ImpSimple Imperative Programs
Z ::= X;;
Y ::= 1;;
WHILE not (Z = 0) DO
Y ::= Y * Z;;
Z ::= Z - 1
END
Y ::= 1;;
WHILE not (Z = 0) DO
Y ::= Y * Z;;
Z ::= Z - 1
END
Require Import Coq.Bool.Bool.
Require Import Coq.Arith.Arith.
Require Import Coq.Arith.EqNat.
Require Import Coq.omega.Omega.
Require Import Coq.Lists.List.
Import ListNotations.
Require Import Maps.
Arithmetic and Boolean Expressions
These two definitions specify the abstract syntax of
arithmetic and boolean expressions.
Inductive aexp : Type :=
| ANum : nat → aexp
| APlus : aexp → aexp → aexp
| AMinus : aexp → aexp → aexp
| AMult : aexp → aexp → aexp.
Inductive bexp : Type :=
| BTrue : bexp
| BFalse : bexp
| BEq : aexp → aexp → bexp
| BLe : aexp → aexp → bexp
| BNot : bexp → bexp
| BAnd : bexp → bexp → bexp.
In this chapter, we'll elide the translation from the
concrete syntax that a programmer would actually write to these
abstract syntax trees — the process that, for example, would
translate the string "1+2*3" to the AST APlus (ANum
1) (AMult (ANum 2) (ANum 3)). The optional chapter ImpParser
develops a simple implementation of a lexical analyzer and parser
that can perform this translation. You do not need to
understand that chapter to understand this one, but if you haven't
taken a course where these techniques are covered (e.g., a
compilers course) you may want to skim it.
For comparison, here's a conventional BNF (Backus-Naur Form)
grammar defining the same abstract syntax:
Compared to the Coq version above...
It's good to be comfortable with both sorts of notations:
informal ones for communicating between humans and formal ones for
carrying out implementations and proofs.
a ::= nat
| a + a
| a - a
| a * a
b ::= true
| false
| a = a
| a ≤ a
| not b
| b and b
| a + a
| a - a
| a * a
b ::= true
| false
| a = a
| a ≤ a
| not b
| b and b
- The BNF is more informal — for example, it gives some
suggestions about the surface syntax of expressions (like the
fact that the addition operation is written + and is an
infix symbol) while leaving other aspects of lexical analysis
and parsing (like the relative precedence of +, -, and
*, the use of parens to explicitly group subexpressions,
etc.) unspecified. Some additional information (and human
intelligence) would be required to turn this description into
a formal definition, for example when implementing a
compiler.
- On the other hand, the BNF version is lighter and easier to
read. Its informality makes it flexible, which is a huge
advantage in situations like discussions at the blackboard,
where conveying general ideas is more important than getting
every detail nailed down precisely.
Fixpoint aeval (a : aexp) : nat :=
match a with
| ANum n ⇒ n
| APlus a_{1} a_{2} ⇒ (aeval a_{1}) + (aeval a_{2})
| AMinus a_{1} a_{2} ⇒ (aeval a_{1}) - (aeval a_{2})
| AMult a_{1} a_{2} ⇒ (aeval a_{1}) * (aeval a_{2})
end.
Example test_aeval1:
aeval (APlus (ANum 2) (ANum 2)) = 4.
Proof. reflexivity. Qed.
Similarly, evaluating a boolean expression yields a boolean.
Fixpoint beval (b : bexp) : bool :=
match b with
| BTrue ⇒ true
| BFalse ⇒ false
| BEq a_{1} a_{2} ⇒ beq_nat (aeval a_{1}) (aeval a_{2})
| BLe a_{1} a_{2} ⇒ leb (aeval a_{1}) (aeval a_{2})
| BNot b_{1} ⇒ negb (beval b_{1})
| BAnd b_{1} b_{2} ⇒ andb (beval b_{1}) (beval b_{2})
end.
Optimization
Fixpoint optimize_0plus (a:aexp) : aexp :=
match a with
| ANum n ⇒
ANum n
| APlus (ANum 0) e_{2} ⇒
optimize_0plus e_{2}
| APlus e_{1} e_{2} ⇒
APlus (optimize_0plus e_{1}) (optimize_0plus e_{2})
| AMinus e_{1} e_{2} ⇒
AMinus (optimize_0plus e_{1}) (optimize_0plus e_{2})
| AMult e_{1} e_{2} ⇒
AMult (optimize_0plus e_{1}) (optimize_0plus e_{2})
end.
To make sure our optimization is doing the right thing we
can test it on some examples and see if the output looks OK.
Example test_optimize_0plus:
optimize_0plus (APlus (ANum 2)
(APlus (ANum 0)
(APlus (ANum 0) (ANum 1))))
= APlus (ANum 2) (ANum 1).
Proof. reflexivity. Qed.
But if we want to be sure the optimization is correct —
i.e., that evaluating an optimized expression gives the same
result as the original — we should prove it.
Theorem optimize_0plus_sound: ∀a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a. induction a.
- (* ANum *) reflexivity.
- (* APlus *) destruct a_{1}.
+ (* a_{1} = ANum n *) destruct n.
* (* n = 0 *) simpl. apply IHa2.
* (* n <> 0 *) simpl. rewrite IHa2. reflexivity.
+ (* a_{1} = APlus a1_1 a1_2 *)
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
+ (* a_{1} = AMinus a1_1 a1_2 *)
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
+ (* a_{1} = AMult a1_1 a1_2 *)
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
- (* AMinus *)
simpl. rewrite IHa1. rewrite IHa2. reflexivity.
- (* AMult *)
simpl. rewrite IHa1. rewrite IHa2. reflexivity. Qed.
Coq Automation
Tacticals
The try Tactical
Theorem silly1 : ∀ae, aeval ae = aeval ae.
Proof. try reflexivity. (* this just does reflexivity *) Qed.
Theorem silly2 : ∀(P : Prop), P → P.
Proof.
intros P HP.
try reflexivity. (* just reflexivity would have failed *)
apply HP. (* we can still finish the proof in some other way *)
Qed.
There is no real reason to use try in completely manual
proofs like these, but we'll see below that it is very useful for
doing automated proofs in conjunction with the ; tactical.
The ; Tactical (Simple Form)
Lemma foo : ∀n, leb 0 n = true.
Proof.
intros.
destruct n.
(* Leaves two subgoals, which are discharged
identically... *)
- (* n=0 *) simpl. reflexivity.
- (* n=Sn' *) simpl. reflexivity.
Qed.
We can simplify this proof using the ; tactical:
Lemma foo' : ∀n, leb 0 n = true.
Proof.
intros.
(* destruct the current goal *)
destruct n;
(* then simpl each resulting subgoal *)
simpl;
(* and do reflexivity on each resulting subgoal *)
reflexivity.
Qed.
Using try and ; together, we can get rid of the repetition in
the proof that was bothering us a little while ago.
Theorem optimize_0plus_sound': ∀a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a.
induction a;
(* Most cases follow directly by the IH... *)
try (simpl; rewrite IHa1; rewrite IHa2; reflexivity).
(* ... but the remaining cases -- ANum and APlus --
are different: *)
- (* ANum *) reflexivity.
- (* APlus *)
destruct a_{1};
(* Again, most cases follow directly by the IH: *)
try (simpl; simpl in IHa1; rewrite IHa1;
rewrite IHa2; reflexivity).
(* The interesting case, on which the try...
does nothing, is when e_{1} = ANum n. In this
case, we have to destruct n (to see whether
the optimization applies) and rewrite with the
induction hypothesis. *)
+ (* a_{1} = ANum n *) destruct n;
simpl; rewrite IHa2; reflexivity. Qed.
Coq experts often use this "...; try... " idiom after a tactic
like induction to take care of many similar cases all at once.
Naturally, this practice has an analog in informal proofs.
Here is an informal proof of this theorem that matches the
structure of the formal one:
Theorem: For all arithmetic expressions a,
This proof can still be improved: the first case (for a = ANum
n) is very trivial — even more trivial than the cases that we
said simply followed from the IH — yet we have chosen to write it
out in full. It would be better and clearer to drop it and just
say, at the top, "Most cases are either immediate or direct from
the IH. The only interesting case is the one for APlus..." We
can make the same improvement in our formal proof too. Here's how
it looks:
aeval (optimize_0plus a) = aeval a.
Proof: By induction on a. Most cases follow directly from the IH.
The remaining cases are as follows:
- Suppose a = ANum n for some n. We must show
aeval (optimize_0plus (ANum n)) = aeval (ANum n).This is immediate from the definition of optimize_0plus.
- Suppose a = APlus a_{1} a_{2} for some a_{1} and a_{2}. We
must show
aeval (optimize_0plus (APlus a_{1} a_{2}))Consider the possible forms of a_{1}. For most of them, optimize_0plus simply calls itself recursively for the subexpressions and rebuilds a new expression of the same form as a_{1}; in these cases, the result follows directly from the IH.
= aeval (APlus a_{1} a_{2}).optimize_0plus (APlus a_{1} a_{2}) = optimize_0plus a_{2}and the IH for a_{2} is exactly what we need. On the other hand, if n = S n' for some n', then again optimize_0plus simply calls itself recursively, and the result follows from the IH. ☐
Theorem optimize_0plus_sound'': ∀a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a.
induction a;
(* Most cases follow directly by the IH *)
try (simpl; rewrite IHa1; rewrite IHa2; reflexivity);
(* ... or are immediate by definition *)
try reflexivity.
(* The interesting case is when a = APlus a_{1} a_{2}. *)
- (* APlus *)
destruct a_{1}; try (simpl; simpl in IHa1; rewrite IHa1;
rewrite IHa2; reflexivity).
+ (* a_{1} = ANum n *) destruct n;
simpl; rewrite IHa2; reflexivity. Qed.
The ; Tactical (General Form)
T; [T_{1} | T_{2} | ... | Tn]
is a tactic that first performs T and then performs T_{1} on the
first subgoal generated by T, performs T_{2} on the second
subgoal, etc.
T; [T' | T' | ... | T']
The repeat Tactical
Theorem In_{10} : In 10 [1;2;3;4;5;6;7;8;9;10].
Proof.
repeat (try (left; reflexivity); right).
Qed.
(* Print In_{10}. *)
The repeat T tactic never fails: if the tactic T doesn't apply
to the original goal, then repeat still succeeds without changing
the original goal (i.e., it repeats zero times).
Theorem In_{10}' : In 10 [1;2;3;4;5;6;7;8;9;10].
Proof.
repeat (left; reflexivity).
repeat (right; try (left; reflexivity)).
Qed.
The repeat T tactic also does not have any upper bound on the
number of times it applies T. If T is a tactic that always
succeeds, then repeat T will loop forever (e.g., repeat simpl
loops forever, since simpl always succeeds). While Coq's term
language is guaranteed to terminate, Coq's tactic language is
not!
Exercise: 3 stars (optimize_0plus_b)
Since the optimize_0plus transformation doesn't change the value of aexps, we should be able to apply it to all the aexps that appear in a bexp without changing the bexp's value. Write a function which performs that transformation on bexps, and prove it is sound. Use the tacticals we've just seen to make the proof as elegant as possible.Fixpoint optimize_0plus_b (b : bexp) : bexp
(* REPLACE THIS LINE WITH := _your_definition_ . *) . Admitted.
Theorem optimize_0plus_b_sound : ∀b,
beval (optimize_0plus_b b) = beval b.
Proof.
(* FILL IN HERE *) Admitted.
☐
(* FILL IN HERE *)
☐
Exercise: 4 stars, optional (optimizer)
Design exercise: The optimization implemented by our optimize_0plus function is only one of many possible optimizations on arithmetic and boolean expressions. Write a more sophisticated optimizer and prove it correct.☐
Defining New Tactic Notations
- The Tactic Notation idiom illustrated below gives a handy way to
define "shorthand tactics" that bundle several tactics into a
single command.
- For more sophisticated programming, Coq offers a small built-in
programming language called Ltac with primitives that can
examine and modify the proof state. The details are a bit too
complicated to get into here (and it is generally agreed that
Ltac is not the most beautiful part of Coq's design!), but they
can be found in the reference manual and other books on Coq, and
there are many examples of Ltac definitions in the Coq standard
library that you can use as examples.
- There is also an OCaml API, which can be used to build tactics that access Coq's internal structures at a lower level, but this is seldom worth the trouble for ordinary Coq users.
Tactic Notation "simpl_and_try" tactic(c) :=
simpl;
try c.
This defines a new tactical called simpl_and_try that takes one
tactic c as an argument and is defined to be equivalent to the
tactic simpl; try c. For example, writing "simpl_and_try
reflexivity." in a proof would be the same as writing "simpl;
try reflexivity."
The omega Tactic
- numeric constants, addition (+ and S), subtraction (-
and pred), and multiplication by constants (this is what
makes it Presburger arithmetic),
- equality (= and ≠) and inequality (≤), and
- the logical connectives ∧, ∨, ¬, and →,
Require Import Coq.omega.Omega.
Example silly_presburger_example : ∀m n o p,
m + n ≤ n + o ∧ o + 3 = p + 3 →
m ≤ p.
Proof.
intros. omega.
Qed.
Leibniz wrote, "It is unworthy of excellent men to lose
hours like slaves in the labor of calculation which could be
relegated to anyone else if machines were used." We recommend
that excellent people of all genders use the omega tactic whenever
possible.
A Few More Handy Tactics
- clear H: Delete hypothesis H from the context.
- subst x: Find an assumption x = e or e = x in the
context, replace x with e throughout the context and
current goal, and clear the assumption.
- subst: Substitute away all assumptions of the form x = e
or e = x.
- rename... into...: Change the name of a hypothesis in the
proof context. For example, if the context includes a variable
named x, then rename x into y will change all occurrences
of x to y.
- assumption: Try to find a hypothesis H in the context that
exactly matches the goal; if one is found, behave just like
apply H.
- contradiction: Try to find a hypothesis H in the current
context that is logically equivalent to False. If one is
found, solve the goal.
- constructor: Try to find a constructor c (from some Inductive definition in the current environment) that can be applied to solve the current goal. If one is found, behave like apply c.
Evaluation as a Relation
Module aevalR_first_try.
Inductive aevalR : aexp → nat → Prop :=
| E_ANum : ∀(n: nat),
aevalR (ANum n) n
| E_APlus : ∀(e_{1} e_{2}: aexp) (n_{1} n_{2}: nat),
aevalR e_{1} n_{1} →
aevalR e_{2} n_{2} →
aevalR (APlus e_{1} e_{2}) (n_{1} + n_{2})
| E_AMinus: ∀(e_{1} e_{2}: aexp) (n_{1} n_{2}: nat),
aevalR e_{1} n_{1} →
aevalR e_{2} n_{2} →
aevalR (AMinus e_{1} e_{2}) (n_{1} - n_{2})
| E_AMult : ∀(e_{1} e_{2}: aexp) (n_{1} n_{2}: nat),
aevalR e_{1} n_{1} →
aevalR e_{2} n_{2} →
aevalR (AMult e_{1} e_{2}) (n_{1} * n_{2}).
As is often the case with relations, we'll find it
convenient to define infix notation for aevalR. We'll write e
⇓ n to mean that arithmetic expression e evaluates to value
n. (This notation is one place where the limitation to ASCII
symbols becomes a little bothersome. The standard notation for
the evaluation relation is a double down-arrow. We'll typeset it
like this in the HTML version of the notes and use a double slash
as the closest approximation in .v files.)
Notation "e '⇓' n"
:= (aevalR e n)
(at level 50, left associativity)
: type_scope.
End aevalR_first_try.
In fact, Coq provides a way to use this notation in the definition
of aevalR itself. This avoids situations where we're working on
a proof involving statements in the form e ⇓ n but we have to
refer back to a definition written using the form aevalR e n.
We do this by first "reserving" the notation, then giving the
definition together with a declaration of what the notation
means.
Reserved Notation "e '⇓' n" (at level 50, left associativity).
Inductive aevalR : aexp → nat → Prop :=
| E_ANum : ∀(n:nat),
(ANum n) ⇓ n
| E_APlus : ∀(e_{1} e_{2}: aexp) (n_{1} n_{2} : nat),
(e_{1} ⇓ n_{1}) → (e_{2} ⇓ n_{2}) → (APlus e_{1} e_{2}) ⇓ (n_{1} + n_{2})
| E_AMinus : ∀(e_{1} e_{2}: aexp) (n_{1} n_{2} : nat),
(e_{1} ⇓ n_{1}) → (e_{2} ⇓ n_{2}) → (AMinus e_{1} e_{2}) ⇓ (n_{1} - n_{2})
| E_AMult : ∀(e_{1} e_{2}: aexp) (n_{1} n_{2} : nat),
(e_{1} ⇓ n_{1}) → (e_{2} ⇓ n_{2}) → (AMult e_{1} e_{2}) ⇓ (n_{1} * n_{2})
where "e '⇓' n" := (aevalR e n) : type_scope.
Inference Rule Notation
| E_APlus : ∀(e_{1} e_{2}: aexp) (n_{1} n_{2}: nat),
aevalR e_{1} n_{1} →
aevalR e_{2} n_{2} →
aevalR (APlus e_{1} e_{2}) (n_{1} + n_{2})
...would be written like this as an inference rule:
aevalR e_{1} n_{1} →
aevalR e_{2} n_{2} →
aevalR (APlus e_{1} e_{2}) (n_{1} + n_{2})
e_{1} ⇓ n_{1} | |
e_{2} ⇓ n_{2} | (E_APlus) |
APlus e_{1} e_{2} ⇓ n_{1}+n_{2} |
(E_ANum) | |
ANum n ⇓ n |
e_{1} ⇓ n_{1} | |
e_{2} ⇓ n_{2} | (E_APlus) |
APlus e_{1} e_{2} ⇓ n_{1}+n_{2} |
e_{1} ⇓ n_{1} | |
e_{2} ⇓ n_{2} | (E_AMinus) |
AMinus e_{1} e_{2} ⇓ n_{1}-n_{2} |
e_{1} ⇓ n_{1} | |
e_{2} ⇓ n_{2} | (E_AMult) |
AMult e_{1} e_{2} ⇓ n_{1}*n_{2} |
Equivalence of the Definitions
Theorem aeval_iff_aevalR : ∀a n,
(a ⇓ n) ↔ aeval a = n.
Proof.
split.
- (* -> *)
intros H.
induction H; simpl.
+ (* E_ANum *)
reflexivity.
+ (* E_APlus *)
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
+ (* E_AMinus *)
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
+ (* E_AMult *)
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
- (* <- *)
generalize dependent n.
induction a;
simpl; intros; subst.
+ (* ANum *)
apply E_ANum.
+ (* APlus *)
apply E_APlus.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
+ (* AMinus *)
apply E_AMinus.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
+ (* AMult *)
apply E_AMult.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
Qed.
split.
- (* -> *)
intros H.
induction H; simpl.
+ (* E_ANum *)
reflexivity.
+ (* E_APlus *)
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
+ (* E_AMinus *)
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
+ (* E_AMult *)
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
- (* <- *)
generalize dependent n.
induction a;
simpl; intros; subst.
+ (* ANum *)
apply E_ANum.
+ (* APlus *)
apply E_APlus.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
+ (* AMinus *)
apply E_AMinus.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
+ (* AMult *)
apply E_AMult.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
Qed.
We can make the proof quite a bit shorter by making more
use of tacticals...
Theorem aeval_iff_aevalR' : ∀a n,
(a ⇓ n) ↔ aeval a = n.
Proof.
(* WORKED IN CLASS *)
split.
- (* -> *)
intros H; induction H; subst; reflexivity.
- (* <- *)
generalize dependent n.
induction a; simpl; intros; subst; constructor;
try apply IHa1; try apply IHa2; reflexivity.
Qed.
Exercise: 3 stars (bevalR)
Write a relation bevalR in the same style as aevalR, and prove that it is equivalent to beval.(*
Inductive bevalR:
(* FILL IN HERE *)
*)
☐
Computational vs. Relational Definitions
For example, suppose that we wanted to extend the arithmetic
operations by considering also a division operation:
Inductive aexp : Type :=
| ANum : nat → aexp
| APlus : aexp → aexp → aexp
| AMinus : aexp → aexp → aexp
| AMult : aexp → aexp → aexp
| ADiv : aexp → aexp → aexp. (* <--- new *)
Extending the definition of aeval to handle this new operation
would not be straightforward (what should we return as the result
of ADiv (ANum 5) (ANum 0)?). But extending aevalR is
straightforward.
Reserved Notation "e '⇓' n"
(at level 50, left associativity).
Inductive aevalR : aexp → nat → Prop :=
| E_ANum : ∀(n:nat),
(ANum n) ⇓ n
| E_APlus : ∀(a_{1} a_{2}: aexp) (n_{1} n_{2} : nat),
(a_{1} ⇓ n_{1}) → (a_{2} ⇓ n_{2}) → (APlus a_{1} a_{2}) ⇓ (n_{1} + n_{2})
| E_AMinus : ∀(a_{1} a_{2}: aexp) (n_{1} n_{2} : nat),
(a_{1} ⇓ n_{1}) → (a_{2} ⇓ n_{2}) → (AMinus a_{1} a_{2}) ⇓ (n_{1} - n_{2})
| E_AMult : ∀(a_{1} a_{2}: aexp) (n_{1} n_{2} : nat),
(a_{1} ⇓ n_{1}) → (a_{2} ⇓ n_{2}) → (AMult a_{1} a_{2}) ⇓ (n_{1} * n_{2})
| E_ADiv : ∀(a_{1} a_{2}: aexp) (n_{1} n_{2} n_{3}: nat),
(a_{1} ⇓ n_{1}) → (a_{2} ⇓ n_{2}) → (n_{2} > 0) →
(mult n_{2} n_{3} = n_{1}) → (ADiv a_{1} a_{2}) ⇓ n_{3}
where "a '⇓' n" := (aevalR a n) : type_scope.
End aevalR_division.
Module aevalR_extended.
Adding Nondeterminism
Reserved Notation "e '⇓' n" (at level 50, left associativity).
Inductive aexp : Type :=
| AAny : aexp (* <--- NEW *)
| ANum : nat → aexp
| APlus : aexp → aexp → aexp
| AMinus : aexp → aexp → aexp
| AMult : aexp → aexp → aexp.
Again, extending aeval would be tricky, since now evaluation is
not a deterministic function from expressions to numbers, but
extending aevalR is no problem:
Inductive aevalR : aexp → nat → Prop :=
| E_Any : ∀(n:nat),
AAny ⇓ n (* <--- new *)
| E_ANum : ∀(n:nat),
(ANum n) ⇓ n
| E_APlus : ∀(a_{1} a_{2}: aexp) (n_{1} n_{2} : nat),
(a_{1} ⇓ n_{1}) → (a_{2} ⇓ n_{2}) → (APlus a_{1} a_{2}) ⇓ (n_{1} + n_{2})
| E_AMinus : ∀(a_{1} a_{2}: aexp) (n_{1} n_{2} : nat),
(a_{1} ⇓ n_{1}) → (a_{2} ⇓ n_{2}) → (AMinus a_{1} a_{2}) ⇓ (n_{1} - n_{2})
| E_AMult : ∀(a_{1} a_{2}: aexp) (n_{1} n_{2} : nat),
(a_{1} ⇓ n_{1}) → (a_{2} ⇓ n_{2}) → (AMult a_{1} a_{2}) ⇓ (n_{1} * n_{2})
where "a '⇓' n" := (aevalR a n) : type_scope.
End aevalR_extended.
Expressions With Variables
States
Syntax
Inductive aexp : Type :=
| ANum : nat → aexp
| AId : id → aexp (* <----- NEW *)
| APlus : aexp → aexp → aexp
| AMinus : aexp → aexp → aexp
| AMult : aexp → aexp → aexp.
Defining a few variable names as notational shorthands will make
examples easier to read:
(This convention for naming program variables (X, Y,
Z) clashes a bit with our earlier use of uppercase letters for
types. Since we're not using polymorphism heavily in this part of
the course, this overloading should not cause confusion.)
The definition of bexps is unchanged (except for using the new
aexps):
Inductive bexp : Type :=
| BTrue : bexp
| BFalse : bexp
| BEq : aexp → aexp → bexp
| BLe : aexp → aexp → bexp
| BNot : bexp → bexp
| BAnd : bexp → bexp → bexp.
Evaluation
Fixpoint aeval (st : state) (a : aexp) : nat :=
match a with
| ANum n ⇒ n
| AId x ⇒ st x (* <----- NEW *)
| APlus a_{1} a_{2} ⇒ (aeval st a_{1}) + (aeval st a_{2})
| AMinus a_{1} a_{2} ⇒ (aeval st a_{1}) - (aeval st a_{2})
| AMult a_{1} a_{2} ⇒ (aeval st a_{1}) * (aeval st a_{2})
end.
Fixpoint beval (st : state) (b : bexp) : bool :=
match b with
| BTrue ⇒ true
| BFalse ⇒ false
| BEq a_{1} a_{2} ⇒ beq_nat (aeval st a_{1}) (aeval st a_{2})
| BLe a_{1} a_{2} ⇒ leb (aeval st a_{1}) (aeval st a_{2})
| BNot b_{1} ⇒ negb (beval st b_{1})
| BAnd b_{1} b_{2} ⇒ andb (beval st b_{1}) (beval st b_{2})
end.
Example aexp1 :
aeval (t_update empty_state X 5)
(APlus (ANum 3) (AMult (AId X) (ANum 2)))
= 13.
Proof. reflexivity. Qed.
Example bexp1 :
beval (t_update empty_state X 5)
(BAnd BTrue (BNot (BLe (AId X) (ANum 4))))
= true.
Proof. reflexivity. Qed.
Commands
Syntax
c ::= SKIP | x ::= a | c ;; c | IFB b THEN c ELSE c FI
| WHILE b DO c END
| WHILE b DO c END
Z ::= X;;
Y ::= 1;;
WHILE not (Z = 0) DO
Y ::= Y * Z;;
Z ::= Z - 1
END
When this command terminates, the variable Y will contain the
factorial of the initial value of X.
Y ::= 1;;
WHILE not (Z = 0) DO
Y ::= Y * Z;;
Z ::= Z - 1
END
Inductive com : Type :=
| CSkip : com
| CAss : id → aexp → com
| CSeq : com → com → com
| CIf : bexp → com → com → com
| CWhile : bexp → com → com.
As usual, we can use a few Notation declarations to make things
more readable. To avoid conflicts with Coq's built-in notations,
we keep this light — in particular, we don't introduce any
notations for aexps and bexps to avoid confusion with the
numeric and boolean operators we've already defined.
Notation "'SKIP'" :=
CSkip.
Notation "x '::=' a" :=
(CAss x a) (at level 60).
Notation "c_{1} ;; c_{2}" :=
(CSeq c_{1} c_{2}) (at level 80, right associativity).
Notation "'WHILE' b 'DO' c 'END'" :=
(CWhile b c) (at level 80, right associativity).
Notation "'IFB' c_{1} 'THEN' c_{2} 'ELSE' c_{3} 'FI'" :=
(CIf c_{1} c_{2} c_{3}) (at level 80, right associativity).
For example, here is the factorial function again, written as a
formal definition to Coq:
Definition fact_in_coq : com :=
Z ::= AId X;;
Y ::= ANum 1;;
WHILE BNot (BEq (AId Z) (ANum 0)) DO
Y ::= AMult (AId Y) (AId Z);;
Z ::= AMinus (AId Z) (ANum 1)
END.
Definition plus2 : com :=
X ::= (APlus (AId X) (ANum 2)).
Definition XtimesYinZ : com :=
Z ::= (AMult (AId X) (AId Y)).
Definition subtract_slowly_body : com :=
Z ::= AMinus (AId Z) (ANum 1) ;;
X ::= AMinus (AId X) (ANum 1).
Definition subtract_slowly : com :=
WHILE BNot (BEq (AId X) (ANum 0)) DO
subtract_slowly_body
END.
Definition subtract_3_from_5_slowly : com :=
X ::= ANum 3 ;;
Z ::= ANum 5 ;;
subtract_slowly.
Evaluation
Evaluation as a Function (Failed Attempt)
Fixpoint ceval_fun_no_while (st : state) (c : com)
: state :=
match c with
| SKIP ⇒
st
| x ::= a_{1} ⇒
t_update st x (aeval st a_{1})
| c_{1} ;; c_{2} ⇒
let st' := ceval_fun_no_while st c_{1} in
ceval_fun_no_while st' c_{2}
| IFB b THEN c_{1} ELSE c_{2} FI ⇒
if (beval st b)
then ceval_fun_no_while st c_{1}
else ceval_fun_no_while st c_{2}
| WHILE b DO c END ⇒
st (* bogus *)
end.
In a traditional functional programming language like OCaml or
Haskell we could add the WHILE case as follows:
Thus, because it doesn't terminate on all inputs, the full version
of ceval_fun cannot be written in Coq — at least not without
additional tricks (see chapter ImpCEvalFun if you're curious
about what those might be).
Fixpoint ceval_fun (st : state) (c : com) : state := match c with ... | WHILE b DO c END => if (beval st b) then ceval_fun st (c; WHILE b DO c END) else st end.Coq doesn't accept such a definition ("Error: Cannot guess decreasing argument of fix") because the function we want to define is not guaranteed to terminate. Indeed, it doesn't always terminate: for example, the full version of the ceval_fun function applied to the loop program above would never terminate. Since Coq is not just a functional programming language, but also a consistent logic, any potentially non-terminating function needs to be rejected. Here is an (invalid!) Coq program showing what would go wrong if Coq allowed non-terminating recursive functions:
Fixpoint loop_false (n : nat) : False := loop_false n.That is, propositions like False would become provable (e.g., loop_false 0 would be a proof of False), which would be a disaster for Coq's logical consistency.
Evaluation as a Relation
Operational Semantics
Here is an informal definition of evaluation, presented as inference rules for the sake of readability:(E_Skip) | |
SKIP / st ⇓ st |
aeval st a_{1} = n | (E_Ass) |
x := a_{1} / st ⇓ (t_update st x n) |
c_{1} / st ⇓ st' | |
c_{2} / st' ⇓ st'' | (E_Seq) |
c_{1};;c_{2} / st ⇓ st'' |
beval st b_{1} = true | |
c_{1} / st ⇓ st' | (E_IfTrue) |
IF b_{1} THEN c_{1} ELSE c_{2} FI / st ⇓ st' |
beval st b_{1} = false | |
c_{2} / st ⇓ st' | (E_IfFalse) |
IF b_{1} THEN c_{1} ELSE c_{2} FI / st ⇓ st' |
beval st b = false | (E_WhileEnd) |
WHILE b DO c END / st ⇓ st |
beval st b = true | |
c / st ⇓ st' | |
WHILE b DO c END / st' ⇓ st'' | (E_WhileLoop) |
WHILE b DO c END / st ⇓ st'' |
Reserved Notation "c_{1} '/' st '⇓' st'"
(at level 40, st at level 39).
Inductive ceval : com → state → state → Prop :=
| E_Skip : ∀st,
SKIP / st ⇓ st
| E_Ass : ∀st a_{1} n x,
aeval st a_{1} = n →
(x ::= a_{1}) / st ⇓ (t_update st x n)
| E_Seq : ∀c_{1} c_{2} st st' st'',
c_{1} / st ⇓ st' →
c_{2} / st' ⇓ st'' →
(c_{1} ;; c_{2}) / st ⇓ st''
| E_IfTrue : ∀st st' b c_{1} c_{2},
beval st b = true →
c_{1} / st ⇓ st' →
(IFB b THEN c_{1} ELSE c_{2} FI) / st ⇓ st'
| E_IfFalse : ∀st st' b c_{1} c_{2},
beval st b = false →
c_{2} / st ⇓ st' →
(IFB b THEN c_{1} ELSE c_{2} FI) / st ⇓ st'
| E_WhileEnd : ∀b st c,
beval st b = false →
(WHILE b DO c END) / st ⇓ st
| E_WhileLoop : ∀st st' st'' b c,
beval st b = true →
c / st ⇓ st' →
(WHILE b DO c END) / st' ⇓ st'' →
(WHILE b DO c END) / st ⇓ st''
where "c_{1} '/' st '⇓' st'" := (ceval c_{1} st st').
The cost of defining evaluation as a relation instead of a
function is that we now need to construct proofs that some
program evaluates to some result state, rather than just letting
Coq's computation mechanism do it for us.
Example ceval_example1:
(X ::= ANum 2;;
IFB BLe (AId X) (ANum 1)
THEN Y ::= ANum 3
ELSE Z ::= ANum 4
FI)
/ empty_state
⇓ (t_update (t_update empty_state X 2) Z 4).
Proof.
(* We must supply the intermediate state *)
apply E_Seq with (t_update empty_state X 2).
- (* assignment command *)
apply E_Ass. reflexivity.
- (* if command *)
apply E_IfFalse.
reflexivity.
apply E_Ass. reflexivity. Qed.
Example ceval_example2:
(X ::= ANum 0;; Y ::= ANum 1;; Z ::= ANum 2) / empty_state ⇓
(t_update (t_update (t_update empty_state X 0) Y 1) Z 2).
Proof.
(* FILL IN HERE *) Admitted.
(X ::= ANum 0;; Y ::= ANum 1;; Z ::= ANum 2) / empty_state ⇓
(t_update (t_update (t_update empty_state X 0) Y 1) Z 2).
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 3 stars, advanced (pup_to_n)
Write an Imp program that sums the numbers from 1 to X (inclusive: 1 + 2 + ... + X) in the variable Y. Prove that this program executes as intended for X = 2 (the latter is trickier than you might expect).Definition pup_to_n : com
(* REPLACE THIS LINE WITH := _your_definition_ . *) . Admitted.
Theorem pup_to_2_ceval :
pup_to_n / (t_update empty_state X 2) ⇓
t_update (t_update (t_update (t_update (t_update (t_update empty_state
X 2) Y 0) Y 2) X 1) Y 3) X 0.
Proof.
(* FILL IN HERE *) Admitted.
☐
Determinism of Evaluation
Theorem ceval_deterministic: ∀c st st_{1} st_{2},
c / st ⇓ st_{1} →
c / st ⇓ st_{2} →
st_{1} = st_{2}.
Proof.
intros c st st_{1} st_{2} E_{1} E_{2}.
generalize dependent st_{2}.
induction E_{1};
intros st_{2} E_{2}; inversion E_{2}; subst.
- (* E_Skip *) reflexivity.
- (* E_Ass *) reflexivity.
- (* E_Seq *)
assert (st' = st'0) as EQ_{1}.
{ (* Proof of assertion *) apply IHE1_1; assumption. }
subst st'0.
apply IHE1_2. assumption.
- (* E_IfTrue, b_{1} evaluates to true *)
apply IHE1. assumption.
- (* E_IfTrue, b_{1} evaluates to false (contradiction) *)
rewrite H in H_{5}. inversion H_{5}.
- (* E_IfFalse, b_{1} evaluates to true (contradiction) *)
rewrite H in H_{5}. inversion H_{5}.
- (* E_IfFalse, b_{1} evaluates to false *)
apply IHE1. assumption.
- (* E_WhileEnd, b_{1} evaluates to false *)
reflexivity.
- (* E_WhileEnd, b_{1} evaluates to true (contradiction) *)
rewrite H in H_{2}. inversion H_{2}.
- (* E_WhileLoop, b_{1} evaluates to false (contradiction) *)
rewrite H in H_{4}. inversion H_{4}.
- (* E_WhileLoop, b_{1} evaluates to true *)
assert (st' = st'0) as EQ_{1}.
{ (* Proof of assertion *) apply IHE1_1; assumption. }
subst st'0.
apply IHE1_2. assumption. Qed.
intros c st st_{1} st_{2} E_{1} E_{2}.
generalize dependent st_{2}.
induction E_{1};
intros st_{2} E_{2}; inversion E_{2}; subst.
- (* E_Skip *) reflexivity.
- (* E_Ass *) reflexivity.
- (* E_Seq *)
assert (st' = st'0) as EQ_{1}.
{ (* Proof of assertion *) apply IHE1_1; assumption. }
subst st'0.
apply IHE1_2. assumption.
- (* E_IfTrue, b_{1} evaluates to true *)
apply IHE1. assumption.
- (* E_IfTrue, b_{1} evaluates to false (contradiction) *)
rewrite H in H_{5}. inversion H_{5}.
- (* E_IfFalse, b_{1} evaluates to true (contradiction) *)
rewrite H in H_{5}. inversion H_{5}.
- (* E_IfFalse, b_{1} evaluates to false *)
apply IHE1. assumption.
- (* E_WhileEnd, b_{1} evaluates to false *)
reflexivity.
- (* E_WhileEnd, b_{1} evaluates to true (contradiction) *)
rewrite H in H_{2}. inversion H_{2}.
- (* E_WhileLoop, b_{1} evaluates to false (contradiction) *)
rewrite H in H_{4}. inversion H_{4}.
- (* E_WhileLoop, b_{1} evaluates to true *)
assert (st' = st'0) as EQ_{1}.
{ (* Proof of assertion *) apply IHE1_1; assumption. }
subst st'0.
apply IHE1_2. assumption. Qed.
Reasoning About Imp Programs
Theorem plus2_spec : ∀st n st',
st X = n →
plus2 / st ⇓ st' →
st' X = n + 2.
Proof.
intros st n st' HX Heval.
Inverting Heval essentially forces Coq to expand one step of
the ceval computation — in this case revealing that st'
must be st extended with the new value of X, since plus2
is an assignment
(* FILL IN HERE *)
Theorem loop_never_stops : ∀st st',
~(loop / st ⇓ st').
Proof.
intros st st' contra. unfold loop in contra.
remember (WHILE BTrue DO SKIP END) as loopdef
eqn:Heqloopdef.
~(loop / st ⇓ st').
Proof.
intros st st' contra. unfold loop in contra.
remember (WHILE BTrue DO SKIP END) as loopdef
eqn:Heqloopdef.
Proceed by induction on the assumed derivation showing that
loopdef terminates. Most of the cases are immediately
contradictory (and so can be solved in one step with
inversion).
(* FILL IN HERE *) Admitted.
Fixpoint no_whiles (c : com) : bool :=
match c with
| SKIP ⇒
true
| _ ::= _ ⇒
true
| c_{1} ;; c_{2} ⇒
andb (no_whiles c_{1}) (no_whiles c_{2})
| IFB _ THEN ct ELSE cf FI ⇒
andb (no_whiles ct) (no_whiles cf)
| WHILE _ DO _ END ⇒
false
end.
This predicate yields true just on programs that have no while
loops. Using Inductive, write a property no_whilesR such that
no_whilesR c is provable exactly when c is a program with no
while loops. Then prove its equivalence with no_whiles.
Inductive no_whilesR: com → Prop :=
(* FILL IN HERE *)
.
Theorem no_whiles_eqv:
∀c, no_whiles c = true ↔ no_whilesR c.
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 4 stars (no_whiles_terminating)
Imp programs that don't involve while loops always terminate. State and prove a theorem no_whiles_terminating that says this. Use either no_whiles or no_whilesR, as you prefer.(* FILL IN HERE *)
☐
Additional Exercises
Exercise: 3 stars (stack_compiler)
HP Calculators, programming languages like Forth and Postscript, and abstract machines like the Java Virtual Machine all evaluate arithmetic expressions using a stack. For instance, the expression(2*3)+(3*(4-2))would be entered as
2 3 * 3 4 2 - * +and evaluated like this (where we show the program being evaluated on the right and the contents of the stack on the left):
[] | 2 3 * 3 4 2 - * + [2] | 3 * 3 4 2 - * + [3, 2] | * 3 4 2 - * + [6] | 3 4 2 - * + [3, 6] | 4 2 - * + [4, 3, 6] | 2 - * + [2, 4, 3, 6] | - * + [2, 3, 6] | * + [6, 6] | + [12] |The task of this exercise is to write a small compiler that translates aexps into stack machine instructions.
- SPush n: Push the number n on the stack.
- SLoad x: Load the identifier x from the store and push it on the stack
- SPlus: Pop the two top numbers from the stack, add them, and push the result onto the stack.
- SMinus: Similar, but subtract.
- SMult: Similar, but multiply.
Inductive sinstr : Type :=
| SPush : nat → sinstr
| SLoad : id → sinstr
| SPlus : sinstr
| SMinus : sinstr
| SMult : sinstr.
Write a function to evaluate programs in the stack language. It
should take as input a state, a stack represented as a list of
numbers (top stack item is the head of the list), and a program
represented as a list of instructions, and it should return the
stack after executing the program. Test your function on the
examples below.
Note that the specification leaves unspecified what to do when
encountering an SPlus, SMinus, or SMult instruction if the
stack contains less than two elements. In a sense, it is
immaterial what we do, since our compiler will never emit such a
malformed program.
Fixpoint s_execute (st : state) (stack : list nat)
(prog : list sinstr)
: list nat
(* REPLACE THIS LINE WITH := _your_definition_ . *) . Admitted.
Example s_execute1 :
s_execute empty_state []
[SPush 5; SPush 3; SPush 1; SMinus]
= [2; 5].
(* FILL IN HERE *) Admitted.
Example s_execute2 :
s_execute (t_update empty_state X 3) [3;4]
[SPush 4; SLoad X; SMult; SPlus]
= [15; 4].
(* FILL IN HERE *) Admitted.
Next, write a function that compiles an aexp into a stack
machine program. The effect of running the program should be the
same as pushing the value of the expression on the stack.
Fixpoint s_compile (e : aexp) : list sinstr
(* REPLACE THIS LINE WITH := _your_definition_ . *) . Admitted.
After you've defined s_compile, prove the following to test
that it works.
Example s_compile1 :
s_compile (AMinus (AId X) (AMult (ANum 2) (AId Y)))
= [SLoad X; SPush 2; SLoad Y; SMult; SMinus].
(* FILL IN HERE *) Admitted.
☐
Prove the following theorem. You will need to start by stating a
more general lemma to get a usable induction hypothesis; the main
theorem will then be a simple corollary of this lemma.
Exercise: 4 stars, advanced (stack_compiler_correct)
Now we'll prove the correctness of the compiler implemented in the previous exercise. Remember that the specification left unspecified what to do when encountering an SPlus, SMinus, or SMult instruction if the stack contains less than two elements. (In order to make your correctness proof easier you might find it helpful to go back and change your implementation!)Theorem s_compile_correct : ∀(st : state) (e : aexp),
s_execute st [] (s_compile e) = [ aeval st e ].
Proof.
(* FILL IN HERE *) Admitted.
Imperative languages like C and Java often include a break or
similar statement for interrupting the execution of loops. In this
exercise we consider how to add break to Imp. First, we need to
enrich the language of commands with an additional case.
Inductive com : Type :=
| CSkip : com
| CBreak : com (* <-- new *)
| CAss : id → aexp → com
| CSeq : com → com → com
| CIf : bexp → com → com → com
| CWhile : bexp → com → com.
Notation "'SKIP'" :=
CSkip.
Notation "'BREAK'" :=
CBreak.
Notation "x '::=' a" :=
(CAss x a) (at level 60).
Notation "c_{1} ;; c_{2}" :=
(CSeq c_{1} c_{2}) (at level 80, right associativity).
Notation "'WHILE' b 'DO' c 'END'" :=
(CWhile b c) (at level 80, right associativity).
Notation "'IFB' c_{1} 'THEN' c_{2} 'ELSE' c_{3} 'FI'" :=
(CIf c_{1} c_{2} c_{3}) (at level 80, right associativity).
Next, we need to define the behavior of BREAK. Informally,
whenever BREAK is executed in a sequence of commands, it stops
the execution of that sequence and signals that the innermost
enclosing loop should terminate. (If there aren't any
enclosing loops, then the whole program simply terminates.) The
final state should be the same as the one in which the BREAK
statement was executed.
One important point is what to do when there are multiple loops
enclosing a given BREAK. In those cases, BREAK should only
terminate the innermost loop. Thus, after executing the
following...
One way of expressing this behavior is to add another parameter to
the evaluation relation that specifies whether evaluation of a
command executes a BREAK statement:
X ::= 0;;
Y ::= 1;;
WHILE 0 ≠ Y DO
WHILE TRUE DO
BREAK
END;;
X ::= 1;;
Y ::= Y - 1
END
... the value of X should be 1, and not 0.
Y ::= 1;;
WHILE 0 ≠ Y DO
WHILE TRUE DO
BREAK
END;;
X ::= 1;;
Y ::= Y - 1
END
Inductive status : Type :=
| SContinue : status
| SBreak : status.
Reserved Notation "c_{1} '/' st '⇓' s '/' st'"
(at level 40, st, s at level 39).
Intuitively, c / st ⇓ s / st' means that, if c is started in
state st, then it terminates in state st' and either signals
that the innermost surrounding loop (or the whole program) should
exit immediately (s = SBreak) or that execution should continue
normally (s = SContinue).
The definition of the "c / st ⇓ s / st'" relation is very
similar to the one we gave above for the regular evaluation
relation (c / st ⇓ st') — we just need to handle the
termination signals appropriately:
Based on the above description, complete the definition of the
ceval relation.
- If the command is SKIP, then the state doesn't change, and
execution of any enclosing loop can continue normally.
- If the command is BREAK, the state stays unchanged, but we
signal a SBreak.
- If the command is an assignment, then we update the binding for
that variable in the state accordingly and signal that execution
can continue normally.
- If the command is of the form IFB b THEN c_{1} ELSE c_{2} FI, then
the state is updated as in the original semantics of Imp, except
that we also propagate the signal from the execution of
whichever branch was taken.
- If the command is a sequence c_{1} ;; c_{2}, we first execute
c_{1}. If this yields a SBreak, we skip the execution of c_{2}
and propagate the SBreak signal to the surrounding context;
the resulting state is the same as the one obtained by
executing c_{1} alone. Otherwise, we execute c_{2} on the state
obtained after executing c_{1}, and propagate the signal
generated there.
- Finally, for a loop of the form WHILE b DO c END, the semantics is almost the same as before. The only difference is that, when b evaluates to true, we execute c and check the signal that it raises. If that signal is SContinue, then the execution proceeds as in the original semantics. Otherwise, we stop the execution of the loop, and the resulting state is the same as the one resulting from the execution of the current iteration. In either case, since BREAK only terminates the innermost loop, WHILE signals SContinue.
Inductive ceval : com → state → status → state → Prop :=
| E_Skip : ∀st,
CSkip / st ⇓ SContinue / st
(* FILL IN HERE *)
where "c_{1} '/' st '⇓' s '/' st'" := (ceval c_{1} st s st').
Now prove the following properties of your definition of ceval:
Theorem break_ignore : ∀c st st' s,
(BREAK;; c) / st ⇓ s / st' →
st = st'.
Proof.
(* FILL IN HERE *) Admitted.
Theorem while_continue : ∀b c st st' s,
(WHILE b DO c END) / st ⇓ s / st' →
s = SContinue.
Proof.
(* FILL IN HERE *) Admitted.
Theorem while_stops_on_break : ∀b c st st',
beval st b = true →
c / st ⇓ SBreak / st' →
(WHILE b DO c END) / st ⇓ SContinue / st'.
Proof.
(* FILL IN HERE *) Admitted.
Theorem while_break_true : ∀b c st st',
(WHILE b DO c END) / st ⇓ SContinue / st' →
beval st' b = true →
∃st'', c / st'' ⇓ SBreak / st'.
Proof.
(* FILL IN HERE *) Admitted.
(WHILE b DO c END) / st ⇓ SContinue / st' →
beval st' b = true →
∃st'', c / st'' ⇓ SBreak / st'.
Proof.
(* FILL IN HERE *) Admitted.
Theorem ceval_deterministic: ∀(c:com) st st_{1} st_{2} s_{1} s_{2},
c / st ⇓ s_{1} / st_{1} →
c / st ⇓ s_{2} / st_{2} →
st_{1} = st_{2} ∧ s_{1} = s_{2}.
Proof.
(* FILL IN HERE *) Admitted.
End BreakImp.
c / st ⇓ s_{1} / st_{1} →
c / st ⇓ s_{2} / st_{2} →
st_{1} = st_{2} ∧ s_{1} = s_{2}.
Proof.
(* FILL IN HERE *) Admitted.
End BreakImp.
☐
Write an alternate version of beval that performs short-circuit
evaluation of BAnd in this manner, and prove that it is
equivalent to beval.
Exercise: 3 stars, optional (short_circuit)
Most modern programming languages use a "short-circuit" evaluation rule for boolean and: to evaluate BAnd b_{1} b_{2}, first evaluate b_{1}. If it evaluates to false, then the entire BAnd expression evaluates to false immediately, without evaluating b_{2}. Otherwise, b_{2} is evaluated to determine the result of the BAnd expression.(* FILL IN HERE *)
☐
A for loop should be parameterized by (a) a statement executed
initially, (b) a test that is run on each iteration of the loop to
determine whether the loop should continue, (c) a statement
executed at the end of each loop iteration, and (d) a statement
that makes up the body of the loop. (You don't need to worry
about making up a concrete Notation for for loops, but feel free
to play with this too if you like.)
Exercise: 4 stars, optional (add_for_loop)
Add C-style for loops to the language of commands, update the ceval definition to define the semantics of for loops, and add cases for for loops as needed so that all the proofs in this file are accepted by Coq.(* FILL IN HERE *)
☐
(* $Date: 2016-07-14 17:02:35 -0400 (Thu, 14 Jul 2016) $ *)