HoareHoare Logic, Part I


Set Warnings "-notation-overridden,-parsing".
From PLF Require Import Maps.
From Coq Require Import Bool.Bool.
From Coq Require Import Arith.Arith.
From Coq Require Import Arith.EqNat.
From Coq Require Import Arith.PeanoNat. Import Nat.
From Coq Require Import Lia.
From PLF Require Export Imp.
In the final chaper of Logical Foundations (Software Foundations, volume 1), we began 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 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 (sometimes also subtly wrong!).
We'll return to the theme of metatheoretic properties of whole languages later in this volume when we discuss types and type soundness. In this chapter, though, we turn to a different set of issues.
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 := state Prop.
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.

Exercise: 1 star, standard, optional (assertions)

Paraphrase the following assertions in English (or your favorite natural language).

Module ExAssertions.
Definition assn1 : Assertion := fun stst X st Y.
Definition assn2 : Assertion :=
  fun stst X = 3 st X st Y.
Definition assn3 : Assertion :=
  fun stst Z × st Z st X
            ¬ (((S (st Z)) × (S (st Z))) st X).
Definition assn4 : Assertion :=
  fun stst Z = max (st X) (st Y).
(* 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 in assertions (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 stst X = m
we'll write just
      X = m
This example also illustrates a convention that we'll use throughout the Hoare Logic chapters: in informal assertions, capital letters like X, Y, and Z are Imp variables, while lowercase letters like x, y, m, and n are ordinary Coq variables (of type nat). This is why, when translating from informal to formal, we replace X with st X but leave m alone.
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 st Q st.

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

Notation "P <<->> Q" :=
  (P ->> Q Q ->> 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 := state nat.

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 /.

Declare Scope assertion_scope.
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 st assert Q st) : assertion_scope.
Notation "P \/ Q" := (fun stassert P st assert Q st) : assertion_scope.
Notation "P -> Q" := (fun stassert P st assert Q st) : assertion_scope.
Notation "P <-> Q" := (fun stassert P st assert Q 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.
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.
Notation "a * b" := (fun stmkAexp a st × mkAexp b st) : assertion_scope.
One small limitation of this approach is that we don't have an automatic way to coerce function applications that appear within an assertion to make appropriate use of the state. Instead, we use an explicit ap operator to lift the function.

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

Definition ap2 {X} (f : nat nat X) (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 := X Y.
Definition assn2 : Assertion := X = 3 X Y.
Definition assn3 : Assertion :=
  Z × Z X ¬ (((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.

Exercise: 1 star, standard, optional (triples)

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

     2) m, {{X = m}} c {{X = m + 5)}}

     3) {{XY}} c {{YX}}

     4) {{True}} c {{False}}

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

     6) m,
          {{X = m}}
          c
          {{(Z × Z) ≤ m ∧ ¬(((S Z) × (S Z)) ≤ m)}}
(* FILL IN HERE *)

Exercise: 1 star, standard, 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 *)

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 custom com at level 99)
  : hoare_spec_scope.
Check ({{True}} X := 0 {{True}}).

Exercise: 1 star, standard (hoare_post_true)

Prove that if Q holds in every state, then any triple with Q as its postcondition 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 P holds in no state, then any triple with P as its precondition is valid.

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

Proof Rules

The goal of Hoare logic is to provide a compositional method for proving the validity of specific Hoare triples. That is, we want the structure of a program's correctness proof to mirror the structure of the program itself. To this end, in the sections below, we'll introduce a 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 for gluing things together. We will then be able to 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 incomplete Hoare triple:
       {{ ??? }} X := Y {{ X = 1 }}
We want to assign Y to X and finish in a state where X is 1. What could the precondition be?
One possibility is Y = 1, because if Y is already 1 then assigning it to X causes X to be 1. That leads to a valid Hoare triple:
       {{ Y = 1 }} X := Y {{ X = 1 }}
It may seem as though coming up with that precondition must have taken some clever thought. But there is a mechanical way we could have done it: if we take the postcondition X = 1 and in it replace X with Y---that is, replace the left-hand side of the assignment statement with the right-hand side---we get the precondition, Y = 1.
That same technique works in more complicated cases. For example,
       {{ ??? }} X := X + Y {{ X = 1 }}
If we replace the X in X = 1 with X + Y, we get X + Y = 1. That again leads to a valid Hoare triple:
       {{ X + Y = 1 }} X := X + Y {{ X = 1 }}
Why does this technique work? The postcondition identifies some property P that we want to hold of the variable X being assigned. In this case, P is "equals 1". To complete the triple and make it valid, we need to identify a precondition that guarantees that property will hold of X. Such a precondition must ensure that the same property holds of whatever is being assigned to X. So, in the example, we need "equals 1" to hold of X + Y. That's exactly what the technique guarantees.
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 this idea of replacing occurrences: Define Q [X > a] to mean "Q where a is substituted in place of 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 also 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, a custom com).
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.
  (* WORKED IN CLASS *)
  apply hoare_asgn. Qed.
(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.

Exercise: 2 stars, standard, optional (hoare_asgn_examples)

Complete these Hoare triples...
    1) {{ ??? }}
       X ::= 2 × X
       {{ X ≤ 10 }}

    2) {{ ??? }}
       X := 3
       {{ 0 ≤ XX ≤ 5 }}
...using the names assn_sub_ex1 and assn_sub_ex2, and prove both with just apply hoare_asgn. If you find that tactic doesn't suffice, double check that you have completed the triple properly.

(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_hoare_asgn_examples : option (nat×string) := None.

Exercise: 2 stars, standard, especially useful (hoare_asgn_wrong)

The assignment rule looks backward to almost everyone the first time they see it. If it still seems puzzling, 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 and argue informally that it is really a counterexample. (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 *)

(* Do not modify the following line: *)
Definition manual_grade_for_hoare_asgn_wrong : option (nat×string) := None.

Exercise: 3 stars, advanced (hoare_asgn_fwd)

However, by using a parameter m (a Coq number) 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 => P st /\ st X = m}}
X := a
{{fun st => P st' /\ st X = aeval st' a }}
(where st' = (X !-> m ; st))
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. (Also note that this rule is more complicated than hoare_asgn.)

Theorem hoare_asgn_fwd :
   m a P,
  {{fun stP st st X = m}}
    X := a
  {{fun stP (X !-> m ; st)
            st X = aeval (X !-> m ; st) a }}.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, advanced, optional (hoare_asgn_fwd_exists)

Another way to define a forward rule for assignment is to existentially quantify over the previous value of the assigned variable. Prove that it is correct.
   (hoare_asgn_fwd_exists)  

{{fun st => P st}}
X := a
{{fun st => exists m, P (X !-> m ; st) /\
st X = aeval (X !-> m ; st) a }}

Theorem hoare_asgn_fwd_exists :
   a P,
  {{fun stP st}}
    X := a
  {{fun st m, P (X !-> m ; st)
                st X = aeval (X !-> m ; st) a }}.
Proof.
  (* 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 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.
  (* WORKED IN CLASS *)
  apply hoare_consequence_pre with (P' := (X = 1) [X > 1]).
  - apply hoare_asgn.
  - unfold "->>", assn_sub, t_update.
    intros st _. simpl. reflexivity.
Qed.
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.
  (* WORKED IN CLASS *)
  apply hoare_consequence_pre with (P' := (X < 5) [X > X + 1]).
  - apply hoare_asgn.
  - unfold "->>", assn_sub, t_update.
    intros st H. simpl in ×. lia.
Qed.
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 hint for the definitions and coercions we're using:

Hint Unfold assert_implies hoare_triple assn_sub t_update : core.
Hint Unfold assert_of_Prop Aexp_of_nat Aexp_of_aexp : core.
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.
The proof of hoare_consequence_pre, repeated below, looks like an opportune place for such automation, because all it does is unfold, intros, and apply. It uses assumption, too, but that's just application of a hypothesis.

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.
Merely using auto, though, doesn't complete the proof.

Theorem hoare_consequence_pre'' : (P P' Q : Assertion) c,
  {{P'}} c {{Q}}
  P ->> P'
  {{P}} c {{Q}}.
Proof.
  auto. (* no progress *)
Abort.
The problem is the apply Hhoare with... part of the proof. Coq isn't able to figure out how to instantiate st without some help from us. Recall, though, that there are versions of many tactics that will use existential variables to make progress even when the regular versions of those tactics would get stuck.
Here, the eapply tactic will introduce an existential variable ?st as a placeholder for st, and eassumption will instantiate ?st with st when it discovers st in assumption Heval. By using eapply we are essentially telling Coq, "Be patient: The missing part is going to be filled in later in the proof."

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.
Of course, it's hard to predict that eauto suffices here without having gone through the original proof of hoare_consequence_pre to see the tactics it used. But now that we know eauto works, it's a good bet that it will also work for hoare_consequence_post.

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 simply identifies the Hoare rules that need to be used and leaves the remaining low-level details up to Coq to figure out.
By now it might be apparent that the entire proof could be automated if we added hoare_consequence_pre and hoare_asgn to the hint database. We won't do that in this chapter, so that we can get a better understanding of when and how the Hoare rules are used. In the next chapter, Hoare2, we'll dive deeper into automating entire proofs of Hoare triples.
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 ×. lia.
Qed.
Let's introduce our own tactic to handle both that bullet and the bullet from example 1:

Ltac assn_auto :=
  try auto; (* as in example 1, above *)
  try (unfold "->>", assn_sub, t_update;
       intros; simpl in *; lia). (* 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 we could possibly want to know about assertions -- there's no magic here! But it's good enough so far.

Exercise: 2 stars, standard (hoare_asgn_examples_2)

Prove these triples. Try to make your proof scripts as nicely automated as those above.

Example assn_sub_ex1' :
  {{ X 5 }}
  X := 2 × X
  {{ X 10 }}.
Proof.
  (* FILL IN HERE *) Admitted.

Example assn_sub_ex2' :
  {{ 0 3 3 5 }}
  X := 3
  {{ 0 X X 5 }}.
Proof.
  (* FILL IN HERE *) Admitted.

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.
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, since 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.
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 }}

Exercise: 2 stars, standard, especially useful (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 }}
Note the use of "->>" decorations, each marking a use of hoare_consequence_pre.
We've started you off by providing a use of hoare_seq that explicitly identifies X = 1 as the intermediate assertion.

Example hoare_asgn_example4 :
  {{ True }}
  X := 1; Y := 2
  {{ X = 1 Y = 2 }}.
Proof.
  apply hoare_seq with (Q := (X = 1)%assertion).
  (* The annotation %assertion is needed here to help Coq parse correctly. *)
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars, standard (swap_exercise)

Write an Imp program c that swaps the values of X and Y and show that it satisfies the following specification:
      {{XY}} c {{YX}}
Your proof should not need to use unfold hoare_triple. (Hint: Remember that the assignment rule works best when it's applied "back to front," from the postcondition to the precondition. So your proof will want to start at the end and work back to the beginning of your program.)

Definition swap_program : com
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Theorem swap_exercise :
  {{X Y}}
  swap_program
  {{Y X}}.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 4 stars, standard (invalid_triple)

Show that
    {{ a = n }}
      X := 3;; Y := a
    {{ Y = n }}
is not a valid Hoare triple for some choices of a and n.
Conceptual hint: invent a particular a and n for which the triple in invalid, then use those to complete the proof.
Technical hint: hypothesis H, below, begins a n, .... You'll want to instantiate that for the particular a and n you've invented. You can do that with assert and apply, but Coq offers an even easier tactic: specialize. If you write
    specialize H with (a := your_a) (n := your_n)
the hypothesis will be instantiated on your_a and your_n.

Theorem invalid_triple : ¬ (a : aexp) (n : nat),
    {{ a = n }}
      X := 3; Y := a
    {{ Y = n }}.
Proof.
  unfold hoare_triple.
  intros H.
  (* FILL IN HERE *) Admitted.

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}} if b then c1 else c2 {{Q}}
However, this is rather weak. For example, using this rule, we cannot show
     {{ True }}
     if X = 0
       then Y := 2
       else Y := X + 1
     end
     {{ XY }}
since the rule tells us nothing about the state in which the assignments take place in the "then" and "else" branches.
Fortunately, we can 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}} if b then c1 else c2 end {{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).

Coercion bassn : bexp >-> Assertion.

Arguments bassn /.
Hint Unfold bassn : core.
A couple of useful facts about bassn:

Lemma bexp_eval_true : b st,
  beval st b = true (bassn b) st.
Proof. auto. Qed.

Lemma bexp_eval_false : b st,
  beval st b = false ¬ ((bassn b) st).
Proof. congruence. Qed.

Hint Resolve bexp_eval_false : core.
We mentioned the congruence tactic in passing in Auto when building the find_rwd tactic. Like find_rwd, congruence is able to automatically find that both beval st b = false and beval st b = true are being assumed, notice the contradiction, and discriminate to complete the proof.

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

Theorem hoare_if : P Q (b:bexp) c1 c2,
  {{ P b }} c1 {{Q}}
  {{ P ¬ b}} c2 {{Q}}
  {{P}} if b then c1 else c2 end {{Q}}.
That is (unwrapping the notations):
      Theorem hoare_if : P Q b c1 c2,
        {{fun stP stbassn b st}} c1 {{Q}} →
        {{fun stP st ∧ ¬(bassn b st)}} c2 {{Q}} →
        {{P}} if b then c1 else c2 end {{Q}}.
Proof.
  intros P Q b c1 c2 HTrue HFalse st st' HE HP.
  inversion HE; subst; eauto.
Qed.

Example

Here is a formal proof that the program we used to motivate the rule satisfies the specification we gave.

Example if_example :
    {{True}}
  if (X = 0)
    then Y := 2
    else Y := X + 1
  end
    {{X Y}}.
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. lia.
  - (* 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 lia.
Now the proof is quite streamlined.

Example if_example'' :
  {{True}}
  if X = 0
    then Y := 2
    else Y := X + 1
  end
  {{X Y}}.
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}}
  if X = 0
    then Y := 2
    else Y := X + 1
  end
  {{X Y}}.
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 lia.

Exercise: 2 stars, standard (if_minus_plus)

Prove the theorem below using hoare_if. Do not use unfold hoare_triple.

Theorem if_minus_plus :
  {{True}}
  if (X Y)
    then Z := Y - X
    else Y := X + Z
  end
  {{Y = X + Z}}.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: One-sided conditionals

In this exercise we consider extending Imp with "one-sided conditionals" of the form if1 b then c end. 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 end does nothing.
We recommend that you complete this exercise before attempting 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 : string aexp com
  | CSeq : com com com
  | CIf : bexp com com com
  | CWhile : bexp com com
  | CIf1 : bexp com com.

Notation "'if1' x 'then' y 'end'" :=
         (CIf1 x y)
             (in custom com at level 0, x custom com at level 99).
Notation "'skip'" :=
         CSkip (in custom com at level 0).
Notation "x := y" :=
         (CAss x y)
            (in custom com at level 0, x constr at level 0,
             y at level 85, no associativity).
Notation "x ; y" :=
         (CSeq x y)
           (in custom com at level 90, right associativity).
Notation "'if' x 'then' y 'else' z 'end'" :=
         (CIf x y z)
           (in custom com at level 89, x at level 99,
            y at level 99, z at level 99).
Notation "'while' x 'do' y 'end'" :=
         (CWhile x y)
            (in custom com at level 89, x at level 99, y at level 99).

Exercise: 2 stars, standard (if1_ceval)

Add two new evaluation rules to relation ceval, below, for if1. Let the rules for if guide you.

Reserved Notation "st '=[' c ']=>'' st'"
         (at level 40, c custom com at level 99,
          st constr, st' constr at next level).

Inductive ceval : com state state Prop :=
  | E_Skip : st,
      st =[ skip ]=> st
  | E_Ass : st a1 n x,
      aeval st a1 = n
      st =[ x := a1 ]=> (x !-> n ; st)
  | E_Seq : c1 c2 st st' st'',
      st =[ c1 ]=> st'
      st' =[ c2 ]=> st''
      st =[ c1 ; c2 ]=> st''
  | E_IfTrue : st st' b c1 c2,
      beval st b = true
      st =[ c1 ]=> st'
      st =[ if b then c1 else c2 end ]=> st'
  | E_IfFalse : st st' b c1 c2,
      beval st b = false
      st =[ c2 ]=> st'
      st =[ if b then c1 else c2 end ]=> st'
  | E_WhileFalse : b st c,
      beval st b = false
      st =[ while b do c end ]=> st
  | E_WhileTrue : st st' st'' b c,
      beval st b = true
      st =[ c ]=> st'
      st' =[ while b do c end ]=> st''
      st =[ while b do c end ]=> st''
(* FILL IN HERE *)

where "st '=[' c ']=>' st'" := (ceval c st st').

Hint Constructors ceval : core.
The following unit tests should be provable simply by eauto if you have defined the rules for if1 correctly.

Example if1true_test :
  empty_st =[ if1 X = 0 then X := 1 end ]=> (X !-> 1).
Proof. (* FILL IN HERE *) Admitted.

Example if1false_test :
  (X !-> 2) =[ if1 X = 0 then X := 1 end ]=> (X !-> 2).
Proof. (* FILL IN HERE *) Admitted.
Now we have to repeat the definition and notation of Hoare triples, so that they will use the updated com type.

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

Hint Unfold hoare_triple : core.

Notation "{{ P }} c {{ Q }}" := (hoare_triple P c Q)
                                  (at level 90, c custom com at level 99)
                                  : hoare_spec_scope.

Exercise: 2 stars, standard (hoare_if1)

Invent a Hoare logic proof rule for if1. State and prove a theorem named hoare_if1 that shows the validity of your rule. Use hoare_if as a guide. Try to invent a rule that is complete, meaning it can be used to prove the correctness of as many one-sided conditionals as possible. Also try to keep your rule compositional, meaning that any Imp command that appears in a premise should syntactically be a part of the command in the conclusion.
Hint: if you encounter difficulty getting Coq to parse part of your rule as an assertion, try manually indicating that it should be in the assertion scope. For example, if you want e to be parsed as an assertion, write it as (e)%assertion.

(* FILL IN HERE *)
For full credit, prove formally hoare_if1_good that your rule is precise enough to show the following valid Hoare triple:
  {{ X + Y = Z }}
  if1 ~(Y = 0) then
    X := X + Y
  end
  {{ X = Z }}
(* Do not modify the following line: *)
Definition manual_grade_for_hoare_if1 : option (nat×string) := None.
Before the next exercise, we need to restate the Hoare rules of consequence (for preconditions) and assignment for the new com type.

Theorem hoare_consequence_pre : (P P' Q : Assertion) c,
  {{P'}} c {{Q}}
  P ->> P'
  {{P}} c {{Q}}.
Proof.
  eauto.
Qed.

Theorem hoare_asgn : Q X a,
  {{Q [X > a]}} (X := a) {{Q}}.
Proof.
  intros Q X a st st' Heval HQ.
  inversion Heval; subst.
  auto.
Qed.

Exercise: 2 stars, standard (hoare_if1_good)

Prove that your if1 rule is complete enough for the following valid Hoare triple.
Hint: assn_auto'' once more will get you most but not all the way to a completely automated proof. You can finish manually, or tweak the tactic further.

Lemma hoare_if1_good :
  {{ X + Y = Z }}
  if1 ¬(Y = 0) then
    X := X + Y
  end
  {{ X = Z }}.
Proof. (* FILL IN HERE *) Admitted.

End If1.

While Loops

The Hoare rule for while loops is based on the idea of an invariant: an assertion whose truth is guaranteed before and after executing a command. An assertion P is an invariant of c if
      {{P}} c {{P}}
holds. Note that in the middle of executing c, the invariant might temporarily become false, but by the end of c, it must be restored.
As a first attempt at a while rule, we could try:

             {{P}} c {{P}}
      ---------------------------
      {{P} while b do c end {{P}}
That rule is valid: if P is an invariant of c, as the premise requires, then no matter how many times the loop body executes, P is going to be true when the loop finally finishes.
But the rule also omits two crucial pieces of information. First, the loop terminates when b becomes false. So we can strengthen the postcondition in the conclusion:

              {{P}} c {{P}}
      ---------------------------------
      {{P} while b do c end {{P ∧ ¬b}}
Second, the loop body will be executed only if b is true. So we can also strengthen the precondition in the premise:

            {{Pb}} c {{P}}
      --------------------------------- (hoare_while)
      {{P} while b do c end {{P ∧ ¬b}}
That is the Hoare while rule. Note how it combines aspects of skip and conditionals:
  • If the loop body executes zero times, the rule is like skip in that the precondition survives to become (part of) the postcondition.
  • Like a conditional, we can assume guard b holds on entry to the subcommand.

Theorem hoare_while : P (b:bexp) c,
  {{P b}} 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 involving while. *)

  remember <{while b do c end}> 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 end if P suffices to prove hoare_while for that loop. Being a loop invariant is different from 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.
This is a slightly (but crucially) 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 :
    {{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.
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.
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 specify what happens if commands terminate, without proving that they do, are said to describe a logic of partial correctness. It is also possible to give Hoare rules for total correctness, which additionally specifies that commands must terminate. Total correctness is out of the scope of this textbook.

Exercise: REPEAT

Exercise: 4 stars, advanced (hoare_repeat)

In this exercise, we'll add a new command to our language of commands: REPEAT c until b end. You will write the evaluation rule for REPEAT and add a new Hoare rule to the language for programs involving it. (You may recall that the evaluation rule is given in an example in the Auto chapter. Try to figure it out yourself here rather than peeking.)

Module RepeatExercise.

Inductive com : Type :=
  | CSkip : com
  | CAss : string 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.

Notation "'repeat' e1 'until' b2 'end'" :=
          (CRepeat e1 b2)
              (in custom com at level 0,
               e1 custom com at level 99, b2 custom com at level 99).
Notation "'skip'" :=
         CSkip (in custom com at level 0).
Notation "x := y" :=
         (CAss x y)
            (in custom com at level 0, x constr at level 0,
             y at level 85, no associativity).
Notation "x ; y" :=
         (CSeq x y)
           (in custom com at level 90, right associativity).
Notation "'if' x 'then' y 'else' z 'end'" :=
         (CIf x y z)
           (in custom com at level 89, x at level 99,
            y at level 99, z at level 99).
Notation "'while' x 'do' y 'end'" :=
         (CWhile x y)
            (in custom com at level 89, x at level 99, y at level 99).
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.

Inductive ceval : state com state Prop :=
  | E_Skip : st,
      st =[ skip ]=> st
  | E_Ass : st a1 n x,
      aeval st a1 = n
      st =[ x := a1 ]=> (x !-> n ; st)
  | E_Seq : c1 c2 st st' st'',
      st =[ c1 ]=> st'
      st' =[ c2 ]=> st''
      st =[ c1 ; c2 ]=> st''
  | E_IfTrue : st st' b c1 c2,
      beval st b = true
      st =[ c1 ]=> st'
      st =[ if b then c1 else c2 end ]=> st'
  | E_IfFalse : st st' b c1 c2,
      beval st b = false
      st =[ c2 ]=> st'
      st =[ if b then c1 else c2 end ]=> st'
  | E_WhileFalse : b st c,
      beval st b = false
      st =[ while b do c end ]=> st
  | E_WhileTrue : st st' st'' b c,
      beval st b = true
      st =[ c ]=> st'
      st' =[ while b do c end ]=> st''
      st =[ while b do c end ]=> st''
(* FILL IN HERE *)

where "st '=[' c ']=>' st'" := (ceval st c st').
A couple of definitions from above, copied here so they use the new ceval.

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 custom com at level 99).
To make sure you've got the evaluation rules for repeat right, prove that ex1_repeat evaluates correctly.

Definition ex1_repeat :=
  <{ repeat
       X := 1;
       Y := Y + 1
     until (X = 1) end }>.

Theorem ex1_repeat_works :
  empty_st =[ ex1_repeat ]=> (Y !-> 1 ; X !-> 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.

(* Do not modify the following line: *)
Definition manual_grade_for_hoare_repeat : option (nat×string) := None.

Summary

So far, we've introduced Hoare Logic as a tool for reasoning about Imp programs. The rules of Hoare Logic are:
   (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}} if b then c1 else c2 end {{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.

Additional Exercises

Havoc

In this exercise, we will derive proof rules for a HAVOC command, which is similar to the nondeterministic any expression from the the Imp 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
  | CAss : string aexp com
  | CSeq : com com com
  | CIf : bexp com com com
  | CWhile : bexp com com
  | CHavoc : string com.

Notation "'havoc' l" := (CHavoc l)
                          (in custom com at level 60, l constr at level 0).
Notation "'skip'" :=
         CSkip (in custom com at level 0).
Notation "x := y" :=
         (CAss x y)
            (in custom com at level 0, x constr at level 0,
             y at level 85, no associativity).
Notation "x ; y" :=
         (CSeq x y)
           (in custom com at level 90, right associativity).
Notation "'if' x 'then' y 'else' z 'end'" :=
         (CIf x y z)
           (in custom com at level 89, x at level 99,
            y at level 99, z at level 99).
Notation "'while' x 'do' y 'end'" :=
         (CWhile x y)
            (in custom com at level 89, x at level 99, y at level 99).

Inductive ceval : com state state Prop :=
  | E_Skip : st,
      st =[ skip ]=> st
  | E_Ass : st a1 n x,
      aeval st a1 = n
      st =[ x := a1 ]=> (x !-> n ; st)
  | E_Seq : c1 c2 st st' st'',
      st =[ c1 ]=> st'
      st' =[ c2 ]=> st''
      st =[ c1 ; c2 ]=> st''
  | E_IfTrue : st st' b c1 c2,
      beval st b = true
      st =[ c1 ]=> st'
      st =[ if b then c1 else c2 end ]=> st'
  | E_IfFalse : st st' b c1 c2,
      beval st b = false
      st =[ c2 ]=> st'
      st =[ if b then c1 else c2 end ]=> st'
  | E_WhileFalse : b st c,
      beval st b = false
      st =[ while b do c end ]=> st
  | E_WhileTrue : st st' st'' b c,
      beval st b = true
      st =[ c ]=> st'
      st' =[ while b do c end ]=> st''
      st =[ while b do c end ]=> st''
  | E_Havoc : st X n,
      st =[ havoc X ]=> (X !-> n ; st)

where "st '=[' c ']=>' st'" := (ceval c st st').

Hint Constructors ceval : core.
The definition of Hoare triples is exactly as before.

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

Hint Unfold hoare_triple : core.

Notation "{{ P }} c {{ Q }}" := (hoare_triple P c Q)
                                  (at level 90, c custom com at level 99)
                                  : hoare_spec_scope.
And the precondition consequence rule is exactly as before.

Theorem hoare_consequence_pre : (P P' Q : Assertion) c,
  {{P'}} c {{Q}}
  P ->> P'
  {{P}} c {{Q}}.
Proof. eauto. Qed.

Exercise: 3 stars, standard (hoare_havoc)

Complete the Hoare rule for HAVOC commands below by defining havoc_pre, and prove that the resulting rule is correct.

Definition havoc_pre (X : string) (Q : Assertion) (st : total_map nat) : Prop
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Theorem hoare_havoc : (Q : Assertion) (X : string),
  {{ havoc_pre X Q }} havoc X {{ Q }}.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars, standard (havoc_post)

Complete the following proof without changing any of the provided commands. If you find that it can't be completed, your definition of havoc_pre is probably too strong. Find a way to relax it so that havoc_post can be proved.
Hint: the assn_auto tactics we've built won't help you here. You need to proceed manually.

Theorem havoc_post : (P : Assertion) (X : string),
  {{ P }} havoc X {{ fun st (n:nat), P [X > n] st }}.
Proof.
  intros P X. eapply hoare_consequence_pre.
  - apply hoare_havoc.
  - (* FILL IN HERE *) Admitted.

End Himp.

Assert and Assume

Exercise: 4 stars, standard, optional (assert_vs_assume)

In this exercise, we will extend IMP with two commands, assert and ASSUME. Both commands are ways to indicate that a certain statement should hold any time this part of the program is reached. However they differ as follows:
  • If an assert statement fails, it causes the program to go into an error state and exit.
  • If an ASSUME statement fails, the program fails to evaluate at all. In other words, the program gets stuck and has no final state.
The new set of commands is:

Inductive com : Type :=
  | CSkip : com
  | CAss : string aexp com
  | CSeq : com com com
  | CIf : bexp com com com
  | CWhile : bexp com com
  | CAssert : bexp com
  | CAssume : bexp com.

Notation "'assert' l" := (CAssert l)
                           (in custom com at level 8, l custom com at level 0).
Notation "'assume' l" := (CAssume l)
                          (in custom com at level 8, l custom com at level 0).
Notation "'skip'" :=
         CSkip (in custom com at level 0).
Notation "x := y" :=
         (CAss x y)
            (in custom com at level 0, x constr at level 0,
             y at level 85, no associativity).
Notation "x ; y" :=
         (CSeq x y)
           (in custom com at level 90, right associativity).
Notation "'if' x 'then' y 'else' z 'end'" :=
         (CIf x y z)
           (in custom com at level 89, x at level 99,
            y at level 99, z at level 99).
Notation "'while' x 'do' y 'end'" :=
         (CWhile x y)
            (in custom com at level 89, x at level 99, y at level 99).
To define the behavior of assert and ASSUME, we need to add notation for an error, which indicates that an assertion has failed. We modify the ceval relation, therefore, so that it relates a start state to either an end state or to error. The result type indicates the end value of a program, either a state or an error:

Inductive result : Type :=
  | RNormal : state result
  | RError : result.
Now we are ready to give you the ceval relation for the new language.

Inductive ceval : com state result Prop :=
  (* Old rules, several modified *)
  | E_Skip : st,
      st =[ skip ]=> RNormal st
  | E_Ass : st a1 n x,
      aeval st a1 = n
      st =[ x := a1 ]=> RNormal (x !-> n ; st)
  | E_SeqNormal : c1 c2 st st' r,
      st =[ c1 ]=> RNormal st'
      st' =[ c2 ]=> r
      st =[ c1 ; c2 ]=> r
  | E_SeqError : c1 c2 st,
      st =[ c1 ]=> RError
      st =[ c1 ; c2 ]=> RError
  | E_IfTrue : st r b c1 c2,
      beval st b = true
      st =[ c1 ]=> r
      st =[ if b then c1 else c2 end ]=> r
  | E_IfFalse : st r b c1 c2,
      beval st b = false
      st =[ c2 ]=> r
      st =[ if b then c1 else c2 end ]=> r
  | E_WhileFalse : b st c,
      beval st b = false
      st =[ while b do c end ]=> RNormal st
  | E_WhileTrueNormal : st st' r b c,
      beval st b = true
      st =[ c ]=> RNormal st'
      st' =[ while b do c end ]=> r
      st =[ while b do c end ]=> r
  | E_WhileTrueError : st b c,
      beval st b = true
      st =[ c ]=> RError
      st =[ while b do c end ]=> RError
  (* Rules for Assert and Assume *)
  | E_AssertTrue : st b,
      beval st b = true
      st =[ assert b ]=> RNormal st
  | E_AssertFalse : st b,
      beval st b = false
      st =[ assert b ]=> RError
  | E_Assume : st b,
      beval st b = true
      st =[ assume b ]=> RNormal st

where "st '=[' c ']=>' r" := (ceval c st r).
We redefine hoare triples: Now, {{P}} c {{Q}} means that, whenever c is started in a state satisfying P, and terminates with result r, then r is not an error and the state of r satisfies Q.

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

Notation "{{ P }} c {{ Q }}" :=
  (hoare_triple P c Q) (at level 90, c custom com at level 99)
  : hoare_spec_scope.
To test your understanding of this modification, give an example precondition and postcondition that are satisfied by the ASSUME statement but not by the assert statement. Then prove that any triple for assert also works for ASSUME.

Theorem assert_assume_differ : (P:Assertion) b (Q:Assertion),
       ({{P}} assume b {{Q}})
   ¬ ({{P}} assert b {{Q}}).
(* FILL IN HERE *) Admitted.

Theorem assert_implies_assume : P b Q,
     ({{P}} assert b {{Q}})
   ({{P}} assume b {{Q}}).
Proof.
(* FILL IN HERE *) Admitted.
Your task is now to state Hoare rules for assert and assume, and use them to prove a simple program correct. Name your hoare rule theorems hoare_assert and hoare_assume.
For your benefit, we provide proofs for the old hoare rules adapted to the new semantics.

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.
   (X !-> aeval st a ; st). split; try reflexivity.
  assumption. Qed.

Theorem hoare_consequence_pre : (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 : (P Q Q' : Assertion) c,
  {{P}} c {{Q'}}
  Q' ->> Q
  {{P}} c {{Q}}.
Proof.
  intros P Q Q' c Hhoare Himp.
  intros st r Hc HP.
  unfold hoare_triple in Hhoare.
  assert ( st', r = RNormal st' Q' st').
  { apply (Hhoare st); assumption. }
  destruct H as [st' [Hr HQ'] ].
   st'. split; try assumption.
  apply Himp. assumption.
Qed.

Theorem hoare_seq : 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 r H12 Pre.
  inversion H12; subst.
  - eapply H1.
    + apply H6.
    + apply H2 in H3. apply H3 in Pre.
        destruct Pre as [st'0 [Heq HQ] ].
        inversion Heq; subst. assumption.
  - (* Find contradictory assumption *)
     apply H2 in H5. apply H5 in Pre.
     destruct Pre as [st' [C _] ].
     inversion C.
Qed.
State and prove your hoare rules, hoare_assert and hoare_assume, below.

(* FILL IN HERE *)
Here are the other proof rules (sanity check)
(* NOTATION : IY -- Do we want <{ }> to be printing in here? *)
Theorem hoare_skip : P,
     {{P}} skip {{P}}.
Proof.
  intros P st st' H HP. inversion H. subst.
  eexists. split. reflexivity. assumption.
Qed.

Theorem hoare_if : P Q (b:bexp) c1 c2,
  {{ P b}} c1 {{Q}}
  {{ P ¬ b}} c2 {{Q}}
  {{P}} if b then c1 else c2 end {{Q}}.
Proof.
  intros P Q b c1 c2 HTrue HFalse st st' HE HP.
  inversion HE; subst.
  - (* b is true *)
    apply (HTrue st st').
      assumption.
      split. assumption.
      apply bexp_eval_true. assumption.
  - (* b is false *)
    apply (HFalse st st').
      assumption.
      split. assumption.
      apply bexp_eval_false. assumption. Qed.

Theorem hoare_while : P (b:bexp) c,
  {{P b}} c {{P}}
  {{P}} while b do c end {{ P ¬b}}.
Proof.
  intros P b c Hhoare st st' He HP.
  remember <{while b do c end}> as wcom eqn:Heqwcom.
  induction He;
    try (inversion Heqwcom); subst; clear Heqwcom.
  - (* E_WhileFalse *)
    eexists. split. reflexivity. split.
    assumption. apply bexp_eval_false. assumption.
  - (* E_WhileTrueNormal *)
    clear IHHe1.
    apply IHHe2. reflexivity.
    clear IHHe2 He2 r.
    unfold hoare_triple in Hhoare.
    apply Hhoare in He1.
    + destruct He1 as [st1 [Heq Hst1] ].
        inversion Heq; subst.
        assumption.
    + split; assumption.
  - (* E_WhileTrueError *)
     exfalso. clear IHHe.
     unfold hoare_triple in Hhoare.
     apply Hhoare in He.
     + destruct He as [st' [C _] ]. inversion C.
     + split; assumption.
Qed.

Example assert_assume_example:
  {{True}}
  assume (X = 1);
  X := X + 1;
  assert (X = 2)
  {{True}}.
Proof.
(* FILL IN HERE *) Admitted.

End HoareAssertAssume.

(* 2020-09-09 21:08 *)