Why are logical connectives and booleans separate in Coq?

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南旧
南旧 2020-12-03 12:07

I come from a JavaScript/Ruby programming background and am used to this being how true/false works (in JS):

!true
// false
!false
// true

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  • 2020-12-03 13:05

    Essentially, Coq has both because they are useful for different things: booleans correspond to facts that can be checked mechanically (i.e., with an algorithm), whereas propositions can express more concepts.

    Strictly speaking, the logical and boolean worlds are not separate in Coq: the boolean world is a subset of the logical world. In other words, every statement that you can phrase as a boolean computation can be viewed as a logical proposition (i.e., something of type Prop): if b : bool represents a statement, we can assert that this statement is true by saying b = true, which is of type Prop.

    The reason there's more in Coq to logic than just booleans is that the converse of the previous statement does not hold: not all logical facts can be viewed as boolean computations. Put in a different way, it is not the case that booleans in normal programming languages such as Ruby and JavaScript subsume both bool and Prop in Coq, because Props can express things that booleans in these languages cannot.

    To illustrate this, consider the following Coq predicate:

    Definition commutative {T} (op : T -> T -> T) : Prop :=
      forall x y, op x y = op y x.
    

    As the name suggests, this predicate asserts that an operator op is commutative. Many operators in programming languages are commutative: take multiplication and addition over integers, for instance. Indeed, in Coq we can prove the following statements (and I believe those are examples in the Software Foundations book):

    Lemma plus_comm : commutative plus. Proof. (* ... *) Qed.
    Lemma mult_comm : commutative mult. Proof. (* ... *) Qed.
    

    Now, try to think how you would translate a predicate like commutative in a more conventional language. If this seems difficult, it is not by chance: it is possible to prove that we can't write a program returning a boolean in these languages to test whether an operation is commutative or not. You can certainly write unit tests for checking whether this fact is true for particular inputs, e.g.:

    2 + 3 == 3 + 2
    4 * 5 == 5 * 4
    

    However, if your operator works with an infinite number of inputs, these unit tests can only cover a fraction of all possible cases. Therefore, testing is always necessarily weaker than a complete formal proof.

    You could wonder why we bother having booleans in Coq if Props can express everything that booleans can. The reason for that is that Coq is a constructive logic, which is what Vinz was alluding to in his comment. The most well-known consequence of this fact is that in Coq we cannot prove the following intuitive principle:

    Definition excluded_middle :=
      forall P : Prop, P \/ ~ P.
    

    which essentially says that every proposition is either true or false. "How could this possibly fail?", you might ask yourself. Roughly speaking, in constructive logics (and Coq in particular), every proof corresponds to an algorithm we can execute. In particular, when you prove a statement of the form A \/ B in a constructive logic, you can extract an (always terminating) algorithm from that proof that answers whether A or B holds. Hence, if we were able to prove the above principle, we would have an algorithm that, given some proposition, tells us whether that proposition is valid or not. Computability theory shows, however, that this is not possible in general because of undecidability: if we take P to mean "program p halts on input x", the excluded middle would yield a decider for the halting problem, which cannot exist.

    Now, what's interesting about booleans in Coq is that by construction they allow the use of the excluded middle, because they do correspond to an algorithm we can run. Specifically, we can prove the following:

    Lemma excluded_middle_bool :
      forall b : bool, b = true \/ negb b = true.
    Proof. intros b; destruct b; simpl; auto. Qed.
    

    Thus, in Coq it is useful to consider booleans as a special case of propositions because they allow forms of reasoning that other propositions do not, namely, case analysis.

    Of course, you can think that requiring that every proof correspond to an algorithm is silly, and indeed most logics allow the principle of the excluded middle. Examples of proof assistants that follow this approach by default include Isabelle/HOL and the Mizar system. In these systems, we don't have to have a distinction between booleans and propositions, and they are treated as the same thing. Isabelle, for instance, has just bool, and no Prop. Coq also allows you to blur the distinction between booleans and propostions, by assuming axioms that allow you to perform case analysis on general propositions. On the other hand, in such a setting, when you write a function that returns a boolean, you might not obtain something that you can execute as an algorithm, whereas this is always the case by default in Coq.

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