Logic in Coq

---

Topics:

• propositions and proofs
• Prop and Set
• propositional logic
• implication
• conjunction
• disjunction
• False and True
• negation
• equality and implication revisited
• tautologies

---

Propositions and proofs

Recall that the Check command type checks an expression and causes Coq to output the type in the output window.

The type of list is Type -> Type. It takes a type as input and produces a type as output. Thus, like OCaml's list, Coq's list is a type constructor. But unlike OCaml, the type of Coq's list contains ->, indicating that it is a function. In Coq, list truly is a function that can be applied:

Think of Coq's list as a type-level function: it is a function that takes types as inputs and produces types as outputs. OCaml doesn't have anything exactly equivalent to that. What, then, is the type of a theorem?

Coq responds:

obvious_fact : 1 + 1 = 2

So the type of obvious_fact is 1 + 1 = 2. That might seem rather mysterious. After all, 1 + 1 = 2 is definitely not an OCaml type. But Coq's type system is far richer than OCaml's. In OCaml, we could think of 42 : int as meaning that 42 is a meaningful expression of type int. There are likewise meaningless expressions---for example, 42+true is meaningless in OCaml, and it therefore cannot be given a type. The meaning of int in OCaml is that of a computation producing a value that fits within 2^63 bits and can be interpreted as an element of Z, the mathematical set of integers. So what are meaningful Coq expressions of type 1 + 1 = 2? They are the proofs of 1 + 1 = 2. There might be many such proofs; here are two:

• reduce 1+1 to 2, then note that x=x.
• subtract 1 from both sides, resulting in 1 + 1 - 1 = 2 - 1, reduce both sides to 1, then note that x = x.

Regardless of the exact proof, it is an argument for, or evidence for, the assertion that 1 + 1 = 2. So when we say that obvious_fact : 1 + 1 = 2, what we're really saying is that obvious_fact is a proof of 1 + 1 = 2. Likewise, when we write Definition x := 42. in Coq, and then observe that x : nat, what we're saying is not just that x has type nat, but also that there is evidence for the type nat, i.e., that there do exist values of that type. Many of them, in fact---just like there are many proofs of 1 + 1 = 2. So now we have an explanation for what the type of obvious_fact is. But what is its value? Let's ask Coq.

Coq responds:

obvious_fact = eq_refl : 1 + 1 = 2

So what is eq_refl?

Amidst all the output that produces, you will spy

eq_refl : x = x 

That is, eq_refl has type x = x. In other words, eq_refl asserts something is always equal to itself, a property known as the reflexivity of equality. So the proof of 1 + 1 = 2 in Coq is really just that equality is reflexive. It turns out that Coq evaluates expressions before applying that fact, so it evaluates 1+1 to 2, resulting in 2 = 2, which holds by reflexivity. Thus the proof we found above, using the trivial tactic, corresponds to the first of the two possible proofs we sketched.

Prop and Set

We've now seen that there are programs and proofs in Coq. Let's investigate their types. We already know that 42, a program, has type nat, and that eq_refl : 1 + 1 = 2. But let's now investigate what the types of nat and 1 + 1 = 2 are. First, nat:

Coq says that nat : Set. Here, Set is a predefined keyword in Coq that we can think of as meaning all program data types. Further, we can ask what the type of Set is:

Coq says that Set : Type, so Set is a type. The Coq documentation describes Set as being the type of program specifications, which describe computations. For example,

• 42 specifies a computation that simply returns 42. It's a specification because 42 : nat and nat : Set.
• fun x:nat => x+1 specifies a computation that increments a natural number. It's a specification because it has type nat -> nat, and nat -> nat : Set.
• We could also write more complicated specifications to express computations such as list sorting functions, or binary search tree lookups.

Next, let's investigate what the type of 1 + 1 = 2 is.

Coq says that 1 + 1 = 2 : Prop. This is the type of propositions, which are logical formulas that we can attempt to prove. Note that propositions are not necessarily provable though. For example, 2110 = 3110 has type Prop, even though it obviously does not hold. What is the type of Prop?

Coq says that Prop : Type. So Prop is also a type. So far we have seen one way of creating a proposition, by using the equality operator. We could attempt to learn more about that operator with Check =., but that will result in an error: Check doesn't work on this kind of notation, and there isn't something like in OCaml where we can wrap an operator in parentheses. Instead, we need to find out what function name that operator corresponds to. The command for that is Locate.

Coq tells us two possible meanings for =, and the second is what we want:

"x = y" := eq x y

That means that anywhere x = y appears, Coq understands it as the function eq applied to x and y. Now that we know the name of that function, we can check it:

The output of that is a bit mysterious because of the ?A that shows up in it. What's going on is that eq has an implicit type argument. (Implicit arguments were discussed in the previous set of notes.) If we prefix eq with @ to treat the argument as explicit, we can get more readable output.

Coq says that

@eq : forall A : Type, A -> A -> Prop

In other words, @eq takes a type argument A, a value of type A, another value of type A, and returns a proposition, which is the proposition asserting the equality of those two values. So:

• @eq nat 42 42 asserts the equality of 42 and 42, i.e., 42 = 42. So does eq 42 42, where nat is implicit.
• @eq nat 2110 3110 asserts the equality of 2110 and 3110. So does eq 2110 3110 and 2110=3110. Of course they aren't equal, but we're still allowed to form such a proposition, even if it isn't provable.
• @eq (nat->nat) (fun x => x+1) (fun n => 1+n) asserts the equality of two syntactically different increment functions, as does eq (fun x => x+1) (fun n => 1+n) and (fun x => x+1) = (fun x => 1+n).

Propositional logic

The standard propositional logic connectives (ways of syntactically connecting propositions together) are:

• Implication: P -> Q. In English we usually express this connective as "if P, then Q" or "P implies Q".
• Conjunction: P /\ Q. In English we usually express this connective as "P and Q".
• Disjunction: P \/ Q. In English we usually express this connective as "P or Q". Keep in mind that this is the sense of the word "or" that allows one or both of P and Q to hold, rather than exactly one.
• Negation: ~P. In English we usually express this connective as "not P".

All of these are ways of creating propositions in Coq. Implication is so primitive that it's simply "baked in" to Coq. But the other connectives are defined as part of Coq's standard library under the very natural names of and, or, and not. In addition to those connectives, there are two propositions True and False which always and never hold, respectively. And we can have variables representing propositions. Idiomatically, we usually choose names like P, Q, R, ... for those variables.

and : Prop -> Prop -> Prop
or  : Prop -> Prop -> Prop
not : Prop -> Prop

Both and and or take two propositions as input and return a proposition; not takes one proposition as input and returns a proposition. At this point, you might have noticed that -> seems to be overloaded:

• t1 -> t2 is the type of functions that take an input of type t1 and return an output of type t2.
• P -> Q is an proposition that asserts P implies Q.

There is a uniform way to think about both uses of ->, which is as a transformer. A function of type t1 -> t2 transforms a value of type t1 into a value of type t2. An implication P -> Q in a way transforms P into Q: if P holds, then Q also holds; or better yet, a proof of P -> Q can be thought of as a function that transforms evidence for P into evidence for Q. Let's do some proofs with these connectives to get a better sense of how they work.

Implication

Let's try one of the simplest possible theorems we could prove using implication: P implies P.

That proposition says that for any proposition P, it holds that P implies P. Intuitively, why should be able able to prove this proposition? That is, what is an argument you could give to another human? We always encourage you to try to answer that question before launching into a Coq proof---for the same reason your introductory programming instructor always encouraged you to think about programs before you begin typing them. So why does this proposition hold? It's rather trivial, really. If P holds, then certainly P holds. That is, P implies itself. An example in English could be "if 3110 is fun, then 3110 is fun." If you've already assumed P, then necessarily P follows from your assumptions: that's what it means to be an assumption. The Coq proof below uses that reasoning.

The second step, intros P_assumed, is a new use for the intros tactic. This usage peels off the left-hand side of an implication and puts it into the proof assumptions. The third step, assumption is a new tactic. It finishes a proof of a subgoal G whenever G is already an assumption in the proof. Let's look at the type of p_implies_p.

Coq says p_implies_p : forall P : Prop, P -> P. So p_implies_p is proof of, or evidence for, forall P : Prop, P -> P, as we discussed above. What is that evidence? We can use Print to find out:

Coq responds

p_implies_p = 
fun (P : Prop) (P_assumed : P) => P_assumed
     : forall P : Prop, P -> P

Let's pull that apart. Coq says that p_implies_p is a name that is bound to a value and has a type. The type we already know. The value is perhaps surprising: it is a function! Actually, maybe it shouldn't be surprising given the discussion we had above about transformers. P -> P should transform evidence for P into evidence for P. Trivially, evidence for P already is evidence for P, so there's nothing to be done. And we can see that in the function itself:

• it takes in an argument named P, which is the proposition; and
• it takes in another argument named P_assumed of type P. Since P_assumed has a proposition as type, P_assumed must be evidence for that proposition.
• the function simply returns P_assumed. That is, it returns the evidence for P that was already passed into it as an argument.

Note that the names of the arguments to that function are the names that we chose with the intros tactic. Let's clarify a subtle piece of terminology, "proof". The proof of forall P:Prop, P -> P is the anonymous function fun (P : Prop) (P_assumed : P) => P_assumed . That is, the proof of a proposition P is a program that has type P. On the other hand, the commands we used above

Proof.
    intros P. 
    intros P_assumed. 
    assumption.
Qed.

are how we help Coq find that proof. We provide guidance using tactics, and Coq uses those tactics to figure out how to construct the program. So although it's tempting to refer to intros P. intros P_assumed. assumption. as the "proof"---and we often will as a kind of shorthand terminology---the proof is really the program that is constructed, not the tactics that help do the construction. It is, by the way, possible to just directly tell Coq what the proof is by using the command Definition:

But rarely do we directly write down proofs, because for most propositions they are not so trivial. Tactics are a huge help in constructing complicated proofs. Let's try another theorem. A syllogism is a classical form of argument that typically goes something like this:

• All humans are mortal.
• Socrates is a human.
• Therefore Socrates is mortal.

The first assumption in that proof, "all humans are mortal", is an implication: if X is a human, then X is mortal. The second assumption is that Socrates is a human. Putting those two assumptions together, we conclude that Socrates is mortal. We can formalize that kind of reasoning as follows:

How would you convince a human that this theorem holds? By the same reasoning as above. Assume that P -> Q. Also assume P. Since P holds, and since P implies Q, we know that Q must also hold. The Coq proof below uses that style of argument.

The first line of the proof, intros P Q evPimpQ evP introduces four assumptions. The first two are the variables P and Q. The third is P->Q, which we give the name evPimpQ as a human-readable hint that it is evidence that P implies Q. The fourth is P, which we give the name evP as a hint that we have assumed we have evidence for P. The second line of the proof, apply evPimpQ, uses a new tactic apply to apply the evidence that P -> Q to the goal of the proof, which is Q. This transforms the goal to be P. Think of this as backward reasoning: we know we want to show Q, and since we have evidence that P -> Q, we can work backward to conclude that if we could only show P, then we'd be done. The third line concludes by pointing out to Coq that in fact we do already have evidence for P as an assumption. Let's look at the proof that these tactics cause Coq to create:

We see that

syllogism = 
fun (P Q : Prop) (evPimpQ : P -> Q) (evP : P) => evPimpQ evP
     : forall P Q : Prop, (P -> Q) -> P -> Q

Picking that apart, syllogism is a function that takes four arguments. The third argument evPimpQ is of type P -> Q. Going back to our reading of -> in different ways, we can think of evPimpQ as

• a function that transforms something of type P into something of type Q, or
• evidence for P -> Q, or
• a transformer that takes in evidence of P and produces evidence of Q.

The deep point here is: all of these are really the same interpretation. There's no difference between them. The value evPimpQ is a function, and it is evidence, and it is an evidence transformer. We can see that evPimpQ is being used as a function in the body evPimpQ evP of the anonymous function, above. It is applied to evP, which is the evidence for P, thus producing evidence for Q. So "apply" really is a great name for the tactic: it causes a function application to occur in the proof. Let's try one more proof with implication.

As usual, let's first try to give an argument in English. We assume that P -> Q and Q -> R, and we want to conclude P -> R. So suppose that P did hold. Then from P -> Q we'd conclude Q, and then from Q -> R we'd conclude R. So there's a kind of chain of evidence here from P to Q to R. This kind of chained relationship is called transitive, as you'll recall from CS 2800. So what we're proving here is that implication is transitive, hence the name imp_trans.

The second line, intros evP, wouldn't actually need to be separated from the first line; we could have introduced evP along with the rest. We did it separately just so that we could see that it peels off the P from P -> R. The third line, apply evQimpR applies the evidence that Q -> R to the goal R, causing the goal to become Q. Again, this is backward reasoning: we want to show R, and we know that Q -> R, so if we could just show Q we'd be done. The fourth line, apply evPimpQ, applies the evidence that P -> Q to the goal Q, causing the goal to become P. Finally, the fifth line, assumption, finishes the proof by pointing out that P is already an assumption. Let's look at the resulting proof:

Coq says

imp_trans = 
fun (P Q R : Prop) (evPimpQ : P -> Q) (evQimpR : Q -> R) (evP : P) =>
evQimpR (evPimpQ evP)
     : forall P Q R : Prop, (P -> Q) -> (Q -> R) -> P -> R

Drilling down into the body of that anonymous function we see evQimpR (evPimpQ evP). So we have evPimpQ being applied to evP, which transforms the evidence for P into the evidence for Q. Then we have evQimpR applied to that evidence for Q, thus producing evidence for R. So there are two function applications. If Coq had the OCaml operator |>, we could rewrite that function body as evP |> evPimpQ |> evQimpR, which might make it even clearer what is going on: the chain of reasoning just takes P to Q to R.

Conjunction

Now we turn our attention to the conjunction connective. Here's a first theorem to prove.

Why does that hold, intuitively? Suppose we have evidence for P /\ Q. Then we must have evidence for both P and for Q. The evidence for P alone suffices to conclude P. As an example, if we have evidence that x > 0 /\ y > 0, then we must have evidence that x > 0; we're allowed to forget about the evidence for y > 0. Having established that intuition, let's create a proof with Coq.

The second line of that proof uses a familiar tactic in a new way. Previously we used destruct to do case analysis, splitting apart a day into the seven possible constructors it could have. Here, we use destruct to split the evidence for P /\ Q into its separate components. To give names to those two components, we use a new syntax, as [...], where the names appearing in the ... are used as the names for the components. If we leave off the as clause, Coq will happily choose names for us, but they won't be human-readable, descriptive names. So it's good style to pick the names ourselves. After destructing the evidence for PandQ into evidence for P and evidence for Q, we can easily finish the proof with assumption. Let's look at the proof of and_fst.

Coq says that

and_fst = 
fun (P Q : Prop) (PandQ : P /\ Q) =>
match PandQ with
| conj P_holds _ => P_holds
end
     : forall P Q : Prop, P /\ Q -> P

Let's dissect that. We see that and_fst is a function that takes three arguments. The third is evidence for P /\ Q. The function then pattern matches against that evidence, providing just a single branch in the pattern match. The pattern it uses is conj P_holds _. As in OCaml, _ is a wildcard that matches anything. The identifier conj is a constructor; unlike OCaml, it's fine for constructors to begin with lowercase characters. So the pattern matches against conj applied to two values, extracts the first value with the name P_holds, and doesn't give a name to the second value. The branch then returns that argument P_holds. So we can already see that the function is splitting apart the evidence into two pieces, and forgetting about one piece. But to fully understand this, we need to know more about /\ and conj. First, what is /\?

Coq says "A /\ B" := and A B. That is, /\ is just an infix notation for the and function. What is and?

Coq says

Inductive and (A B : Prop) : Prop :=  
  conj : A -> B -> A /\ B

It might help to compare to lists.

Coq says

Inductive list (A : Type) : Type :=
    nil : list A | cons : A -> list A -> list A

In Coq, Inductive defines a so-called inductive type, and provides its constructors. In OCaml, the type keyword serves a similar purpose. So list in Coq is a type with two constructors named nil and cons. Similarly, and is a type with a single constructor named conj, which is a function that takes in a value of type A, a value of type B, and returns a value of type A /\ B, which is just infix notation for and A B. Another way of putting that is that conj takes in evidence of A, evidence of B, and returns evidence of and A B. So there's only one way of producing evidence of A /\ B, which is to separately produce evidence of A and of B, both of which are passed into conj. That's why when we destructed A /\ B, it had to produce both evidence of A and of B. Going back to and_fst:

and_fst = 
fun (P Q : Prop) (PandQ : P /\ Q) =>
match PandQ with
| conj P_holds _ => P_holds
end
     : forall P Q : Prop, P /\ Q -> P

we have a function that pattern matches against PandQ, extracts the evidence for P, forgets about the evidence for Q, and simply returns the evidence for P. Given all that, the following theorem and its program should be unsurprising.

Coq responds:

and_snd = 
fun (P Q : Prop) (PandQ : P /\ Q) =>
match PandQ with
| conj _ Q_holds => Q_holds
end
     : forall P Q : Prop, P /\ Q -> Q

In that program the pattern match returns the second piece of evidence, which shows Q holds, rather than the first, which would show that P holds. Here is another proof involving and.

Why does that hold, intuitively? Because equality is reflexive, regardless of how many times we connect that fact with /\.

The first line of that proof, split, is a new tactic. It splits a goal of the form P /\ Q into two separate subgoals, one for P, and another for Q. Both must be proved individually. In the proof above, trivial suffices to prove them, because they are both trivial equalitiies. What is and_ex?

Coq responds:

and_ex = conj eq_refl eq_refl
     : 42 = 42 /\ 43 = 43

So and_ex is conj applied to eq_refl as its first argument and eq_refl as its second argument. As another example of conjunction, let's prove that it is commutative.

Why does this hold? If we assume that P /\ Q holds, then separately we must have evidence for P as well as Q. But the we could just assemble that evidence in the opposite order, producing evidence for Q /\ P. That's what the proof below does.

There's nothing new in that proof: we split the evidence into two pieces using pattern matching, then reassemble them in a different order. We can see that in the program:

Coq responds:

and_comm = 
fun (P Q : Prop) (PandQ : P /\ Q) =>
match PandQ with
| conj P_holds Q_holds => conj Q_holds P_holds
end
     : forall P Q : Prop, P /\ Q -> Q /\ P

Note how the pattern match binds two variables, then returns the conj constructor applied to those variables in the opposite order. Here's one more proof involving implication and conjunction.

Intuitively, why does this hold? Because we can assume P, Q and R. as well as that P /\ Q -> R, and that we have evidence for P and Q already. We can combine those two pieces of evidence for P and Q into a single piece of evidence for P /\ Q, which then yields the desired evidence for R.

There are no new tactics in the proof above. In line 2, we use apply to once again do backwards reasoning, transforming the goal of R into P /\ Q. Then we split that goal into two pieces, each of which can be solved by assumption. Let's look at the resulting program:

and_to_imp = 
fun (P Q R : Prop) (evPandQimpR : P /\ Q -> R) (evP : P) (evQ : Q) =>
evPandQimpR (conj evP evQ)
     : forall P Q R : Prop, (P /\ Q -> R) -> P -> Q -> R

That program uses conj to combine the evidence for P with the evidence for Q and get evidence for P /\ Q, then passes that evidence to evPandQimpR, thus obtaining evidence for R.

Disjunction

Let's start with disjunction by proving a theorem very similar to the first theorem we proved for conjunction:

As always, what's the intuition? Well, if we have evidence for P, then we have evidence for P \/ Q, because we have evidence already for the left-hand side of that connective. Let's formalize that argument in Coq.

The second line of that proof, left, uses a new tactic that tells Coq we want to prove the left-hand side of a disjunction. Specifically, the goal at that point is P \/ Q, and left tells Coq the throw out Q and just focus on proving P. That's easy, because P is already an assumption. Let's investigate the resulting program.

Coq responds

or_left = 
fun (P Q : Prop) (P_holds : P) => or_introl P_holds
     : forall P Q : Prop, P -> P \/ Q

That function's arguments are no mystery by now, but what is its body? We need to find out more about or_introl.

We learn that "\/" is infix notation for or A B, and or_introl is one of two constructors of the type or:

Inductive or (A B : Prop) : Prop :=
    or_introl : A -> A \/ B 
  | or_intror : B -> A \/ B

Those constructors take evidence for either the left-hand side or right-hand side of the disjuncation. So the body of or_left is just taking evidence for A and using or_introl to construct a value with that evidence. Similarly, the following theorem constructs proof using evidence for the right-hand side of a disjunction.

Much like with or_left, this intuitively holds because evidence for Q suffices as evidence for P \/ Q.

The resulting program uses the constructor or_intror:

or_right = 
fun (P Q : Prop) (Q_holds : Q) => or_intror Q_holds
     : forall P Q : Prop, Q -> P \/ Q

We could use those theorems to prove some related theorems. For example, 3110 = 3110 implies that 3110 = 3110 \/ 2110 = 3110.

Why does that hold? Because the left-hand side holds---even though the right hand side does not.

Coq responds that

or_thm = or_introl eq_refl
     : 3110 = 3110 \/ 2110 = 3110

In other words, the theorem is proved by applying the or_introl constructor to eq_refl. It matters, though, that we provided Coq with the guidance to prove the left-hand side. If we had chosen the right-hand side, we would have been stuck trying to prove that 2110 = 3110, which of course it does not. Next, let's prove that disjunction is commutative, as we did for conjunction above.

Why does this hold? If you assume you have evidence for P \/ Q, then you either have evidence for P or evidence for Q. If it's P, then you can prove Q \/ P by providing evidence for the right-hand side; or if it's Q, for the left-hand side. That's what the Coq proof below does.

In the second line of that proof, we destruct the disjunction PorQ with a slightly different syntax than when we destructed conjunction above. There's now a vertical bar. The reason for that has to do with the definitions of and and or. The definition of and had just a single constructor:

conj : A -> B -> A /\ B

So when we wrote destruct PandQ as [P_holds Q_holds], we were essentially writing a pattern match against that single constructor conj, and binding its A argument to the name P_holds, and its B argument to the name Q_holds. But the definition of or has two constructors:

    or_introl : A -> A \/ B 
  | or_intror : B -> A \/ B

So when we write destruct PorQ as [P_holds | Q_holds], we're providing two patterns: the first matches against the first constructor or_introl, binding its A argument to the name P_holds; and the second against the second constructor or_intror, binding its B argument to Q_holds. What makes this confusing to an OCaml programmer is that the as clause doesn't actually name the constructors. It might be clearer if Coq's destruct tactic used the following hypothetical syntax:

  destruct PandQ as [conj P_holds Q_holds].
  destruct PorQ as [or_introl P_holds | or_intror Q_holds].

but that's just not how destruct..as works. After the destruction occurs, we get two new subgoals to prove, corresponding to whether PorQ matched or_introl or or_intror. In the third line of the proof, which corresponds to PorQ matching or_introl, we have evidence for P, so we choose to prove the right side of Q \/ P by assumption. The fourth line does a similar thing, but using evidence for Q hence proving the left side of Q \/ P. Let's look at the resulting proof.

Coq responds

or_comm = 
fun (P Q : Prop) (PorQ : P \/ Q) =>
match PorQ with
| or_introl P_holds => or_intror P_holds
| or_intror Q_holds => or_introl Q_holds
end
     : forall P Q : Prop, P \/ Q -> Q \/ P)

That function takes an argument PorQ is evidence for P \/ Q, pattern matches against that argument, extracts the evidence for either P or Q, then returns that evidence wrapped in the "opposite" constructor: evidence that came in in the left constructor is returned with the right constructor, and vice versa. Next, let's prove a theorem involving both conjunction and disjunction.

This theorem says that "or" distributes over "and". Intuitively, if we have evidence for P \/ (Q /\ R), then one of two things holds: either we have evidence for P, or we have evidence for Q /\ R.

• If we have evidence for P, then we use that evidence to create evidence both for P \/ Q and for P /\ R.
• If we have evidence for Q /\ R, then we split that evidence apart into evidence separately for Q and for R. We use the evidence for Q to create evidence for P \/ Q, and likewise the evidence for R to create evidence for P \/ R.

The Coq proof below follows that intuition.

We used nested bullets in that proof to keep the structure of the proof clear to the reader, and to make it easier to follow the proof in the proof window when we step through it. But you'll notice there's a lot of repetition in the proof. We can eliminate some by using the ; tactical (discussed in the previous notes) to chain together some proof steps.

Either way, the resulting proof is

or_distr_and = 
fun (P Q R : Prop) (PorQR : P \/ Q /\ R) =>
match PorQR with
| or_introl P_holds => conj (or_introl P_holds) (or_introl P_holds)
| or_intror QR_holds =>
    match QR_holds with
    | conj Q_holds R_holds => conj (or_intror Q_holds) (or_intror R_holds)
    end
end
     : forall P Q R : Prop, P \/ Q /\ R -> (P \/ Q) /\ (P \/ R)

The nested pattern match in that proof corresponds to the nested destruct above. Note that Coq omits some parentheses around conjunction, because it is defined to have higher precedence than disjunction---just like * has higher precedence than +.

False and True

False is the proposition that can never hold: we can never actually have evidence for it. If we did somehow have evidence for False, our entire system of reasoning would be broken. There's a Latin phrase often used for that idea: ex falso quodlibet, meaning "from false, anything". In English it's sometimes known as the Principle of Explosion: if you're able to show False, then everything just explodes, and in fact anything at all becomes provable. Here's the definition of False:

Coq responds

Inductive False : Prop :=  

No, that isn't a typo above. There is nothing on the right-hand side of the :=. The definition is saying that False is an inductive type, like and and or, but it has zero constructors. Since it has no constructors, we can never create a value of type False---and that means we can never create evidence that False holds. Let's prove the Principle of Explosion.

Why should this theorem hold? Because of the intuition we gave above regarding ex falso.

The second line of the proof uses a new tactic, contradiction. This tactic looks for any contradictions or assumptions of False, and uses those to conclude the proof. In this case, it immediately finds False as an assumption named false_holds. The proof that Coq finds for explosion given the above tactics is a little bit hard to follow because it uses two functions False_ind and False_rect that are already defined for us. Their definitions are given below, but it's okay if you want to skip over reading them at first.

explosion = 
fun (P : Prop) (false_holds : False) => False_ind P false_holds
     : forall P : Prop, False -> P

False_ind = fun P : Prop => False_rect P
     : forall P : Prop, False -> P

False_rect = 
fun (P : Type) (f : False) => match f return P with
                              end
     : forall P : Type, False -> P

(The return P in the pattern match above is a type annotation: it says that the return type of the entire match expression is P.) Note that False_ind just calls False_rect, and False_rect immediately does a pattern match. So we could simplify the proof of explosion by just directly writing that pattern match ourselves. Let's do that using the Definition command we saw above.

The proof of explosion' is a function that takes two inputs, a proposition P and a value f of type False. As usual, we should understand that second argument as being evidence for False. But it should be impossible to construct such evidence! And indeed it is, because False has no constructors. So we have a function that we could never actually apply. In the body of that function is a pattern match against f. That pattern match has zero branches in it, exactly because False has zero constructors. There is nothing that could ever possibly match f. The return type of the function is P, meaning it purportedly is evidence for P. But of course no such evidence is ever constructed by the function, nor coud it be. So it's a good thing that it's impossible to apply the function. Were it possible, it would have to fabricate evidence for any proposition P whatsoever---which is not possible. We can never prove P -> P /\ False, because there is no way to construct the evidence for the right side.

But we can always prove P -> P \/ False by just focusing on the non-false part of the disjunction.

True is the proposition that is always true, i.e., for which we can always provide. It is defined by Coq as an inductive type:

Coq responds:

Inductive True : Prop :=  I : True

So True has a single constructor I, and anywhere we write I, that provides evidence for True. Let's redo the two theorem we just did for False, but with True in place of False.

The final tactic in that proof, exact, can be used whenever you already know exactly the program expression you want to write to provide evidence for the goal you are trying to prove. In this case, we know that I always provides evidence for true. Instead of exact we could also have used trivial, which is capable of proving trivial propositions like True.

Negation

The negation connective can be the trickiest to work with in Coq. Let's start by seeing how it is defined.

Coq responds

not = fun A : Prop => A -> False
     : Prop -> Prop

Unlike conjunction and disjunction, negation is defined as a function, rather than an inductive type. Anywhere we write ~P, it really means not P, which is (fun A => A -> False) P, which reduces to P -> False. In short, ~P is effectively syntactic sugar for P -> False. Here are two proofs involving negation, so that we can get used to this definition of it.

Intuition: anything implies True, regardless of what that anything is.

The first line of that proof, unfold not, is a new tactic, which replaces the ~ in the goal with its definition and simplifies it. So ~False becomes False -> False. The second line uses a familiar tactic in a new way. We don't provide any names to intros, which causes Coq to choose its own names. Normally we consider it good style to choose them ourselves, but here, we don't care what they are, because we are never going to use them. Finally, the third line uses I to prove True. Looking at the actual program produced, we see that it's very simple:

Coq responds

notFalse = fun _ : False -> False => I

which is a function that takes an argument that is never used, and simply returns I, which is the evidence for True. Here's a second proof involving negation.

Intuition: if True implies False, and if True, then False.

The first line of the proof replaces ~True with True -> False. The rest of the proof proceeds by moving True -> False into the assumptions, then applying it to do backward reasoning, leaving us with needing to prove True. That holds by the I constructor. The program that proof produces is interesting:

Coq responds

notTrue = fun t_imp_f : True -> False => t_imp_f I
     : ~ True -> False

That proof is actually a higher-order function that takes in a function t_imp_f and applies that function to I, thus transforming evidence for True into evidence for False, and returning that evidence. If that seems impossible, it is! We'll never be able to apply this function, because we'll never be able to construct a value to pass to it whose type is ~True, i.e., True -> False. Next, let's return to the idea of explosion. From a contradiction we should be able to derive anything at all. One kind of contradiction is for a proposition and its negation to hold simultaneously, e.g., P /\ ~P.

Intuition: principle of explosion

This proof proceeds along familiar lines: We destruct the evidence of P /\ ~P into two pieces. That leaves us with P as an assumption as well as ~P. The contradiction tactic detects those contradictory assumptions and finished the proof. By the way, we could have left out the as clause in the destruct here, since we are never going to use the names we chose, but they will help make the following program more readable:

Coq responds:

contra_implies_anything = 
fun (P : Prop) (Q : Type) (PandnotP : P /\ (P -> False)) =>
match PandnotP with
| conj P_holds notP_holds => False_rect Q (notP_holds P_holds)
end
     : forall (P : Prop) (Q : Type), P /\ ~ P -> Q

The really interesting part of this is the body of the pattern match, False_rect Q (notP_holds P_holds). It applies notP_holds to P_holds, thus transforming evidence for P into evidence for False. It passes that hypothetical evidence for False to False_rect, which as we've seen before uses such evidence to produce evidence for anything at all that we would like---in this case, Q, its first argument. Next, let's try a proof involving all the connectives we've seen: negation, conjunction, disjunction, and implication. The following theorem shows how negation distributes over disjunction, and in so doing, produces a conjunction. The name we choose for this theorem is traditional and gives credit to Augustus De Morgan, a 19th century logician, even though the theorem was known far earlier in history.

Intuition: if evidence for P or Q would lead to an explosion (i.e., False), then evidence for P would lead to an explosion, and evidence for Q would also lead to an explosion.

Nothing we've done in that proof is new, but it is longer than any of our proofs so far. The same is true of the program it produces:

Coq responds:

deMorgan = 
fun (P Q : Prop) (PorQ_imp_false : P \/ Q -> False) =>
conj (fun P_holds : P => PorQ_imp_false (or_introl P_holds))
     (fun Q_holds : Q => PorQ_imp_false (or_intror Q_holds))
       : forall P Q : Prop, ~ (P \/ Q) -> ~ P /\ ~ Q

There is a second "De Morgan's Law", which says that negation distributes over conjunction, thus producing a disjunction. But something seemingly goes very wrong when we try to prove it in Coq:

Intuition: if evidence for P and Q would produce an explosion, then either evidence for P would produce an explosion, or evidence for Q would.

When we get to the point in the proof above where we need to prove Q, we have no means to do so. (The same problem would occur for P if instead of left we had gone right.) Why does this happen? One reason is that the intuition we gave above is wrong! There's no reason that "if evidence for P and Q would produce an explosion" implies "either evidence for P would produce an explosion, or evidence for Q would". It's the combined evidence for P and Q that produces the explosion in that assumption---nothing is said about evidence for them individually. But a deeper reason is that this theorem simply doesn't hold in Coq's logic: there isn't any way to prove it. That might surprise you if you've studied some logic before, and you were taught that both De Morgan laws are sound reasoning principles. Indeed they are in some logics, but not in others. In classical logic, which is what you almost certainly would have studied before (e.g., in CS 2800), we are permitted to think of every proposition as having a truth value, which is either True or False. And one way to prove a theorem is to construct a truth table showing what truth value a proposition has for any assignment of True or False to its variables. For example (writing T and F as abbreviations):

P  Q  ~(P/\Q) (~P \/ ~Q) (~(P/\Q) -> (~P \/ ~Q))
T  T  F       F          T
T  F  T       T          T
F  T  T       T          T
F  F  T       T          T

For every possible assignment of truth values to P and Q, we get that the truth value of ~(P/\Q) -> (~P \/ ~Q) is True. So in classical logic, ~(P/\Q) -> (~P \/ ~Q) is a theorem. But Coq uses a different logic called constructive logic. Constructive logic is more convervative than classical logic, in that it always requires evidence to be produced as part of a proof. As we saw above when we tried to prove deMorgan2, there just isn't a way to construct evidence for ~P \/ ~Q out of evidence for ~(P/\Q). There are many other propositions that are provable in classical logic but not in constructive logic. Another pair of such propositions involves double negation: P -> ~~P is provable in both logics, but ~~P -> P is provable in classical logic and not provable in constructive logic. Let's try proving both just to see what happens.

Intuition: ~~P is (P -> False) -> False. So the theorem could be restated as P -> (P -> False) -> False. That's really just a syllogism, with the first two arguments in the opposite order:

• P implies False.
• P holds.
• Therefore False holds.

We could make that intuition even more apparent by proving a version of the syllogism theorem with its first two arguments swapped:

Now we use syllogism' to prove P -> ~~P:

In that proof, we have a new use for the apply tactic: we use it with the name of a theorem we've already proved, because our goal is in fact that theorem. In the resulting program, we saw this indeed becomes an application of the syllogism' function:

Coq responds:

p_imp_nnp' = 
fun P : Prop => syllogism' P False
     : forall P : Prop, P -> ~ ~ P

Now let's try the other direction for double negation:

Once we get past introducing assumptions in that proof, we're stuck. There's nothing we can do with (P -> False) -> False to prove P. Why? Because in constructive logic, to prove P, we must produce evidence for P. Nothing in (P -> False) -> False gives us such evidence. That's very different from classical logic, where we could just construct a truth table:

P ~~P  (~~P -> P)
T T    T
F F    T

Here's another even more brutally perplexing proposition that's provable in classical logic but not in constructive logic. It's called excluded middle, because it says that every proposition or its negation must hold; there is no "middle ground":

Whether we go left or right in the second step of that proof, we immediately get stuck. If we go left, Coq challenges us to construct evidence for P, but we don't have any. If we go right, Coq challenges us to construct evidence for P -> False, but we don't have any. Yet in classical logic, excluded middle is easily proved by a truth table:

P  ~P  (P \/ ~P)
T  F   T
F  T   T

Why does Coq use constructive logic rather than classical logic? The reason goes back to why we started looking at Coq, namely, program verification. We'd like to be able to extract verified programs. Well, there simply is no program whose type is P \/ ~P, because such a program would have to use either or_introl or or_intror to construct its result, and it would have to magically guess which one to use, then magically somehow produce the appropriate evidence. Nonetheless, if all you want to do is reasoning in classical logic, and you don't care about extracting verified programs, Coq does support that in a library Coq.Logic.Classical. We'll load that library now in a nested Module, which restricts its influence just to that module so that we don't pollute the rest of this file.

Coq responds:

*** [ classic : forall P : Prop, P \/ ~ P ]

The *** indicates that classic is an axiom that the library simply asserts without proof. Using that axiom, all the usual theorems of classical logic can be proved, such as double negation:

Coq responds:

NNPP = 
fun (p : Prop) (H : (p -> False) -> False) =>
or_ind (fun H0 : p => H0) (fun NP : ~ p => False_ind p (H NP)) (classic p)
     : forall p : Prop, ~ ~ p -> p

You can see on the last line that NNPP proves ~~p -> p, which we couldn't prove above. And you'll see that classic shows up in the program to magically construct evidence of p \/ ~p.

Equality and implication

Let's return to the two connectives with which we started, equality and implication. Earlier we didn't explain them fully, but now we're equipped to appreciate their definitions. Recall how Coq defines equality:

Coq responds:

Inductive eq (A : Type) (x : A) : A -> Prop :=  
  eq_refl : x = x

For eq: Argument A is implicit ...

Reading that carefully, we see that eq is parameterized on

• a type A, and
• a value x whose type is A.

We also see that A is implicit, so let's use @eq from now on in our discussion just to be clear about A. When we apply @eq to a type A and a value x, we get back a function of type A -> Prop. The idea is that function will take its argument, let's call it y, and construct the proposition that asserts that x and y are equal. For example:

eq42 : nat -> Prop
eq42 42 : Prop
eq42 43 : Prop

There's only one way to construct a value of type eq, though, and that's with the eq_refl constructor. If we "desugar" the = notation, that constructor has type @eq A x x, where x must be of type A.

@eq_refl nat 42 : 42 = 42

Note how the constructor above takes just a single argument of type nat, not two arguments: it will only ever show that argument is equal to itself, never to anything else. There's literally no way to write an expression using eq_refl to construct evidence that (e.g.) 42 and 43 are equal. So instead of using trivial, we could directly use eq_refl as a constructor to prove equalities, much like we directly used I as a constructor to prove True earlier in this file:

Equality, therefore, is not something that has to be "baked in" to Coq, but rather something that is definable as an inductive type---much like and and or. And now back to implication. It, too, is defined.

Coq responds:

"A -> B" := forall _ : A, B 

That is, A -> B is really just syntactic sugar for forall (_:A), B, where we've added some parentheses for a little bit more clarity. Still, that expression is tricky to read because of the wildcard in it. It might help if we made A and B concrete, for example, if A is P /\ Q and B is Q, where P and Q are propositions. Then we could think of the type forall (_ : P /\ Q), Q as follows:

• (_ : P /\ Q) means an unnamed value of type P /\ Q. Using the evidence interpretation we've been developing throughout this file, that value would be evidence that P /\ Q holds. It's unnamed because it's not used on the right-hand side.
• A value of type Q would be evidence for Q.
• So a value of type forall (_ : P /\ Q), Q would be a "thing" that for any piece of evidence that P /\ Q holds can produce evidence that Q holds.

What is such a "thing"? A function! It transforms evidence for one proposition into evidence for another proposition. So of all the logic we've coded up in Coq in this file, the only truly primitive pieces are Inductive definitions and forall types. Everything else---equality, implication, conjunction, disjunction, true, false, negation---can all be expressed in terms of those two primitives. And although we had to learn many tactics, every proof we constructed with them was really just a program that used functions, application, and pattern matching.

Tautologies

(Or, "How to make the computer do your CS 2800 logic homework for you".) We needed to learn the tactics and definitions above for the more complicated proofs we will later want to do in Coq. But if all you care about is proving simple logical propositions, there's a tactic for that: tauto. It will succeed in finding a proof of any propositional tautology, which is a formula that must always hold regardless of the values of variables in it. For example, one of the most complicated proofs we did above was De Morgan's law. Here it is in just one tactic:

Coq responds:

deMorgan' = 
fun (P Q : Prop) (H : ~ (P \/ Q)) =>
let H0 := (fun H0 : P => H (or_introl H0)) : P -> False in
let H1 := (fun H1 : Q => H (or_intror H1)) : Q -> False in
conj (fun H2 : P => let H3 := H0 H2 : False in False_ind False H3)
  (fun H2 : Q => let H3 := H1 H2 : False in False_ind False H3)
     : forall P Q : Prop, ~ (P \/ Q) -> ~ P /\ ~ Q

That's a bit more complicated of a proof than the one we constructed ourselves, mainly because of the let expressions, but it still is correct. So if your CS 2800 prof wants you to prove a propositional tautology, and if it's a proposition that holds in constructive logic, Coq's got your back: just use tauto to do it. But don't tell your 2800 prof I told you that.

Summary

Coq's built-in logic is constructive: it isn't sufficient to argue that a proposition must be true or false; rather, we have to construct evidence for the proposition. Programs are how we construct and transform evidence. All of the propositional connectives, except implication, are "coded up" in Coq using inductive types, and proofs about them routinely use pattern matching and function application.

Terms and concepts

• assumption
• classical logic
• conjunction
• constructive logic
• constructor
• contradiction
• De Morgan's laws
• disjunction
• double negation
• evidence
• excluded middle
• implication
• inductive type
• negation
• Principle of Explosion
• Prop
• proposition
• reflexivity
• Set
• syllogism
• tautology
• transitivity
• truth table

Tactics

• apply
• assumption
• contradiction
• destruct..as
• exact
• left
• right
• split
• tauto
• tacticals: nested bullets -, *, +

Further reading

• Software Foundations, Volume 1: Logical Foundations. Chapter 6: Logic.
• Interactive Theorem Proving and Program Development. Chapters 5 and 8.2. Available online from the Cornell library.