CODEWITHSUNDEEP
coq
V. Semeria
V. Semeria

Reputation: 3256

Decide disjunctions in sort Prop

I am interested in the probably false lemma :

Lemma decideOr : forall (P Q : Prop),
    (P \/ Q) -> {P} + {Q}.

that asserts we can algorithmically decide any proof of an or in sort Prop. Of course, Coq does not let us destruct the input to extract it in sort Set. However, a proof of P \/ Q is a lambda-term that Coq accepts to print, so external tools can process it.

First question : can this lambda-term be decided outside of Coq (assuming the term uses no axioms, only plain Coq) ? It might be, because the rules of constructive logic demand that all disjunctions be explicitely chosen, without cheating by a proof by contradiction. So can we code a parser of Coq proof terms, and try to decide whether the first or the second operand of the or was proved ? If the term starts with or_introl or or_intror we are done. So I guess the problems are when the term is a lambda-application. But then Coq terms are strongly normalizing, so we reduce it to a normal form and it seems it will start with either or_introl or or_intror.

Second question : if this problem can be decided outside of Coq, what prevents us from internalizing it within Coq, ie proving lemma decideOr above ?

Upvotes: 1

Views: 67

Answers (1)

Arthur Azevedo De Amorim
Arthur Azevedo De Amorim

Reputation: 23622

First question

Yes, you can write a program that takes as input a Coq proof of A \/ B and outputs true or false depending on which side was used to prove the disjunction. Indeed, if you write

Compute P.

in Coq, where P : A \/ B, Coq will normalize the proof P and print which constructor was used. This will not work if P uses proofs that end in Qed (because those are not unfolded by the evaluator), but in principle it is possible to replace Qed by Defined everywhere and make it work.

Second question

What prevents us from proving decideOr is that the designers of Coq wanted to have a type of propositions that supports the excluded middle (using an axiom) while allowing programs to execute. If decideOr were a theorem and we wanted to use the excluded middle (classical : forall A : Prop, A \/ ~ A), it would not be possible to execute programs that branch on the result of decideOr (classical A). This does not mean that decideOr is false: it is perfectly possible to admit it as an axiom. There is a difference between not being provable ("there does not exist a proof of A") and being refutable ("there exists a proof of ~ A").

Upvotes: 1

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