# HoareHoare Logic, Part I

Our goal is to carry out some simple examples of program verification -- i.e., to use 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 generic "proof rule" that can be used to reason compositionally about the correctness of programs involving this construct.
Hoare Logic originated 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 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

An assertion is a claim about the current state of memory. We will use assertions to write program specifications.

Definition Assertion := stateProp.
For example,
• fun st st X = 3 holds if the value of X according to st is 3,
• fun st True always holds, and
• fun st False never holds.

Paraphrase the following assertions in English
(1) fun st st X st Y
(2) fun st st X = 3 st X st Y
(3) fun st st Z × st Z st X ¬ (((S (st Z)) × (S (st Z))) st X)

fun stst X = m
we'll write just
X = m
Given two assertions P and Q, we say that P implies Q, written P ->> Q, if, whenever P holds in some state st, Q also holds.

Definition assert_implies (P Q : Assertion) : Prop :=
st, P stQ st.

Notation "P ->> Q" := (assert_implies P Q)
(at level 80) : hoare_spec_scope.
Open Scope hoare_spec_scope.
We'll also want the "iff" variant of implication between assertions:

Notation "P <<->> Q" :=
(P ->> QQ ->> P) (at level 80) : hoare_spec_scope.
Our convention can be implemented uses Coq coercions and anotation scopes (much as we did with %imp in Imp) to automatically lift aexps, numbers, and Props into Assertions when they appear in the %assertion scope or when Coq knows the type of an expression is Assertion.

Definition Aexp : Type := statenat.

Definition assert_of_Prop (P : Prop) : Assertion := fun _P.
Definition Aexp_of_nat (n : nat) : Aexp := fun _n.

Definition Aexp_of_aexp (a : aexp) : Aexp := fun staeval st a.

Coercion assert_of_Prop : Sortclass >-> Assertion.
Coercion Aexp_of_nat : nat >-> Aexp.
Coercion Aexp_of_aexp : aexp >-> Aexp.

Arguments assert_of_Prop /.
Arguments Aexp_of_nat /.
Arguments Aexp_of_aexp /.

Bind Scope assertion_scope with Assertion.
Bind Scope assertion_scope with Aexp.
Delimit Scope assertion_scope with assertion.

Notation assert P := (P%assertion : Assertion).
Notation mkAexp a := (a%assertion : Aexp).

Notation "~ P" := (fun st ⇒ ¬assert P st) : assertion_scope.
Notation "P /\ Q" := (fun stassert P stassert Q st) : assertion_scope.
Notation "P \/ Q" := (fun stassert P stassert Q st) : assertion_scope.
Notation "P -> Q" := (fun stassert P stassert Q st) : assertion_scope.
Notation "P <-> Q" := (fun stassert P stassert Q st) : assertion_scope.
Notation "a = b" := (fun stmkAexp a st = mkAexp b st) : assertion_scope.
Notation "a <> b" := (fun stmkAexp a stmkAexp b st) : assertion_scope.
Notation "a <= b" := (fun stmkAexp a stmkAexp b st) : assertion_scope.
Notation "a < b" := (fun stmkAexp a st < mkAexp b st) : assertion_scope.
Notation "a >= b" := (fun stmkAexp a stmkAexp b st) : assertion_scope.
Notation "a > b" := (fun stmkAexp a st > mkAexp b st) : assertion_scope.
Notation "a + b" := (fun stmkAexp a st + mkAexp b st) : assertion_scope.
Notation "a - b" := (fun stmkAexp a st - mkAexp b st) : assertion_scope.
Notation "a * b" := (fun stmkAexp a st × mkAexp b st) : assertion_scope.
Lift functions to operate on assertion expressions.

Definition ap {X} (f : natX) (x : Aexp) :=
fun stf (x st).

Definition ap2 {X} (f : natnatX) (x : Aexp) (y : Aexp) (st : state) :=
f (x st) (y st).

Module ExPrettyAssertions.
Definition ex1 : Assertion := X = 3.
Definition ex2 : Assertion := True.
Definition ex3 : Assertion := False.

Definition assn1 : Assertion := XY.
Definition assn2 : Assertion := X = 3 ∨ XY.
Definition assn3 : Assertion :=
Z × ZX ∧ ¬(((ap S Z) × (ap S Z)) ≤ X).
Definition assn4 : Assertion :=
Z = ap2 max X Y.
End ExPrettyAssertions.

# Hoare Triples, Informally

A Hoare triple is a claim about the state before and after executing a command. A standard notation is
{P} c {Q}
meaning:
• If command c begins execution in a state satisfying assertion P,
• and if c eventually terminates in some final state,
• then that final state will satisfy the assertion Q.
Assertion P is called the precondition of the triple, and Q is the postcondition.
Because single braces are already used in other ways in Coq, we'll write Hoare triples with double braces:
{{P}} c {{Q}}
For example,
• {{X = 0}} X ::= X + 1 {{X = 1}} is a valid Hoare triple, stating that command X ::= X + 1 would transform a state in which X = 0 to a state in which X = 1.
• {{X = m}} X ::= X + 1 {{X = m + 1}}, is also a valid Hoare triple. It's even more descriptive of the exact behavior of that command than the previous example.

Paraphrase the following Hoare triples in English.
1) {{True}} c {{X = 5}}

2) {{X = m}} c {{X = m + 5)}}

3) {{XY}} c {{YX}}

4) {{True}} c {{False}}

5) {{X = m}}
c
{{Y = real_fact m}}

6) {{X = m}}
c
{{(Z × Z) ≤ m ∧ ¬(((S Z) × (S Z)) ≤ m)}}

Is the following Hoare triple valid -- i.e., is the claimed relation between P, c, and Q true?
{{True}} X ::= 5 {{X = 5}}
(1) Yes
(2) No
{{X = 2}} X ::= X + 1 {{X = 3}}
(1) Yes
(2) No
{{True}} X ::= 5;; Y ::= 0 {{X = 5}}
(1) Yes
(2) No
{{X = 2 ∧ X = 3}} X ::= 5 {{X = 0}}
(1) Yes
(2) No
{{True}} SKIP {{False}}
(1) Yes
(2) No
{{False}} SKIP {{True}}
(1) Yes
(2) No
{{True}} WHILE true DO SKIP END {{False}}
(1) Yes
(2) No
This one?
{{X = 0}}
WHILE X = 0 DO X ::= X + 1 END
{{X = 1}}
(1) Yes
(2) No
This one?
{{X = 1}}
WHILE ~(X = 0) DO X ::= X + 1 END
{{X = 100}}
(1) Yes
(2) No

# Hoare Triples, Formally

We can formalize Hoare triples and their notation in Coq as follows:

Definition hoare_triple
(P : Assertion) (c : com) (Q : Assertion) : Prop :=
st st',
st =[ c ]=> st'
P st
Q st'.

Notation "{{ P }} c {{ Q }}" :=
(hoare_triple P c Q) (at level 90, c at next level)
: hoare_spec_scope.

#### Exercise: 1 star, standard (hoare_post_true)

Prove that if postcondition Q holds in every state, then any triple with it is valid.

Theorem hoare_post_true : (P Q : Assertion) c,
( st, Q st) →
{{P}} c {{Q}}.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 1 star, standard (hoare_pre_false)

Prove that if precondition P holds in no state, then any triple with it is valid.

Theorem hoare_pre_false : (P Q : Assertion) c,
( st, ¬(P st)) →
{{P}} c {{Q}}.
Proof.
(* FILL IN HERE *) Admitted.

# Proof Rules

Plan:
• introduce one "proof rule" for each Imp syntactic form
• plus a couple of "structural rules" that help glue proofs together
• prove programs correct using these proof rules, without ever unfolding the definition of hoare_triple

## Assignment

How can we complete this triple?
{{ ??? }} X ::= Y {{ X = 1 }}
One possibility is:
{{ Y = 1 }} X ::= Y {{ X = 1 }}
The precondition is just the postcondition, but with X replaced by Y.
Another example:
{{ ??? }} X ::= X + Y {{ X = 1 }}
Replace X with X + Y:
{{ X + Y = 1 }} X ::= X + Y {{ X = 1 }}
This works because "equals 1" holding of X is guaranteed by "equals 1" holding of whatever is being assigned to X.
In general, the postcondition could be some arbitrary assertion Q, and the right-hand side of the assignment could be some arithmetic expression a:
{{ ??? }} X ::= a {{ Q }}
The precondition would then be Q, but with any occurrences of X in it replaced by a. Let's introduce a notation for that idea of replacing occurrences: let Q [X > a] mean "Q where a is substituted for X".
That yields the Hoare logic rule for assignment:
{{ Q [X > a] }} X ::= a {{ Q }}
One way of reading that rule is: if you want statement X ::= a to terminate in a state that satisfies assertion Q, then it suffices to start in a state that also satisfies Q, except where a is substituted for every occurrence of X.
To many people, this rule seems "backwards" at first, because it proceeds from the postcondition to the precondition. Actually it makes good sense to go in this direction: the postcondition is often what is more important, because it characterizes what we can assume afer running the code.
Nonetheless, it's possible to formulate a "forward" assignment rule. We'll do that later in some exercises.
Here are some valid instances of the assignment rule:
{{ (X ≤ 5) [X > X + 1] }} (that is, X + 1 ≤ 5)
X ::= X + 1
{{ X ≤ 5 }}

{{ (X = 3) [X > 3] }} (that is, 3 = 3)
X ::= 3
{{ X = 3 }}

{{ (0 ≤ XX ≤ 5) [X > 3] (that is, 0 ≤ 3 ∧ 3 ≤ 5)
X ::= 3
{{ 0 ≤ XX ≤ 5 }}
To formalize the rule, we must first formalize the idea of "substituting an expression for an Imp variable in an assertion", which we refer to as assertion substitution, or assn_sub. 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 P' should mention a wherever P mentions X.
Since P is an arbitrary Coq assertion, we can't directly "edit" its text. However, we can achieve the same effect by evaluating P in an updated state:

Definition assn_sub X a (P:Assertion) : Assertion :=
fun (st : state) ⇒
P (X !-> aeval st a ; st).

Notation "P [ X > a ]" := (assn_sub X a P)
(at level 10, X at next level).
That is, P [X > a] stands for an assertion -- let's call it 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)
(X !-> aeval st 3 ; st),
which simplifies to
fun st
(fun st'st' X ≤ 5)
(X !-> 3 ; st)
and further simplifies to
fun st
((X !-> 3 ; st) X) ≤ 5
and finally 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)
(X !-> aeval st (X + 1) ; st),
which simplifies to
fun st
(X !-> aeval st (X + 1) ; st) X ≤ 5
and further simplifies to
fun st
(aeval st (X + 1)) ≤ 5.
That is, P' is the assertion that X + 1 is at most 5.
Now, using the concept of substitution, 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 : 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 :
{{(X < 5) [X > X + 1]}}
X ::= X + 1
{{X < 5}}.
Proof.
(* WORK IN CLASS *) Admitted.
Of course, what would be even more helpful is to prove this simpler triple:
{{X < 4}} X ::= X + 1 {{X < 5}}
We will see how to do so in the next section.

Here is the assignment rule again:
{{ Q [X > a] }} X ::= a {{ Q }}
Is the following triple a valid instance of this rule?
{{ X = 5 }} X ::= X + 1 {{ X = 6 }}
(1) Yes
(2) No

Here is the assignment rule again:
{{ Q [X > a] }} X ::= a {{ Q }}
Is the following triple a valid instance of this rule?
{{ Y < Z }} X ::= Y {{ X < Z }}
(1) Yes
(2) No

The assignment rule again:
{{ Q [X > a] }} X ::= a {{ Q }}
Is the following triple a valid instance of this rule?
{{ X+1 < Y }} X ::= X + 1 {{ X < Y }}
(1) Yes
(2) No

The assignment rule again:
{{ Q [X > a] }} X ::= a {{ Q }}
Is the following triple a valid instance of this rule?
{{ X < Y }} X ::= X + 1 {{ X+1 < Y }}
(1) Yes
(2) No

The assignment rule again:
{{ Q [X > a] }} X ::= a {{ Q }}
Is the following triple a valid instance of this rule?
{{ X < Y }} X ::= X + 1 {{ X < Y+1 }}
(1) Yes
(2) No

The assignment rule again:
{{ Q [X > a] }} X ::= a {{ Q }}
Is the following triple a valid instance of this rule?
{{ True }} X ::= 3 {{ X = 3 }}
(1) Yes
(2) No

## 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 can 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 : (P P' Q : Assertion) c,
{{P'}} c {{Q}} →
P ->> P'
{{P}} c {{Q}}.
Proof.
unfold hoare_triple, "->>".
intros P P' Q c Hhoare Himp st st' Heval Hpre.
apply Hhoare with (st := st).
- assumption.
- apply Himp. assumption.
Qed.

Theorem hoare_consequence_post : (P Q Q' : Assertion) c,
{{P}} c {{Q'}} →
Q' ->> Q
{{P}} c {{Q}}.
Proof.
unfold hoare_triple, "->>".
intros P Q Q' c Hhoare Himp st st' Heval Hpre.
apply Himp.
apply Hhoare with (st := st).
- assumption.
- assumption.
Qed.
For example, we can use the first consequence rule like this:
{{ True }} ->>
{{ (X = 1) [X > 1] }}
X ::= 1
{{ X = 1 }}
Or, formally...

Example hoare_asgn_example1 :
{{True}} X ::= 1 {{X = 1}}.
Proof.
(* WORK IN CLASS *) Admitted.
We can also use it to prove the example mentioned earlier.
{{ X < 4 }} ->>
{{ (X < 5)[X > X + 1] }}
X ::= X + 1
{{ X < 5 }}
Or, formally ...

Example assn_sub_example2 :
{{X < 4}}
X ::= X + 1
{{X < 5}}.
Proof.
(* WORK IN CLASS *) Admitted.
Finally, here is 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 : (P P' Q Q' : Assertion) c,
{{P'}} c {{Q'}} →
P ->> P'
Q' ->> Q
{{P}} c {{Q}}.
Proof.
intros P P' Q Q' c Htriple Hpre Hpost.
apply hoare_consequence_pre with (P' := P').
- apply hoare_consequence_post with (Q' := Q').
+ assumption.
+ assumption.
- assumption.
Qed.

## Automation

Many of the proofs we have done so far with Hoare triples can be streamlined using the automation techniques that we introduced in the Auto chapter of Logical Foundations.
Recall that the auto tactic can be told to unfold definitions as part of its proof search. Let's give that it hint for the definitions and coercions we're using:

Hint Unfold assert_implies hoare_triple assn_sub t_update.
Hint Unfold assert_of_Prop Aexp_of_nat Aexp_of_aexp.
Also recall that auto will search for a proof involving intros and apply. By default, the theorems that it will apply include any of the local hypotheses, as well as theorems in a core database.
A good candidate for automation:

Theorem hoare_consequence_pre' : (P P' Q : Assertion) c,
{{P'}} c {{Q}} →
P ->> P'
{{P}} c {{Q}}.
Proof.
unfold hoare_triple, "->>".
intros P P' Q c Hhoare Himp st st' Heval Hpre.
apply Hhoare with (st := st).
- assumption.
- apply Himp. assumption.
Qed.
Tactic eapply will find st for us.

Theorem hoare_consequence_pre''' : (P P' Q : Assertion) c,
{{P'}} c {{Q}} →
P ->> P'
{{P}} c {{Q}}.
Proof.
unfold hoare_triple, "->>".
intros P P' Q c Hhoare Himp st st' Heval Hpre.
eapply Hhoare.
- eassumption.
- apply Himp. assumption.
Qed.
Tactic eauto will use eapply as part of its proof search. So, the entire proof can be done in just one line.

Theorem hoare_consequence_pre'''' : (P P' Q : Assertion) c,
{{P'}} c {{Q}} →
P ->> P'
{{P}} c {{Q}}.
Proof.
eauto.
Qed.
As can the same proof for the postcondition consequence rule.

Theorem hoare_consequence_post' : (P Q Q' : Assertion) c,
{{P}} c {{Q'}} →
Q' ->> Q
{{P}} c {{Q}}.
Proof.
eauto.
Qed.
We can also use eapply to streamline a proof, hoare_asgn_example1, that we did earlier as an example of using the consequence rule:

Example hoare_asgn_example1' :
{{True}} X ::= 1 {{X = 1}}.
Proof.
eapply hoare_consequence_pre. (* no need to state an assertion *)
- apply hoare_asgn.
- unfold "->>", assn_sub, t_update.
intros st _. simpl. reflexivity.
Qed.
The final bullet of that proof also looks like a candidate for automation.

Example hoare_asgn_example1'' :
{{True}} X ::= 1 {{X = 1}}.
Proof.
eapply hoare_consequence_pre.
- apply hoare_asgn.
- auto.
Qed.
Now we have quite a nice proof script: it identifies the Hoare rules that need to be used, and leaves the remaining low-level details up to Coq to figure out.
The other example of using consequence that we did earlier, hoare_asgn_example2, requires a little more work to automate. We can streamline the first line with eapply, but we can't just use auto for the final bullet, since it needs omega.

Example assn_sub_example2' :
{{X < 4}}
X ::= X + 1
{{X < 5}}.
Proof.
eapply hoare_consequence_pre.
- apply hoare_asgn.
- auto. (* no progress *)
unfold "->>", assn_sub, t_update.
intros st H. simpl in ×. omega.
Qed.
Let's introduce our own tactic to handle both that bullet, as well as the bullet from example 1:

Ltac assn_auto :=
try auto; (* as in example 1, above *)
try (unfold "->>", assn_sub, t_update;
intros; simpl in *; omega). (* as in example 2 *)
Example assn_sub_example2'' :
{{X < 4}}
X ::= X + 1
{{X < 5}}.
Proof.
eapply hoare_consequence_pre.
- apply hoare_asgn.
- assn_auto.
Qed.

Example hoare_asgn_example1''':
{{True}} X ::= 1 {{X = 1}}.
Proof.
eapply hoare_consequence_pre.
- apply hoare_asgn.
- assn_auto.
Qed.
Again, we have quite a nice proof script. All the low-level details of proof about assertions have been taken care of automatically. Of course, assn_auto isn't able to prove everything about assertions! But it's good enough so far.

## Skip

Since SKIP doesn't change the state, it preserves any assertion P:
-------------------- (hoare_skip)
{{ P }} SKIP {{ P }}

Theorem hoare_skip : P,
{{P}} SKIP {{P}}.
Proof.
intros P st st' H HP. inversion H; subst. assumption.
Qed.

## Sequencing

If 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 : P Q R c1 c2,
{{Q}} c2 {{R}} →
{{P}} c1 {{Q}} →
{{P}} c1;;c2 {{R}}.
Proof.
unfold hoare_triple.
intros P Q R c1 c2 H1 H2 st st' H12 Pre.
inversion H12; subst.
eauto.
Qed.
Here's an example of a program involving sequencing. Note the use of hoare_seq in conjunction with hoare_consequence_pre and the eapply tactic.

Example hoare_asgn_example3 : (a:aexp) (n:nat),
{{a = n}}
X ::= a;; SKIP
{{X = n}}.
Proof.
intros a n. eapply hoare_seq.
- (* right part of seq *)
apply hoare_skip.
- (* left part of seq *)
eapply hoare_consequence_pre.
+ apply hoare_asgn.
+ assn_auto.
Qed.
Informally, a nice way of displaying 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 }}

## Conditionals

What sort of rule do we want for reasoning about conditional commands?
Certainly, if the same assertion Q holds after executing either of the branches, then it holds after the whole conditional. So we might be tempted to write:
 {{P}} c1 {{Q}} {{P}} c2 {{Q}} {{P}} TEST b THEN c1 ELSE c2 {{Q}}
However, this is rather weak. For example, using this rule, we cannot show
{{ True }}
TEST X = 0
THEN Y ::= 2
ELSE Y ::= X + 1
FI
{{ XY }}
since the rule tells us nothing about the state in which the assignments take place in the "then" and "else" branches.
Better:
 {{P /\   b}} c1 {{Q}} {{P /\ ~ b}} c2 {{Q}} (hoare_if) {{P}} TEST b THEN c1 ELSE c2 FI {{Q}}
To make this formal, 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."

Definition bassn b : Assertion :=
fun st ⇒ (beval st b = true).

Now we can formalize the Hoare proof rule for conditionals and prove it correct.

Theorem hoare_if : P Q (b:bexp) c1 c2,
{{ Pb }} c1 {{Q}} →
{{ P ∧ ¬b}} c2 {{Q}} →
{{P}} TEST b THEN c1 ELSE c2 FI {{Q}}.
Proof.
intros P Q b c1 c2 HTrue HFalse st st' HE HP.
inversion HE; subst; eauto.
Qed.

### Example

Example if_example :
{{True}}
TEST X = 0
THEN Y ::= 2
ELSE Y ::= X + 1
FI
{{XY}}.
Proof.
apply hoare_if.
- (* Then *)
eapply hoare_consequence_pre.
+ apply hoare_asgn.
+ assn_auto. (* no progress *)
unfold "->>", assn_sub, t_update, bassn.
simpl. intros st [_ H].
apply eqb_eq in H.
rewrite H. omega.
- (* Else *)
eapply hoare_consequence_pre.
+ apply hoare_asgn.
+ assn_auto.
Qed.
As we did earlier, it would be nice to eliminate all the low-level proof script that isn't about the Hoare rules. Unfortunately, the assn_auto tactic we wrote wasn't quite up to the job. Looking at the proof of if_example, we can see why. We had to unfold a definition (bassn) and use a theorem (eqb_eq) that we didn't need in earlier proofs. So, let's add those into our tactic, and clean it up a little in the process.

Ltac assn_auto' :=
unfold "->>", assn_sub, t_update, bassn;
intros; simpl in *;
try rewriteeqb_eq in *; (* for equalities *)
auto; try omega.
Now the proof is quite streamlined.

Example if_example'' :
{{True}}
TEST X = 0
THEN Y ::= 2
ELSE Y ::= X + 1
FI
{{XY}}.
Proof.
apply hoare_if.
- eapply hoare_consequence_pre.
+ apply hoare_asgn.
+ assn_auto'.
- eapply hoare_consequence_pre.
+ apply hoare_asgn.
+ assn_auto'.
Qed.
We can even shorten it a little bit more.

Example if_example''' :
{{True}}
TEST X = 0
THEN Y ::= 2
ELSE Y ::= X + 1
FI
{{XY}}.
Proof.
apply hoare_if; eapply hoare_consequence_pre;
try apply hoare_asgn; try assn_auto'.
Qed.
For later proofs, it will help to extend assn_auto' to handle inequalities, too.

Ltac assn_auto'' :=
unfold "->>", assn_sub, t_update, bassn;
intros; simpl in *;
try rewriteeqb_eq in *;
try rewriteleb_le in *; (* for inequalities *)
auto; try omega.

## While Loops

Assertion P is an invariant of c if
{{P}} c {{P}}
holds.
The Hoare while rule combines the idea of an invariant with information about when guard b does or does not hold.

{{Pb}} c {{P}}
----------------------------- (hoare_while)
{{P} WHILE b DO c {{P ∧ ¬b}}

Theorem hoare_while : P (b:bexp) c,
{{Pb}} c {{P}} →
{{P}} WHILE b DO c END {{P ∧ ¬b}}.
Proof.
intros P b c Hhoare st st' Heval HP.
(* We proceed by induction on Heval, because, in the "keep looping" case,
its hypotheses talk about the whole loop instead of just c. The
remember is used to keep the original command in the hypotheses;
otherwise, it would be lost in the induction. By using inversion
we clear away all the cases except those involing while. *)

remember (WHILE b DO c END)%imp as original_command eqn:Horig.
induction Heval;
try (inversion Horig; subst; clear Horig);
eauto.
Qed.
We say that P is a loop invariant of WHILE b DO c if P suffices to prove hoare_while for that loop. Being a loop invariant is different than being an invariant of the body, because it means being able to prove correctness of the loop. For example, X = 0 is a loop invariant of
WHILE X = 2 DO X ::= 1 END
even though X = 0 is not an invariant of X ::= 1.

Is the assertion
Y = 0
a loop invariant of the following?
WHILE X < 100 DO X ::= X + 1 END
(1) Yes
(2) No
Is the assertion
X = 0
a loop invariant of the following?
WHILE X < 100 DO X ::= X + 1 END
(1) Yes
(2) No
Is the assertion
X < Y
a loop invariant of the following?
WHILE true DO X ::= X + 1;; Y ::= Y + 1 END
(1) Yes
(2) No
Is the assertion
X = Y + Z
a loop invariant of the following?
WHILE Y > 10 DO Y := Y - 1;; Z ::= Z + 1 END
(1) Yes
(2) No

Example while_example :
{{X ≤ 3}}
WHILE X ≤ 2 DO
X ::= X + 1
END
{{X = 3}}.
Proof.
eapply hoare_consequence_post.
- apply hoare_while.
eapply hoare_consequence_pre.
+ apply hoare_asgn.
+ assn_auto''.
- assn_auto''.
Qed.
Is the assertion
X > 0
a loop invariant of the following?
WHILE X = 0 DO X ::= X - 1 END
(1) Yes
(2) No
Is the assertion
X < 100
a loop invariant of the following?
WHILE X < 100 DO X ::= X + 1 END
(1) Yes
(2) No
Is the assertion
X > 10
a loop invariant of the following?
WHILE X > 10 DO X ::= X + 1 END
(1) Yes
(2) No

If the loop never terminates, any postcondition will work.

Theorem always_loop_hoare : Q,
{{True}} WHILE true DO SKIP END {{Q}}.
Proof.
intros Q.
eapply hoare_consequence_post.
- apply hoare_while. apply hoare_post_true. auto.
- simpl. intros st [Hinv Hguard]. congruence.
Qed.