16 October 2015

prerequisites: The post An interpreter, a compiler and a virtual machine

1. Introduction

In this post, we prove an equivalence relation between interpretation of an arithmetic expression and compilation of an arithmetic expression followed by execution of the bytecode program resulting from compilation.

First, we derive the equivalence relation in Section 2, start the proof in Section 3, take a detour in Section 4, and finish the proof in Section 5. Finally, we conclude in Section 4.

2. Equivalence relation

Before we can prove an equivalence relation between interpretation of an arithmetic expression and compilation of an arithmetic expression followed by execution of the compiled bytecode program, we first have to capture the nature of this relation. If we look at the signatures of interpret, compile, and execute_bytecode_program:

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Fixpoint interpret
           (e : arithmetic_expression) : nat := ...
Fixpoint compile
           (e : arithmetic_expression) : bytecode_program := ...
Fixpoint execute_bytecode_program
           (s : data_stack)
           (bcp : bytecode_program) : data_stack := ...

we note that while interpret returns a nat, when given an arithmetic_expression, compile returns a bytecode_program, when given an arithmetic_expression, which we can then pass to execute_bytecode_program, along with a data_stack, and finally obtain a data_stack.

Given that interpret does not use a data_stack, or similar, a possible candidate for an equivalence relation would not depend on the state of the stack. As such, we would expect that any arithmetic_expression given to interpret results in the same value as found at the top of the data_stack, when applying execute_bytecode_program on the output of compile, when given the same arithmetic_expression. Thus, if we let our example expression be , we can capture the equivalence relation with the following snippet of code:

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Compute let e := Plus (Lit 5) (Mult (Lit 3) (Lit 2))
        in (interpret e :: nil,
            execute_bytecode_program nil (compile e)).

where both expressions found in the body of the let expression evaluate to the same result, 11 : nat. If we turn the above let expression into a theorem, we get the following candidate:

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Theorem equality_of_interpret_and_compile_candidate :
  forall (e : arithmetic_expression),
    (interpret e) :: nil =
    execute_bytecode_program nil (compile e).

However, since we just established earlier that the state of the data_stack was irrelevant, we can actually generalize the equivalence relation to hold for all possible stacks, and not just the empty stack. This gives us the following revised version of the theorem:

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Theorem equality_of_interpret_and_compile :
  forall (e : arithmetic_expression) (s : data_stack),
    (interpret e) :: s =
    execute_bytecode_program s (compile e).

which perfectly captures the equivalence relation between interpretation and compilation we were after.

Having derived our equivalence relation, we are ready to move on to proving the relation.

3. Equivalence proof

Just as in the introductory post on the Coq Proof Assistant, we start our proof by writing the Proof keyword, which gives us the following goal:

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============================
forall (e : arithmetic_expression) (s : data_stack),
  interpret e :: s = execute_bytecode_program s (compile e)

corresponding to the statement of the final theorem in the previous section.

Since the objective of our proof is to show that interpretation and compilation yields the same result for all possible arithmetic expressions, we suspect that doing structural induction on the arithmetic expression, e, would be a fruitful approach for getting there. With this approach, we first introduce the arithmetic expression e and then use induction on its structure,

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intro e.
induction e as [ n |
                 e1' IH_e1' e2' IH_e2' |
                 e1' IH_e1' e2' IH_e2' ].

which gives us three subgoals:

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n : nat
============================
forall s : data_stack,
  interpret (Lit n) :: s =
  execute_bytecode_program s (compile (Lit n))

subgoal 2 is:
 forall s : data_stack,
 interpret (Plus e1' e2') :: s =
 execute_bytecode_program s (compile (Plus e1' e2'))

subgoal 3 is:
 forall s : data_stack,
 interpret (Mult e1' e2') :: s =
 execute_bytecode_program s (compile (Mult e1' e2'))

one for each of the constructors of the arithmetic_expression type. Starting with the case of the literal expression, Lit:

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n : nat
============================
forall s : data_stack,
  interpret (Lit n) :: s =
  execute_bytecode_program s (compile (Lit n))

Here, we note that both interpretation and compilation of a Lit expression involves no recursive calls and therefore we can just unfold the definitions present in both expressions:

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intro s.
unfold interpret.
unfold compile.
unfold execute_bytecode_program.
unfold execute_bytecode_instruction.

and obtain the goal n :: s = n :: s which we can prove with reflexivity.

When we look at the next subgoal:

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e1' : arithmetic_expression
e2' : arithmetic_expression
IH_e1' : forall s : data_stack,
         interpret e1' :: s =
         execute_bytecode_program s (compile e1')
IH_e2' : forall s : data_stack,
         interpret e2' :: s =
         execute_bytecode_program s (compile e2')
============================
forall s : data_stack,
  interpret (Plus e1' e2') :: s =
  execute_bytecode_program s (compile (Plus e1' e2'))

things become a bit more challenging as our arithmetic expression, (Plus e1' e2'), now has two sub expressions, e1 and e2, which means we have to bring our goal in a position where we can use the two induction hypotheses, IH_e1' and IH_e2'. If we repeat the steps of the previous proof and try to unfold the definitions at hand:

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intro s.
unfold interpret; fold interpret.
unfold compile; fold compile.

we arrive at the following goal:

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============================
interpret e1' + interpret e2' :: s =
execute_bytecode_program
  s (compile e2' ++ compile e1' ++ ADD :: nil)

However, now we are not able to do anymore unfolding - we do not known how to unfold execute_bytecode_program when given a concatenated list - so we have to somehow restate the right-hand side of the equation,

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execute_bytecode_program
  s (compile e2' ++ compile e1' ++ ADD :: nil)

such that we can use our induction hypotheses on it.

4. Lists and bytecode programs

If we look at the last statement of the previous section, it says something about the execution of several concatenated bytecode programs, compile e2', compile e1' and ADD :: nil. Since we know that executing a bytecode program, with respect to a stack, results in a new stack, which again can be used to execute a new bytecode program, we propose the following statement:

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Lemma execute_bytecode_program_is_associative :
  forall (p1 p2 : bytecode_program) (s : data_stack),
    execute_bytecode_program s (p1 ++ p2) =
    execute_bytecode_program (execute_bytecode_program s p1) p2.

which states that executing the concatenation of two bytecode program, p1 and p2, on an initial stack, s, is equivalent to executing the first bytecode program, p1, on the initial stack, s, and then executing the second bytecode program, p2, on the resulting stack of the previous execution.

For this proof, we use structural induction on the first bytecode program, p1, consisting of a list of bytecode instructions, which means we first have to prove that the statement holds for the empty program, nil, and then for all non-empty programs, bci :: bcis':

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intro p1.
induction p1 as [ | bci' bcis' IH_bcis' ].

In the case of the empty program, nil, the proof is trivial and follows from unfolding the definitions of the statement:

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intros p2 s.
unfold app.
unfold execute_bytecode_program; fold execute_bytecode_program.
reflexivity.

For the inductive case, bci :: bcis', we start by introducing the other bytecode program, p2, and the stack, s:

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intros p2 s.

which gives us the following goal:

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============================
execute_bytecode_program
  s ((bci' :: bcis') ++ p2) =
execute_bytecode_program
  (execute_bytecode_program s (bci' :: bcis')) 

Here, we would like to unfold execute_bytecode_program on the left-hand side of the equation, but in order to do so we need to grab the following lemma from the Coq library:

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app_comm_cons
  : forall (A : Type) (x y : list A) (a : A),
     a :: x ++ y = (a :: x) ++ 

which allows us to shift the inner parentheses in expression on the left-hand side of the equality, ((bci' :: bcis') ++ p2),

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rewrite <- app_comm_cons.

such that we can unfold execute_bytecode_program and apply the induction hypothesis, IH_bcis':

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unfold execute_bytecode_program; fold execute_bytecode_program.
rewrite <- IH_bcis'.

at which point we can apply reflexivity to finish the proof, as both sides of the equality have the exact same value.

The complete proof of execute_bytecode_program_is_associative ends up looking like so:

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Lemma execute_bytecode_program_is_associative
  : forall (p1 p2 : bytecode_program) (s : data_stack),
    execute_bytecode_program s (p1 ++ p2) =
    execute_bytecode_program (execute_bytecode_program s p1) p2.
Proof.
  intro p1.
  induction p1 as [ | p1' p1s' IH_p1s' ].

  (* case: p1 = nil *)
  intros p2 s.
  unfold app.
  unfold execute_bytecode_program; fold execute_bytecode_program.
  reflexivity.

  (* case: p1 = p1' :: p1s' *)
  intros p2 s.
  rewrite <- app_comm_cons.
  unfold execute_bytecode_program; fold execute_bytecode_program.
  rewrite -> IH_p1s'.
  reflexivity.
Qed.

At which point we are now ready to return to the proof of the original equivalence relation.

5. Equivalence proof, continued

Having proved execute_bytecode_program_is_associative, we return to the Plus e1' e2' case of equality_of_interpret_and_compile,

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============================
interpret e1' + interpret e2' :: s =
execute_bytecode_program
  s (compile e2' ++ compile e1' ++ ADD :: nil)

where we can now use execute_bytecode_program_is_associative to rewrite the bytecode program expression from a concatenated list to a nested structure,

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rewrite ->2 execute_bytecode_program_is_associative.

giving us the following goal:

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============================
interpret e1' + interpret e2' :: s =
  execute_bytecode_program
    (execute_bytecode_program
      (execute_bytecode_program s (compile e2'))
      (compile e1'))
    (ADD :: nil)

Now, we are able to rewrite the execute_bytecode_program expression with both of our induction hypotheses,

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rewrite <- IH_e1'.
rewrite <- IH_e2'.
unfold execute_bytecode_program.
unfold execute_bytecode_instruction.

resulting in the following equality,

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============================
interpret e1' + interpret e2' :: s =
interpret e1' + interpret e2' :: 

which is again proved with reflexivity.

The proof of the last subgoal,

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e1' : arithmetic_expression
e2' : arithmetic_expression
IH_e1' : forall s : data_stack,
         interpret e1' :: s =
         execute_bytecode_program s (compile e1')
IH_e2' : forall s : data_stack,
         interpret e2' :: s =
         execute_bytecode_program s (compile e2')
============================
forall s : data_stack,
interpret (Mult e1' e2') :: s =
execute_bytecode_program s (compile (Mult e1' e2'))

is completely identical to the proof of the previous subgoal, except that Plus and ADD have been substituted with Mult and MUL. This brings us to the final version of the proof for equality_of_interpret_and_compile:

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Theorem equality_of_interpret_and_compile :
  forall (e : arithmetic_expression) (s : data_stack),
    (interpret e) :: s = execute_bytecode_program s (compile e).
Proof.
  intro e.
  induction e as [ n |
                   e1' IH_e1' e2' IH_e2' |
                   e1' IH_e1' e2' IH_e2' ].

  (* case: e = Lit n *)
  intro s.
  unfold interpret.
  unfold compile.
  unfold execute_bytecode_program.
  unfold execute_bytecode_instruction.
  reflexivity.

  (* case: e = Plus e1' e2' *)
  intro s.
  unfold interpret; fold interpret.
  unfold compile; fold compile.
  rewrite ->2 execute_bytecode_program_is_associative.
  rewrite <- IH_e1'.
  rewrite <- IH_e2'.
  unfold execute_bytecode_program.
  unfold execute_bytecode_instruction.
  reflexivity.

  (* case: e = Mult e1' e2' *)
  intro s.
  unfold interpret; fold interpret.
  unfold compile; fold compile.
  rewrite ->2 execute_bytecode_program_is_associative.
  rewrite <- IH_e1'.
  rewrite <- IH_e2'.
  unfold execute_bytecode_program.
  unfold execute_bytecode_instruction.
  reflexivity.
Qed.

with which we have now proved an equivalence relation between interpreting an arithmetic expression and compiling it and then executing it on a virtual machine.

6. Conclusion

In this post, we have proved an equivalence relation between interpretation of an arithmetic expression and compilation of an arithmetic expression followed by execution of the bytecode program resulting from compilation.

With the above result, we have actually proved that we can think of interpretation as the act of compiling an expression and immediately executing it on a virtual machine, which is a pretty cool result.