.
(** In the past couple of chapters, we've begun applying the
mathematical tools developed in the first part of the course to
studying the theory of a small programming language, Imp.
- We defined a type of _abstract syntax trees_ for Imp, together
with an _evaluation relation_ (a partial function on states)
that specifies the _operational semantics_ of programs.
The language we defined, though small, captures some of the key
features of full-blown languages like C, C++, and Java,
including the fundamental notion of mutable state and some
common control structures.
- We proved a number of _metatheoretic properties_ -- "meta" in
the sense that they are properties of the language as a whole,
rather than properties of particular programs in the language.
These included:
- determinism of evaluation
- equivalence of some different ways of writing down the
definitions (e.g. functional and relational definitions of
arithmetic expression evaluation)
- guaranteed termination of certain classes of programs
- correctness (in the sense of preserving meaning) of a number
of useful program transformations
- behavioral equivalence of programs (in the [Equiv] chapter).
If we stopped here, we would already have something useful: a set
of tools for defining and discussing programming languages and
language features that are mathematically precise, flexible, and
easy to work with, applied to a set of key properties. All of
these properties are things that language designers, compiler
writers, and users might care about knowing. Indeed, many of them
are so fundamental to our understanding of the programming
languages we deal with that we might not consciously recognize
them as "theorems." But properties that seem intuitively obvious
can sometimes be quite subtle (in some cases, even subtly wrong!).
We'll return to the theme of metatheoretic properties of whole
languages later in the course when we discuss _types_ and _type
soundness_. In this chapter, though, we'll turn to a different
set of issues.
Our goal is to see how to carry out some simple examples of
_program verification_ -- i.e., using the precise definition of
Imp to prove formally that particular programs satisfy particular
specifications of their behavior. We'll develop a reasoning system
called _Floyd-Hoare Logic_ -- often shortened to just _Hoare
Logic_ -- in which each of the syntactic constructs of Imp is
equipped with a single, generic "proof rule" that can be used to
reason compositionally about the correctness of programs involving
this construct.
Hoare Logic originates in the 1960s, and it continues to be the
subject of intensive research right up to the present day. It
lies at the core of a multitude of tools that are being used in
academia and industry to specify and verify real software
systems. *)
(* ####################################################### *)
(** * Hoare Logic *)
(** Hoare Logic combines two beautiful ideas: a natural way of
writing down _specifications_ of programs, and a _compositional
proof technique_ for proving that programs are correct with
respect to such specifications -- where by "compositional" we mean
that the structure of proofs directly mirrors the structure of the
programs that they are about. *)
(* ####################################################### *)
(** ** Assertions *)
(** To talk about specifications of programs, the first thing we
need is a way of making _assertions_ about properties that hold at
particular points during a program's execution -- i.e., claims
about the current state of the memory when program execution
reaches that point. Formally, an assertion is just a family of
propositions indexed by a [state]. *)
Definition Assertion := state -> Prop.
(** **** Exercise: 1 star, optional (assertions) *)
Module ExAssertions.
(** Paraphrase the following assertions in English. *)
Definition as1 : Assertion := fun st => st X = 3.
Definition as2 : Assertion := fun st => st X <= st Y.
Definition as3 : Assertion :=
fun st => st X = 3 \/ st X <= st Y.
Definition as4 : Assertion :=
fun st => st Z * st Z <= st X /\
~ (((S (st Z)) * (S (st Z))) <= st X).
Definition as5 : Assertion := fun st => True.
Definition as6 : Assertion := fun st => False.
(* FILL IN HERE *)
End ExAssertions.
(** [] *)
(** This way of writing assertions can be a little bit heavy,
for two reasons: (1) every single assertion that we ever write is
going to begin with [fun st => ]; and (2) this state [st] is the
only one that we ever use to look up variables (we will never need
to talk about two different memory states at the same time). For
discussing examples informally, we'll adopt some simplifying
conventions: we'll drop the initial [fun st =>], and we'll write
just [X] to mean [st X]. Thus, instead of writing *)
(**
fun st => (st Z) * (st Z) <= m /\
~ ((S (st Z)) * (S (st Z)) <= m)
we'll write just
Z * Z <= m /\ ~((S Z) * (S Z) <= m).
*)
(** Given two assertions [P] and [Q], we say that [P] _implies_ [Q],
written [P ->> Q] (in ASCII, [P -][>][> Q]), if, whenever [P]
holds in some state [st], [Q] also holds. *)
Definition assert_implies (P Q : Assertion) : Prop :=
forall st, P st -> Q st.
Notation "P ->> Q" :=
(assert_implies P Q) (at level 80) : hoare_spec_scope.
Open Scope hoare_spec_scope.
(** We'll also have occasion to use the "iff" variant of implication
between assertions: *)
Notation "P <<->> Q" :=
(P ->> Q /\ Q ->> P) (at level 80) : hoare_spec_scope.
(* ####################################################### *)
(** ** Hoare Triples *)
(** Next, we need a way of making formal claims about the
behavior of commands. *)
(** Since the behavior of a command is to transform one state to
another, it is natural to express claims about commands in terms
of assertions that are true before and after the command executes:
- "If command [c] is started in a state satisfying assertion
[P], and if [c] eventually terminates in some final state,
then this final state will satisfy the assertion [Q]."
Such a claim is called a _Hoare Triple_. The property [P] is
called the _precondition_ of [c], while [Q] is the
_postcondition_. Formally: *)
Definition hoare_triple
(P:Assertion) (c:com) (Q:Assertion) : Prop :=
forall st st',
c / st || st' ->
P st ->
Q st'.
(** Since we'll be working a lot with Hoare triples, it's useful to
have a compact notation:
{{P}} c {{Q}}.
*)
(** (The traditional notation is [{P} c {Q}], but single braces
are already used for other things in Coq.) *)
Notation "{{ P }} c {{ Q }}" :=
(hoare_triple P c Q) (at level 90, c at next level)
: hoare_spec_scope.
(** (The [hoare_spec_scope] annotation here tells Coq that this
notation is not global but is intended to be used in particular
contexts. The [Open Scope] tells Coq that this file is one such
context.) *)
(** **** Exercise: 1 star, optional (triples) *)
(** Paraphrase the following Hoare triples in English.
1) {{True}} c {{X = 5}}
2) {{X = m}} c {{X = m + 5)}}
3) {{X <= Y}} c {{Y <= X}}
4) {{True}} c {{False}}
5) {{X = m}}
c
{{Y = real_fact m}}.
6) {{True}}
c
{{(Z * Z) <= m /\ ~ (((S Z) * (S Z)) <= m)}}
*)
(** [] *)
(** **** Exercise: 1 star, optional (valid_triples) *)
(** Which of the following Hoare triples are _valid_ -- i.e., the
claimed relation between [P], [c], and [Q] is true?
1) {{True}} X ::= 5 {{X = 5}}
2) {{X = 2}} X ::= X + 1 {{X = 3}}
3) {{True}} X ::= 5; Y ::= 0 {{X = 5}}
4) {{X = 2 /\ X = 3}} X ::= 5 {{X = 0}}
5) {{True}} SKIP {{False}}
6) {{False}} SKIP {{True}}
7) {{True}} WHILE True DO SKIP END {{False}}
8) {{X = 0}}
WHILE X == 0 DO X ::= X + 1 END
{{X = 1}}
9) {{X = 1}}
WHILE X <> 0 DO X ::= X + 1 END
{{X = 100}}
*)
(* FILL IN HERE *)
(** [] *)
(** (Note that we're using informal mathematical notations for
expressions inside of commands, for readability, rather than their
formal [aexp] and [bexp] encodings. We'll continue doing so
throughout the chapter.) *)
(** To get us warmed up for what's coming, here are two simple
facts about Hoare triples. *)
Theorem hoare_post_true : forall (P Q : Assertion) c,
(forall st, Q st) ->
{{P}} c {{Q}}.
Proof.
intros P Q c H. unfold hoare_triple.
intros st st' Heval HP.
apply H. Qed.
Theorem hoare_pre_false : forall (P Q : Assertion) c,
(forall st, ~(P st)) ->
{{P}} c {{Q}}.
Proof.
intros P Q c H. unfold hoare_triple.
intros st st' Heval HP.
unfold not in H. apply H in HP.
inversion HP. Qed.
(* ####################################################### *)
(** ** Proof Rules *)
(** The goal of Hoare logic is to provide a _compositional_
method for proving the validity of Hoare triples. That is, the
structure of a program's correctness proof should mirror the
structure of the program itself. To this end, in the sections
below, we'll introduce one rule for reasoning about each of the
different syntactic forms of commands in Imp -- one for
assignment, one for sequencing, one for conditionals, etc. -- plus
a couple of "structural" rules that are useful for gluing things
together. We will prove programs correct using these proof rules,
without ever unfolding the definition of [hoare_triple]. *)
(* ####################################################### *)
(** *** Assignment *)
(** The rule for assignment is the most fundamental of the Hoare logic
proof rules. Here's how it works.
Consider this (valid) Hoare triple:
{{ Y = 1 }} X ::= Y {{ X = 1 }}
In English: if we start out in a state where the value of [Y]
is [1] and we assign [Y] to [X], then we'll finish in a
state where [X] is [1]. That is, the property of being equal
to [1] gets transferred from [Y] to [X].
Similarly, in
{{ Y + Z = 1 }} X ::= Y + Z {{ X = 1 }}
the same property (being equal to one) gets transferred to
[X] from the expression [Y + Z] on the right-hand side of
the assignment.
More generally, if [a] is _any_ arithmetic expression, then
{{ a = 1 }} X ::= a {{ X = 1 }}
is a valid Hoare triple.
This can be made even more general. To conclude that an
_arbitrary_ property [Q] holds after [X ::= a], we need to assume
that [Q] holds before [X ::= a], but _with all occurrences of_ [X]
replaced by [a] in [Q]. This leads to the Hoare rule for
assignment
{{ Q [X |-> a] }} X ::= a {{ Q }}
where "[Q [X |-> a]]" is pronounced "[Q] where [a] is substituted
for [X]".
For example, these are valid applications of the assignment
rule:
{{ (X <= 5) [X |-> X + 1]
i.e., X + 1 <= 5 }}
X ::= X + 1
{{ X <= 5 }}
{{ (X = 3) [X |-> 3]
i.e., 3 = 3}}
X ::= 3
{{ X = 3 }}
{{ (0 <= X /\ X <= 5) [X |-> 3]
i.e., (0 <= 3 /\ 3 <= 5)}}
X ::= 3
{{ 0 <= X /\ X <= 5 }}
*)
(** To formalize the rule, we must first formalize the idea of
"substituting an expression for an Imp variable in an assertion."
That is, given a proposition [P], a variable [X], and an
arithmetic expression [a], we want to derive another proposition
[P'] that is just the same as [P] except that, wherever [P]
mentions [X], [P'] should instead mention [a].
Since [P] is an arbitrary Coq proposition, we can't directly
"edit" its text. Instead, we can achieve the effect we want by
evaluating [P] in an updated state: *)
Definition assn_sub X a P : Assertion :=
fun (st : state) =>
P (update st X (aeval st a)).
Notation "P [ X |-> a ]" := (assn_sub X a P) (at level 10).
(** That is, [P [X |-> a]] is an assertion [P'] that is just like [P]
except that, wherever [P] looks up the variable [X] in the current
state, [P'] instead uses the value of the expression [a].
To see how this works, let's calculate what happens with a couple
of examples. First, suppose [P'] is [(X <= 5) [X |-> 3]] -- that
is, more formally, [P'] is the Coq expression
fun st =>
(fun st' => st' X <= 5)
(update st X (aeval st (ANum 3))),
which simplifies to
fun st =>
(fun st' => st' X <= 5)
(update st X 3)
and further simplifies to
fun st =>
((update st X 3) X) <= 5)
and by further simplification to
fun st =>
(3 <= 5).
That is, [P'] is the assertion that [3] is less than or equal to
[5] (as expected).
For a more interesting example, suppose [P'] is [(X <= 5) [X |->
X+1]]. Formally, [P'] is the Coq expression
fun st =>
(fun st' => st' X <= 5)
(update st X (aeval st (APlus (AId X) (ANum 1)))),
which simplifies to
fun st =>
(((update st X (aeval st (APlus (AId X) (ANum 1))))) X) <= 5
and further simplifies to
fun st =>
(aeval st (APlus (AId X) (ANum 1))) <= 5.
That is, [P'] is the assertion that [X+1] is at most [5].
*)
(** Now we can give the precise proof rule for assignment:
------------------------------ (hoare_asgn)
{{Q [X |-> a]}} X ::= a {{Q}}
*)
(** We can prove formally that this rule is indeed valid. *)
Theorem hoare_asgn : forall Q X a,
{{Q [X |-> a]}} (X ::= a) {{Q}}.
Proof.
unfold hoare_triple.
intros Q X a st st' HE HQ.
inversion HE. subst.
unfold assn_sub in HQ. assumption. Qed.
(** Here's a first formal proof using this rule. *)
Example assn_sub_example :
{{(fun st => st X = 3) [X |-> ANum 3]}}
(X ::= (ANum 3))
{{fun st => st X = 3}}.
Proof.
apply hoare_asgn. Qed.
(** **** Exercise: 2 stars (hoare_asgn_examples) *)
(** Translate these informal Hoare triples...
1) {{ (X <= 5) [X |-> X + 1] }}
X ::= X + 1
{{ X <= 5 }}
2) {{ (0 <= X /\ X <= 5) [X |-> 3] }}
X ::= 3
{{ 0 <= X /\ X <= 5 }}
...into formal statements and use [hoare_asgn] to prove them. *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 2 stars (hoare_asgn_wrong) *)
(** The assignment rule looks backward to almost everyone the first
time they see it. If it still seems backward to you, it may help
to think a little about alternative "forward" rules. Here is a
seemingly natural one:
------------------------------ (hoare_asgn_wrong)
{{ True }} X ::= a {{ X = a }}
Give a counterexample showing that this rule is incorrect
(informally). Hint: The rule universally quantifies over the
arithmetic expression [a], and your counterexample needs to
exhibit an [a] for which the rule doesn't work. *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 3 stars, advanced (hoare_asgn_fwd) *)
(** However, using an auxiliary variable [m] to remember the original
value of [X] we can define a Hoare rule for assignment that does,
intuitively, "work forwards" rather than backwards.
------------------------------------------ (hoare_asgn_fwd)
{{fun st => Q st /\ st X = m}}
X ::= a
{{fun st => Q st' /\ st X = aeval st' a }}
(where st' = update st X m)
Note that we use the original value of [X] to reconstruct the
state [st'] before the assignment took place. Prove that this rule
is correct (the first hypothesis is the functional extensionality
axiom, which you will need at some point). Also note that this
rule is more complicated than [hoare_asgn].
*)
Theorem hoare_asgn_fwd :
(forall {X Y: Type} {f g : X -> Y}, (forall (x: X), f x = g x) -> f = g) ->
forall m a Q,
{{fun st => Q st /\ st X = m}}
X ::= a
{{fun st => Q (update st X m) /\ st X = aeval (update st X m) a }}.
Proof.
intros functional_extensionality v a Q.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ####################################################### *)
(** *** Consequence *)
(** Sometimes the preconditions and postconditions we get from the
Hoare rules won't quite be the ones we want in the particular
situation at hand -- they may be logically equivalent but have a
different syntactic form that fails to unify with the goal we are
trying to prove, or they actually may be logically weaker (for
preconditions) or stronger (for postconditions) than what we need.
For instance, while
{{(X = 3) [X |-> 3]}} X ::= 3 {{X = 3}},
follows directly from the assignment rule,
{{True}} X ::= 3 {{X = 3}}.
does not. This triple is valid, but it is not an instance of
[hoare_asgn] because [True] and [(X = 3) [X |-> 3]] are not
syntactically equal assertions. However, they are logically
equivalent, so if one triple is valid, then the other must
certainly be as well. We might capture this observation with the
following rule:
{{P'}} c {{Q}}
P <<->> P'
----------------------------- (hoare_consequence_pre_equiv)
{{P}} c {{Q}}
Taking this line of thought a bit further, we can see that
strengthening the precondition or weakening the postcondition of a
valid triple always produces another valid triple. This
observation is captured by two _Rules of Consequence_.
{{P'}} c {{Q}}
P ->> P'
----------------------------- (hoare_consequence_pre)
{{P}} c {{Q}}
{{P}} c {{Q'}}
Q' ->> Q
----------------------------- (hoare_consequence_post)
{{P}} c {{Q}}
*)
(** Here are the formal versions: *)
Theorem hoare_consequence_pre : forall (P P' Q : Assertion) c,
{{P'}} c {{Q}} ->
P ->> P' ->
{{P}} c {{Q}}.
Proof.
intros P P' Q c Hhoare Himp.
intros st st' Hc HP. apply (Hhoare st st').
assumption. apply Himp. assumption. Qed.
Theorem hoare_consequence_post : forall (P Q Q' : Assertion) c,
{{P}} c {{Q'}} ->
Q' ->> Q ->
{{P}} c {{Q}}.
Proof.
intros P Q Q' c Hhoare Himp.
intros st st' Hc HP.
apply Himp.
apply (Hhoare st st').
assumption. assumption. Qed.
(** For example, we might use the first consequence rule like this:
{{ True }} ->>
{{ 1 = 1 }}
X ::= 1
{{ X = 1 }}
Or, formally...
*)
Example hoare_asgn_example1 :
{{fun st => True}} (X ::= (ANum 1)) {{fun st => st X = 1}}.
Proof.
apply hoare_consequence_pre
with (P' := (fun st => st X = 1) [X |-> ANum 1]).
apply hoare_asgn.
intros st H. unfold assn_sub, update. simpl. reflexivity.
Qed.
(** Finally, for convenience in some proofs, we can state a "combined"
rule of consequence that allows us to vary both the precondition
and the postcondition.
{{P'}} c {{Q'}}
P ->> P'
Q' ->> Q
----------------------------- (hoare_consequence)
{{P}} c {{Q}}
*)
Theorem hoare_consequence : forall (P P' Q Q' : Assertion) c,
{{P'}} c {{Q'}} ->
P ->> P' ->
Q' ->> Q ->
{{P}} c {{Q}}.
Proof.
intros P P' Q Q' c Hht HPP' HQ'Q.
apply hoare_consequence_pre with (P' := P').
apply hoare_consequence_post with (Q' := Q').
assumption. assumption. assumption. Qed.
(* ####################################################### *)
(** *** Digression: The [eapply] Tactic *)
(** This is a good moment to introduce another convenient feature of
Coq. We had to write "[with (P' := ...)]" explicitly in the proof
of [hoare_asgn_example1] and [hoare_consequence] above, to make
sure that all of the metavariables in the premises to the
[hoare_consequence_pre] rule would be set to specific
values. (Since [P'] doesn't appear in the conclusion of
[hoare_consequence_pre], the process of unifying the conclusion
with the current goal doesn't constrain [P'] to a specific
assertion.)
This is a little annoying, both because the assertion is a bit
long and also because for [hoare_asgn_example1] the very next
thing we are going to do -- applying the [hoare_asgn] rule -- will
tell us exactly what it should be! We can use [eapply] instead of
[apply] to tell Coq, essentially, "Be patient: The missing part is
going to be filled in soon." *)
Example hoare_asgn_example1' :
{{fun st => True}}
(X ::= (ANum 1))
{{fun st => st X = 1}}.
Proof.
eapply hoare_consequence_pre.
apply hoare_asgn.
intros st H. reflexivity. Qed.
(** In general, [eapply H] tactic works just like [apply H] except
that, instead of failing if unifying the goal with the conclusion
of [H] does not determine how to instantiate all of the variables
appearing in the premises of [H], [eapply H] will replace these
variables with so-called _existential variables_ (written [?nnn])
as placeholders for expressions that will be determined (by
further unification) later in the proof. *)
(** In order for [Qed] to succeed, all existential variables need to
be determined by the end of the proof. Otherwise Coq
will (rightly) refuse to accept the proof. Remember that the Coq
tactics build proof objects, and proof objects containing
existential variables are not complete. *)
Lemma silly1 : forall (P : nat -> nat -> Prop) (Q : nat -> Prop),
(forall x y : nat, P x y) ->
(forall x y : nat, P x y -> Q x) ->
Q 42.
Proof.
intros P Q HP HQ. eapply HQ. apply HP.
(** Coq gives a warning after [apply HP]:
No more subgoals but non-instantiated existential variables:
Existential 1 =
?171 : [P : nat -> nat -> Prop
Q : nat -> Prop
HP : forall x y : nat, P x y
HQ : forall x y : nat, P x y -> Q x |- nat]
(dependent evars: ?171 open,)
You can use Grab Existential Variables.
Trying to finish the proof with [Qed] gives an error:
<<
Error: Attempt to save a proof with existential variables still
non-instantiated
>> *)
Abort.
(** An additional constraint is that existential variables cannot be
instantiated with terms containing (ordinary) variables that did
not exist at the time the existential variable was created. *)
Lemma silly2 :
forall (P : nat -> nat -> Prop) (Q : nat -> Prop),
(exists y, P 42 y) ->
(forall x y : nat, P x y -> Q x) ->
Q 42.
Proof.
intros P Q HP HQ. eapply HQ. destruct HP as [y HP'].
(** Doing [apply HP'] above fails with the following error:
Error: Impossible to unify "?175" with "y".
In this case there is an easy fix:
doing [destruct HP] _before_ doing [eapply HQ].
*)
Abort.
Lemma silly2_fixed :
forall (P : nat -> nat -> Prop) (Q : nat -> Prop),
(exists y, P 42 y) ->
(forall x y : nat, P x y -> Q x) ->
Q 42.
Proof.
intros P Q HP HQ. destruct HP as [y HP'].
eapply HQ. apply HP'.
Qed.
(** In the last step we did [apply HP'] which unifies the existential
variable in the goal with the variable [y]. The [assumption]
tactic doesn't work in this case, since it cannot handle
existential variables. However, Coq also provides an [eassumption]
tactic that solves the goal if one of the premises matches the
goal up to instantiations of existential variables. We can use
it instead of [apply HP']. *)
Lemma silly2_eassumption : forall (P : nat -> nat -> Prop) (Q : nat -> Prop),
(exists y, P 42 y) ->
(forall x y : nat, P x y -> Q x) ->
Q 42.
Proof.
intros P Q HP HQ. destruct HP as [y HP']. eapply HQ. eassumption.
Qed.
(** **** Exercise: 2 stars (hoare_asgn_examples_2) *)
(** Translate these informal Hoare triples...
{{ X + 1 <= 5 }} X ::= X + 1 {{ X <= 5 }}
{{ 0 <= 3 /\ 3 <= 5 }} X ::= 3 {{ 0 <= X /\ X <= 5 }}
...into formal statements and use [hoare_asgn] and
[hoare_consequence_pre] to prove them. *)
(* FILL IN HERE *)
(** [] *)
(* ####################################################### *)
(** *** Skip *)
(** Since [SKIP] doesn't change the state, it preserves any
property P:
-------------------- (hoare_skip)
{{ P }} SKIP {{ P }}
*)
Theorem hoare_skip : forall P,
{{P}} SKIP {{P}}.
Proof.
intros P st st' H HP. inversion H. subst.
assumption. Qed.
(* ####################################################### *)
(** *** Sequencing *)
(** More interestingly, if the command [c1] takes any state where
[P] holds to a state where [Q] holds, and if [c2] takes any
state where [Q] holds to one where [R] holds, then doing [c1]
followed by [c2] will take any state where [P] holds to one
where [R] holds:
{{ P }} c1 {{ Q }}
{{ Q }} c2 {{ R }}
--------------------- (hoare_seq)
{{ P }} c1;;c2 {{ R }}
*)
Theorem hoare_seq : forall P Q R c1 c2,
{{Q}} c2 {{R}} ->
{{P}} c1 {{Q}} ->
{{P}} c1;;c2 {{R}}.
Proof.
intros P Q R c1 c2 H1 H2 st st' H12 Pre.
inversion H12; subst.
apply (H1 st'0 st'); try assumption.
apply (H2 st st'0); assumption. Qed.
(** Note that, in the formal rule [hoare_seq], the premises are
given in "backwards" order ([c2] before [c1]). This matches the
natural flow of information in many of the situations where we'll
use the rule: the natural way to construct a Hoare-logic proof is
to begin at the end of the program (with the final postcondition)
and push postconditions backwards through commands until we reach
the beginning. *)
(** Informally, a nice way of recording a proof using the sequencing
rule is as a "decorated program" where the intermediate assertion
[Q] is written between [c1] and [c2]:
{{ a = n }}
X ::= a;;
{{ X = n }} <---- decoration for Q
SKIP
{{ X = n }}
*)
Example hoare_asgn_example3 : forall a n,
{{fun st => aeval st a = n}}
(X ::= a;; SKIP)
{{fun st => st X = n}}.
Proof.
intros a n. eapply hoare_seq.
Case "right part of seq".
apply hoare_skip.
Case "left part of seq".
eapply hoare_consequence_pre. apply hoare_asgn.
intros st H. subst. reflexivity. Qed.
(** You will most often use [hoare_seq] and
[hoare_consequence_pre] in conjunction with the [eapply] tactic,
as done above. *)
(** **** Exercise: 2 stars (hoare_asgn_example4) *)
(** Translate this "decorated program" into a formal proof:
{{ True }} ->>
{{ 1 = 1 }}
X ::= 1;;
{{ X = 1 }} ->>
{{ X = 1 /\ 2 = 2 }}
Y ::= 2
{{ X = 1 /\ Y = 2 }}
*)
Example hoare_asgn_example4 :
{{fun st => True}} (X ::= (ANum 1);; Y ::= (ANum 2))
{{fun st => st X = 1 /\ st Y = 2}}.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars (swap_exercise) *)
(** Write an Imp program [c] that swaps the values of [X] and [Y]
and show (in Coq) that it satisfies the following
specification:
{{X <= Y}} c {{Y <= X}}
*)
Definition swap_program : com :=
(* FILL IN HERE *) admit.
Theorem swap_exercise :
{{fun st => st X <= st Y}}
swap_program
{{fun st => st Y <= st X}}.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars (hoarestate1) *)
(** Explain why the following proposition can't be proven:
forall (a : aexp) (n : nat),
{{fun st => aeval st a = n}}
(X ::= (ANum 3);; Y ::= a)
{{fun st => st Y = n}}.
*)
(* FILL IN HERE *)
(** [] *)
(* ####################################################### *)
(** *** Conditionals *)
(** What sort of rule do we want for reasoning about conditional
commands? Certainly, if the same assertion [Q] holds after
executing either branch, then it holds after the whole
conditional. So we might be tempted to write:
{{P}} c1 {{Q}}
{{P}} c2 {{Q}}
--------------------------------
{{P}} IFB b THEN c1 ELSE c2 {{Q}}
However, this is rather weak. For example, using this rule,
we cannot show that:
{{ True }}
IFB X == 0
THEN Y ::= 2
ELSE Y ::= X + 1
FI
{{ X <= Y }}
since the rule tells us nothing about the state in which the
assignments take place in the "then" and "else" branches. *)
(** But we can actually say something more precise. In the
"then" branch, we know that the boolean expression [b] evaluates to
[true], and in the "else" branch, we know it evaluates to [false].
Making this information available in the premises of the rule gives
us more information to work with when reasoning about the behavior
of [c1] and [c2] (i.e., the reasons why they establish the
postcondition [Q]). *)
(**
{{P /\ b}} c1 {{Q}}
{{P /\ ~b}} c2 {{Q}}
------------------------------------ (hoare_if)
{{P}} IFB b THEN c1 ELSE c2 FI {{Q}}
*)
(** To interpret this rule formally, we need to do a little work.
Strictly speaking, the assertion we've written, [P /\ b], is the
conjunction of an assertion and a boolean expression -- i.e., it
doesn't typecheck. To fix this, we need a way of formally
"lifting" any bexp [b] to an assertion. We'll write [bassn b] for
the assertion "the boolean expression [b] evaluates to [true] (in
the given state)." *)
Definition bassn b : Assertion :=
fun st => (beval st b = true).
(** A couple of useful facts about [bassn]: *)
Lemma bexp_eval_true : forall b st,
beval st b = true -> (bassn b) st.
Proof.
intros b st Hbe.
unfold bassn. assumption. Qed.
Lemma bexp_eval_false : forall b st,
beval st b = false -> ~ ((bassn b) st).
Proof.
intros b st Hbe contra.
unfold bassn in contra.
rewrite -> contra in Hbe. inversion Hbe. Qed.
(** Now we can formalize the Hoare proof rule for conditionals
and prove it correct. *)
Theorem hoare_if : forall P Q b c1 c2,
{{fun st => P st /\ bassn b st}} c1 {{Q}} ->
{{fun st => P st /\ ~(bassn b st)}} c2 {{Q}} ->
{{P}} (IFB b THEN c1 ELSE c2 FI) {{Q}}.
Proof.
intros P Q b c1 c2 HTrue HFalse st st' HE HP.
inversion HE; subst.
Case "b is true".
apply (HTrue st st').
assumption.
split. assumption.
apply bexp_eval_true. assumption.
Case "b is false".
apply (HFalse st st').
assumption.
split. assumption.
apply bexp_eval_false. assumption. Qed.
(* ####################################################### *)
(** * Hoare Logic: So Far *)
(**
Idea: create a _domain specific logic_ for reasoning about properties of Imp programs.
- This hides the low-level details of the semantics of the program
- Leads to a compositional reasoning process
The basic structure is given by _Hoare triples_ of the form:
{{P}} c {{Q}}
]]
- [P] and [Q] are predicates about the state of the Imp program
- "If command [c] is started in a state satisfying assertion
[P], and if [c] eventually terminates in some final state,
then this final state will satisfy the assertion [Q]."
*)
(** ** Hoare Logic Rules (so far) *)
(**
------------------------------ (hoare_asgn)
{{Q [X |-> a]}} X::=a {{Q}}
-------------------- (hoare_skip)
{{ P }} SKIP {{ P }}
{{ P }} c1 {{ Q }}
{{ Q }} c2 {{ R }}
--------------------- (hoare_seq)
{{ P }} c1;;c2 {{ R }}
{{P /\ b}} c1 {{Q}}
{{P /\ ~b}} c2 {{Q}}
------------------------------------ (hoare_if)
{{P}} IFB b THEN c1 ELSE c2 FI {{Q}}
{{P'}} c {{Q'}}
P ->> P'
Q' ->> Q
----------------------------- (hoare_consequence)
{{P}} c {{Q}}
*)
(** *** Example *)
(** Here is a formal proof that the program we used to motivate the
rule satisfies the specification we gave. *)
Example if_example :
{{fun st => True}}
IFB (BEq (AId X) (ANum 0))
THEN (Y ::= (ANum 2))
ELSE (Y ::= APlus (AId X) (ANum 1))
FI
{{fun st => st X <= st Y}}.
Proof.
(* WORKED IN CLASS *)
apply hoare_if.
Case "Then".
eapply hoare_consequence_pre. apply hoare_asgn.
unfold bassn, assn_sub, update, assert_implies.
simpl. intros st [_ H].
apply beq_nat_true in H.
rewrite H. omega.
Case "Else".
eapply hoare_consequence_pre. apply hoare_asgn.
unfold assn_sub, update, assert_implies.
simpl; intros st _. omega.
Qed.
(** **** Exercise: 2 stars (if_minus_plus) *)
(** Prove the following hoare triple using [hoare_if]: *)
Theorem if_minus_plus :
{{fun st => True}}
IFB (BLe (AId X) (AId Y))
THEN (Z ::= AMinus (AId Y) (AId X))
ELSE (Y ::= APlus (AId X) (AId Z))
FI
{{fun st => st Y = st X + st Z}}.
Proof.
(* FILL IN HERE *) Admitted.
(* ####################################################### *)
(** *** Exercise: One-sided conditionals *)
(** **** Exercise: 4 stars (if1_hoare) *)
(** In this exercise we consider extending Imp with "one-sided
conditionals" of the form [IF1 b THEN c FI]. Here [b] is a
boolean expression, and [c] is a command. If [b] evaluates to
[true], then command [c] is evaluated. If [b] evaluates to
[false], then [IF1 b THEN c FI] does nothing.
We recommend that you do this exercise before the ones that
follow, as it should help solidify your understanding of the
material. *)
(** The first step is to extend the syntax of commands and introduce
the usual notations. (We've done this for you. We use a separate
module to prevent polluting the global name space.) *)
Module If1.
Inductive com : Type :=
| CSkip : com
| CAss : id -> aexp -> com
| CSeq : com -> com -> com
| CIf : bexp -> com -> com -> com
| CWhile : bexp -> com -> com
| CIf1 : bexp -> com -> com.
Tactic Notation "com_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "SKIP" | Case_aux c "::=" | Case_aux c ";"
| Case_aux c "IFB" | Case_aux c "WHILE" | Case_aux c "CIF1" ].
Notation "'SKIP'" :=
CSkip.
Notation "c1 ;; c2" :=
(CSeq c1 c2) (at level 80, right associativity).
Notation "X '::=' a" :=
(CAss X a) (at level 60).
Notation "'WHILE' b 'DO' c 'END'" :=
(CWhile b c) (at level 80, right associativity).
Notation "'IFB' e1 'THEN' e2 'ELSE' e3 'FI'" :=
(CIf e1 e2 e3) (at level 80, right associativity).
Notation "'IF1' b 'THEN' c 'FI'" :=
(CIf1 b c) (at level 80, right associativity).
(** Next we need to extend the evaluation relation to accommodate
[IF1] branches. This is for you to do... What rule(s) need to be
added to [ceval] to evaluate one-sided conditionals? *)
Reserved Notation "c1 '/' st '||' st'" (at level 40, st at level 39).
Inductive ceval : com -> state -> state -> Prop :=
| E_Skip : forall st : state, SKIP / st || st
| E_Ass : forall (st : state) (a1 : aexp) (n : nat) (X : id),
aeval st a1 = n -> (X ::= a1) / st || update st X n
| E_Seq : forall (c1 c2 : com) (st st' st'' : state),
c1 / st || st' -> c2 / st' || st'' -> (c1 ;; c2) / st || st''
| E_IfTrue : forall (st st' : state) (b1 : bexp) (c1 c2 : com),
beval st b1 = true ->
c1 / st || st' -> (IFB b1 THEN c1 ELSE c2 FI) / st || st'
| E_IfFalse : forall (st st' : state) (b1 : bexp) (c1 c2 : com),
beval st b1 = false ->
c2 / st || st' -> (IFB b1 THEN c1 ELSE c2 FI) / st || st'
| E_WhileEnd : forall (b1 : bexp) (st : state) (c1 : com),
beval st b1 = false -> (WHILE b1 DO c1 END) / st || st
| E_WhileLoop : forall (st st' st'' : state) (b1 : bexp) (c1 : com),
beval st b1 = true ->
c1 / st || st' ->
(WHILE b1 DO c1 END) / st' || st'' ->
(WHILE b1 DO c1 END) / st || st''
(* FILL IN HERE *)
where "c1 '/' st '||' st'" := (ceval c1 st st').
Tactic Notation "ceval_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "E_Skip" | Case_aux c "E_Ass" | Case_aux c "E_Seq"
| Case_aux c "E_IfTrue" | Case_aux c "E_IfFalse"
| Case_aux c "E_WhileEnd" | Case_aux c "E_WhileLoop"
(* FILL IN HERE *)
].
(** Now we repeat (verbatim) the definition and notation of Hoare triples. *)
Definition hoare_triple (P:Assertion) (c:com) (Q:Assertion) : Prop :=
forall st st',
c / st || st' ->
P st ->
Q st'.
Notation "{{ P }} c {{ Q }}" := (hoare_triple P c Q)
(at level 90, c at next level)
: hoare_spec_scope.
(** Finally, we (i.e., you) need to state and prove a theorem,
[hoare_if1], that expresses an appropriate Hoare logic proof rule
for one-sided conditionals. Try to come up with a rule that is
both sound and as precise as possible. *)
(* FILL IN HERE *)
(** For full credit, prove formally that your rule is precise enough
to show the following valid Hoare triple:
{{ X + Y = Z }}
IF1 Y <> 0 THEN
X ::= X + Y
FI
{{ X = Z }}
*)
(** Hint: Your proof of this triple may need to use the other proof
rules also. Because we're working in a separate module, you'll
need to copy here the rules you find necessary. *)
Lemma hoare_if1_good :
{{ fun st => st X + st Y = st Z }}
IF1 BNot (BEq (AId Y) (ANum 0)) THEN
X ::= APlus (AId X) (AId Y)
FI
{{ fun st => st X = st Z }}.
Proof. (* FILL IN HERE *) Admitted.
End If1.
(** [] *)
(* ####################################################### *)
(** *** Loops *)
(** Finally, we need a rule for reasoning about while loops. *)
(** Suppose we have a loop
WHILE b DO c END
and we want to find a pre-condition [P] and a post-condition
[Q] such that
{{P}} WHILE b DO c END {{Q}}
is a valid triple. *)
(** *** *)
(** First of all, let's think about the case where [b] is false at the
beginning -- i.e., let's assume that the loop body never executes
at all. In this case, the loop behaves like [SKIP], so we might
be tempted to write: *)
(**
{{P}} WHILE b DO c END {{P}}.
*)
(**
But, as we remarked above for the conditional, we know a
little more at the end -- not just [P], but also the fact
that [b] is false in the current state. So we can enrich the
postcondition a little:
*)
(**
{{P}} WHILE b DO c END {{P /\ ~b}}
*)
(**
What about the case where the loop body _does_ get executed?
In order to ensure that [P] holds when the loop finally
exits, we certainly need to make sure that the command [c]
guarantees that [P] holds whenever [c] is finished.
Moreover, since [P] holds at the beginning of the first
execution of [c], and since each execution of [c]
re-establishes [P] when it finishes, we can always assume
that [P] holds at the beginning of [c]. This leads us to the
following rule:
*)
(**
{{P}} c {{P}}
-----------------------------------
{{P}} WHILE b DO c END {{P /\ ~b}}
*)
(**
This is almost the rule we want, but again it can be improved a
little: at the beginning of the loop body, we know not only that
[P] holds, but also that the guard [b] is true in the current
state. This gives us a little more information to use in
reasoning about [c] (showing that it establishes the invariant by
the time it finishes). This gives us the final version of the rule:
*)
(**
{{P /\ b}} c {{P}}
----------------------------------- (hoare_while)
{{P}} WHILE b DO c END {{P /\ ~b}}
The proposition [P] is called an _invariant_ of the loop.
*)
Lemma hoare_while : forall P b c,
{{fun st => P st /\ bassn b st}} c {{P}} ->
{{P}} WHILE b DO c END {{fun st => P st /\ ~ (bassn b st)}}.
Proof.
intros P b c Hhoare st st' He HP.
(* Like we've seen before, we need to reason by induction
on He, because, in the "keep looping" case, its hypotheses
talk about the whole loop instead of just c *)
remember (WHILE b DO c END) as wcom eqn:Heqwcom.
ceval_cases (induction He) Case;
try (inversion Heqwcom); subst; clear Heqwcom.
Case "E_WhileEnd".
split. assumption. apply bexp_eval_false. assumption.
Case "E_WhileLoop".
apply IHHe2. reflexivity.
apply (Hhoare st st'). assumption.
split. assumption. apply bexp_eval_true. assumption.
Qed.
(**
One subtlety in the terminology is that calling some assertion [P]
a "loop invariant" doesn't just mean that it is preserved by the
body of the loop in question (i.e., [{{P}} c {{P}}], where [c] is
the loop body), but rather that [P] _together with the fact that
the loop's guard is true_ is a sufficient precondition for [c] to
ensure [P] as a postcondition.
This is a slightly (but significantly) weaker requirement. For
example, if [P] is the assertion [X = 0], then [P] _is_ an
invariant of the loop
WHILE X = 2 DO X := 1 END
although it is clearly _not_ preserved by the body of the
loop.
*)
Example while_example :
{{fun st => st X <= 3}}
WHILE (BLe (AId X) (ANum 2))
DO X ::= APlus (AId X) (ANum 1) END
{{fun st => st X = 3}}.
Proof.
eapply hoare_consequence_post.
apply hoare_while.
eapply hoare_consequence_pre.
apply hoare_asgn.
unfold bassn, assn_sub, assert_implies, update. simpl.
intros st [H1 H2]. apply ble_nat_true in H2. omega.
unfold bassn, assert_implies. intros st [Hle Hb].
simpl in Hb. destruct (ble_nat (st X) 2) eqn : Heqle.
apply ex_falso_quodlibet. apply Hb; reflexivity.
apply ble_nat_false in Heqle. omega.
Qed.
(** *** *)
(** We can use the while rule to prove the following Hoare triple,
which may seem surprising at first... *)
Theorem always_loop_hoare : forall P Q,
{{P}} WHILE BTrue DO SKIP END {{Q}}.
Proof.
(* WORKED IN CLASS *)
intros P Q.
apply hoare_consequence_pre with (P' := fun st : state => True).
eapply hoare_consequence_post.
apply hoare_while.
Case "Loop body preserves invariant".
apply hoare_post_true. intros st. apply I.
Case "Loop invariant and negated guard imply postcondition".
simpl. intros st [Hinv Hguard].
apply ex_falso_quodlibet. apply Hguard. reflexivity.
Case "Precondition implies invariant".
intros st H. constructor. Qed.
(** Of course, this result is not surprising if we remember that
the definition of [hoare_triple] asserts that the postcondition
must hold _only_ when the command terminates. If the command
doesn't terminate, we can prove anything we like about the
post-condition. *)
(** Hoare rules that only talk about terminating commands are
often said to describe a logic of "partial" correctness. It is
also possible to give Hoare rules for "total" correctness, which
build in the fact that the commands terminate. However, in this
course we will only talk about partial correctness. *)
(* ####################################################### *)
(** *** Exercise: [REPEAT] *)
Module RepeatExercise.
(** **** Exercise: 4 stars, advanced (hoare_repeat) *)
(** In this exercise, we'll add a new command to our language of
commands: [REPEAT] c [UNTIL] a [END]. You will write the
evaluation rule for [repeat] and add a new Hoare rule to
the language for programs involving it. *)
Inductive com : Type :=
| CSkip : com
| CAsgn : id -> aexp -> com
| CSeq : com -> com -> com
| CIf : bexp -> com -> com -> com
| CWhile : bexp -> com -> com
| CRepeat : com -> bexp -> com.
(** [REPEAT] behaves like [WHILE], except that the loop guard is
checked _after_ each execution of the body, with the loop
repeating as long as the guard stays _false_. Because of this,
the body will always execute at least once. *)
Tactic Notation "com_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "SKIP" | Case_aux c "::=" | Case_aux c ";"
| Case_aux c "IFB" | Case_aux c "WHILE"
| Case_aux c "CRepeat" ].
Notation "'SKIP'" :=
CSkip.
Notation "c1 ;; c2" :=
(CSeq c1 c2) (at level 80, right associativity).
Notation "X '::=' a" :=
(CAsgn X a) (at level 60).
Notation "'WHILE' b 'DO' c 'END'" :=
(CWhile b c) (at level 80, right associativity).
Notation "'IFB' e1 'THEN' e2 'ELSE' e3 'FI'" :=
(CIf e1 e2 e3) (at level 80, right associativity).
Notation "'REPEAT' e1 'UNTIL' b2 'END'" :=
(CRepeat e1 b2) (at level 80, right associativity).
(** Add new rules for [REPEAT] to [ceval] below. You can use the rules
for [WHILE] as a guide, but remember that the body of a [REPEAT]
should always execute at least once, and that the loop ends when
the guard becomes true. Then update the [ceval_cases] tactic to
handle these added cases. *)
Inductive ceval : state -> com -> state -> Prop :=
| E_Skip : forall st,
ceval st SKIP st
| E_Ass : forall st a1 n X,
aeval st a1 = n ->
ceval st (X ::= a1) (update st X n)
| E_Seq : forall c1 c2 st st' st'',
ceval st c1 st' ->
ceval st' c2 st'' ->
ceval st (c1 ;; c2) st''
| E_IfTrue : forall st st' b1 c1 c2,
beval st b1 = true ->
ceval st c1 st' ->
ceval st (IFB b1 THEN c1 ELSE c2 FI) st'
| E_IfFalse : forall st st' b1 c1 c2,
beval st b1 = false ->
ceval st c2 st' ->
ceval st (IFB b1 THEN c1 ELSE c2 FI) st'
| E_WhileEnd : forall b1 st c1,
beval st b1 = false ->
ceval st (WHILE b1 DO c1 END) st
| E_WhileLoop : forall st st' st'' b1 c1,
beval st b1 = true ->
ceval st c1 st' ->
ceval st' (WHILE b1 DO c1 END) st'' ->
ceval st (WHILE b1 DO c1 END) st''
(* FILL IN HERE *)
.
Tactic Notation "ceval_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "E_Skip" | Case_aux c "E_Ass"
| Case_aux c "E_Seq"
| Case_aux c "E_IfTrue" | Case_aux c "E_IfFalse"
| Case_aux c "E_WhileEnd" | Case_aux c "E_WhileLoop"
(* FILL IN HERE *)
].
(** A couple of definitions from above, copied here so they use the
new [ceval]. *)
Notation "c1 '/' st '||' st'" := (ceval st c1 st')
(at level 40, st at level 39).
Definition hoare_triple (P:Assertion) (c:com) (Q:Assertion)
: Prop :=
forall st st', (c / st || st') -> P st -> Q st'.
Notation "{{ P }} c {{ Q }}" :=
(hoare_triple P c Q) (at level 90, c at next level).
(** To make sure you've got the evaluation rules for [REPEAT] right,
prove that [ex1_repeat evaluates correctly. *)
Definition ex1_repeat :=
REPEAT
X ::= ANum 1;;
Y ::= APlus (AId Y) (ANum 1)
UNTIL (BEq (AId X) (ANum 1)) END.
Theorem ex1_repeat_works :
ex1_repeat / empty_state ||
update (update empty_state X 1) Y 1.
Proof.
(* FILL IN HERE *) Admitted.
(** Now state and prove a theorem, [hoare_repeat], that expresses an
appropriate proof rule for [repeat] commands. Use [hoare_while]
as a model, and try to make your rule as precise as possible. *)
(* FILL IN HERE *)
(** For full credit, make sure (informally) that your rule can be used
to prove the following valid Hoare triple:
{{ X > 0 }}
REPEAT
Y ::= X;;
X ::= X - 1
UNTIL X = 0 END
{{ X = 0 /\ Y > 0 }}
*)
End RepeatExercise.
(** [] *)
(* ####################################################### *)
(** ** Exercise: [HAVOcC] *)
(** **** Exercise: 3 stars (himp_hoare) *)
(** In this exercise, we will derive proof rules for the [HAVOC] command
which we studied in the last chapter. First, we enclose this work
in a separate module, and recall the syntax and big-step semantics
of Himp commands. *)
Module Himp.
Inductive com : Type :=
| CSkip : com
| CAsgn : id -> aexp -> com
| CSeq : com -> com -> com
| CIf : bexp -> com -> com -> com
| CWhile : bexp -> com -> com
| CHavoc : id -> com.
Tactic Notation "com_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "SKIP" | Case_aux c "::=" | Case_aux c ";"
| Case_aux c "IFB" | Case_aux c "WHILE" | Case_aux c "HAVOC" ].
Notation "'SKIP'" :=
CSkip.
Notation "X '::=' a" :=
(CAsgn X a) (at level 60).
Notation "c1 ;; c2" :=
(CSeq c1 c2) (at level 80, right associativity).
Notation "'WHILE' b 'DO' c 'END'" :=
(CWhile b c) (at level 80, right associativity).
Notation "'IFB' e1 'THEN' e2 'ELSE' e3 'FI'" :=
(CIf e1 e2 e3) (at level 80, right associativity).
Notation "'HAVOC' X" := (CHavoc X) (at level 60).
Reserved Notation "c1 '/' st '||' st'" (at level 40, st at level 39).
Inductive ceval : com -> state -> state -> Prop :=
| E_Skip : forall st : state, SKIP / st || st
| E_Ass : forall (st : state) (a1 : aexp) (n : nat) (X : id),
aeval st a1 = n -> (X ::= a1) / st || update st X n
| E_Seq : forall (c1 c2 : com) (st st' st'' : state),
c1 / st || st' -> c2 / st' || st'' -> (c1 ;; c2) / st || st''
| E_IfTrue : forall (st st' : state) (b1 : bexp) (c1 c2 : com),
beval st b1 = true ->
c1 / st || st' -> (IFB b1 THEN c1 ELSE c2 FI) / st || st'
| E_IfFalse : forall (st st' : state) (b1 : bexp) (c1 c2 : com),
beval st b1 = false ->
c2 / st || st' -> (IFB b1 THEN c1 ELSE c2 FI) / st || st'
| E_WhileEnd : forall (b1 : bexp) (st : state) (c1 : com),
beval st b1 = false -> (WHILE b1 DO c1 END) / st || st
| E_WhileLoop : forall (st st' st'' : state) (b1 : bexp) (c1 : com),
beval st b1 = true ->
c1 / st || st' ->
(WHILE b1 DO c1 END) / st' || st'' ->
(WHILE b1 DO c1 END) / st || st''
| E_Havoc : forall (st : state) (X : id) (n : nat),
(HAVOC X) / st || update st X n
where "c1 '/' st '||' st'" := (ceval c1 st st').
Tactic Notation "ceval_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "E_Skip" | Case_aux c "E_Ass" | Case_aux c "E_Seq"
| Case_aux c "E_IfTrue" | Case_aux c "E_IfFalse"
| Case_aux c "E_WhileEnd" | Case_aux c "E_WhileLoop"
| Case_aux c "E_Havoc" ].
(** The definition of Hoare triples is exactly as before. Unlike our
notion of program equivalence, which had subtle consequences with
occassionally nonterminating commands (exercise [havoc_diverge]),
this definition is still fully satisfactory. Convince yourself of
this before proceeding. *)
Definition hoare_triple (P:Assertion) (c:com) (Q:Assertion) : Prop :=
forall st st', c / st || st' -> P st -> Q st'.
Notation "{{ P }} c {{ Q }}" := (hoare_triple P c Q)
(at level 90, c at next level)
: hoare_spec_scope.
(** Complete the Hoare rule for [HAVOC] commands below by defining
[havoc_pre] and prove that the resulting rule is correct. *)
Definition havoc_pre (X : id) (Q : Assertion) : Assertion :=
(* FILL IN HERE *) admit.
Theorem hoare_havoc : forall (Q : Assertion) (X : id),
{{ havoc_pre X Q }} HAVOC X {{ Q }}.
Proof.
(* FILL IN HERE *) Admitted.
End Himp.
(** [] *)
(* ####################################################### *)
(** ** Complete List of Hoare Logic Rules *)
(** Above, we've introduced Hoare Logic as a tool to reasoning
about Imp programs. In the reminder of this chapter we will
explore a systematic way to use Hoare Logic to prove properties
about programs. The rules of Hoare Logic are the following: *)
(**
------------------------------ (hoare_asgn)
{{Q [X |-> a]}} X::=a {{Q}}
-------------------- (hoare_skip)
{{ P }} SKIP {{ P }}
{{ P }} c1 {{ Q }}
{{ Q }} c2 {{ R }}
--------------------- (hoare_seq)
{{ P }} c1;;c2 {{ R }}
{{P /\ b}} c1 {{Q}}
{{P /\ ~b}} c2 {{Q}}
------------------------------------ (hoare_if)
{{P}} IFB b THEN c1 ELSE c2 FI {{Q}}
{{P /\ b}} c {{P}}
----------------------------------- (hoare_while)
{{P}} WHILE b DO c END {{P /\ ~b}}
{{P'}} c {{Q'}}
P ->> P'
Q' ->> Q
----------------------------- (hoare_consequence)
{{P}} c {{Q}}
In the next chapter, we'll see how these rules are used to prove
that programs satisfy specifications of their behavior.
*)
(* $Date: 2013-11-20 13:03:49 -0500 (Wed, 20 Nov 2013) $ *)