Like its built-in programming language, Coq's built-in logic is extremely small: universal quantification (forall) and implication (->) are primitive, but all the other familiar logical connectives -- conjunction, disjunction, negation, existential quantification, even equality -- can be defined using just these and Inductive.

Conjunction

The logical conjunction of propositions A and B is represented by the following inductively defined proposition.

Note that, like the definition of ev, this definition is parameterized; however, in this case, the parameters are themselves propositions.

The intuition behind this definition is simple: to construct evidence for and A B, we must provide evidence for A and evidence for B. More precisely:

• conj e1 e2 can be taken as evidence for and A B if e1 is evidence for A and e2 is evidence for B; and
  
  
• this is the only way to give evidence for and A B -- that is, if someone gives us evidence for and A B, we know it must have the form conj e1 e2, where e1 is evidence for A and e2 is evidence for B.

Since we'll be using conjunction a lot, let's introduce a more familiar-looking infix notation for it.

(The type_scope annotation tells Coq that this notation will be appearing in propositions, not values.)

Consider the "type" of the constructor conj

Notice that it "takes 4 inputs", namely the propositions A,B and evidence of A, B, and "returns as output", the evidence of A /\ B.

Exercise: 1 star (and_ind_principle)

See if you can predict the induction principle for conjunction.

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Besides the elegance of building everything up from a tiny foundation, what's nice about defining conjunction this way is that we can prove statements involving conjunction using the tactics that we already know. For example, if the goal statement is a conjuction, we can prove it by applying the single constructor conj, which (as can be seen from the type of conj) solves the current goal and leaves the two parts of the conjunction as subgoals to be proved separately.

Let's take a look at the proof object for the above theorem.

Note that the proof is of the form

      conj (ev 0) (ev 4) (Pf of ev 0) (Pf of ev 4)

which is what you'd expect, given the type of conj

The split tactic is a convenient shorthand for apply conj.

Conversely, the inversion tactic can be used to investigate a conjunction hypothesis in the context and calculate what evidence must have been used to build it.

Exercise: 1 star

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Exercise: 2 stars

Notice the nested inversion breaks H : A /\ (B /\ C) down into HA: A, HB : B and HC : C. Finish the proof from there:

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Exercise: 2 stars

Now we can prove the other direction of the equivalence of even and ev. Notice that the left-hand conjunct here is the statement we are actually interested in; the right-hand conjunct is needed in order to make the induction hypothesis strong enough that we can carry out the reasoning in the inductive step. (To see why this is needed, try proving the left conjunct by itself and observe where things get stuck.)

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Iff

The familiar logical "if and only if" is just the conjunction of two implications.

Exercise: 1 star (iff_properties)

Using the above proof that <-> is symmetric (iff_sym) as a guide, prove that it is also reflexive and transitive.

Hint: If you have an iff hypothesis in the context, you can use inversion to break it into two separate implications. (Think about why this works.)

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Exercise: 2 stars

We have seen that the families of propositions MyProp and ev actually characterize the same set of numbers (the even ones). Prove that MyProp n <-> ev n for all n (MyProp_iff_ev in Logic.v). Just for fun, write your proof as an explicit proof object, rather than using tactics.

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Unfortunately, propositions phrased with <-> are a bit inconvenient to use as hypotheses or lemmas, because they have to be deconstructed into their two directed components in order to be applied. (The basic problem is that there's no way to apply an iff proposition directly. If it's a hypothesis, you can invert it, which is tedious; if it's a lemma, you have to destruct it into hypotheses, which is worse.) Consequently, many Coq developments avoid <->, despite its appealing compactness. It can actually be made much more convenient using a Coq feature called "setoid rewriting," but that is a bit beyond the scope of this course.

Disjunction

Disjunction ("logical or") can also be defined as an inductive proposition.

Exercise: 1 star (or_ind_principle)

See if you can predict the induction principle for disjunction.

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Consider the "type" of the constructor or_introl

Notice that it "takes 3 inputs", namely the propositions A,B and evidence of A, and "returns as output", the evidence of A /\ B. Next, look at the "type" of or_intror

It is like or_introl but instead of evidence of A, requires evidence of B

Intuition: There are two ways of giving evidence for A \/ B:

• give evidence for A (and say that it is A you are giving evidence for! -- this is the function of the or_introl constructor)
  
  
• give evidence for B, tagged with the or_intror constructor.

Since A \/ B has two constructors, doing inversion on a hypothesis of type A \/ B yields two subgoals.

From here on, we'll use the handy tactics left and right in place of apply or_introl and apply or_intror.

Exercise: 2 stars

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Exercise: 1 star

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Relating /\ and \/ with andb and orb

We've already seen several places where analogous structures can be found in Coq's computational (Type) and logical (Prop) worlds. Here is one more: the boolean operators andb and orb are obviously analogs, in some sense, of the logical connectives /\ and \/. This analogy can be made more precise by the following theorems, which show how to "translate" knowledge about andb and orb's behaviors on certain inputs into propositional facts about those inputs.

Exercise: 1 star (bool_prop)

Prove these properties relating operations on booleans and logical connectives in Prop.

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Falsehood

Logical falsehood can be represented in Coq as an inductively defined proposition with no constructors.

Intuition: False is a proposition for which there is no way to give evidence.

Exercise: 1 star (False_ind_principle)

Can you predict the induction principle for falsehood?

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Since False has no constructors, inverting it always yields zero subgoals, allowing us to immediately prove any goal.

What happened? inversion breaks contra into each of its possible cases, and yields a subgoal for each case. As contra is evidence for False, it has no possible cases, hence, there are no possible subgoals and the proof is done.

Conversely, the only way to prove False is if there is already something nonsensical or contradictory in the context:

Actually, since the proof of False_implies_nonsense doesn't actually have anything to do with the specific nonsensical thing being proved; it can easily be generalized to work for an arbitrary P:

The Latin ex falso quodlibet means, literally, "from falsehood follows whatever you please." This theorem is also known as the principle of explosion.

Truth

Since we have defined falsehood in Coq, we should also mention that it is, of course, possible to define truth in the same way.

Exercise: 2 stars

Define True as another inductively defined proposition. What induction principle will Coq generate for your definition? (The intution is that True should be a proposition for which it is trivial to give evidence. Alternatively, you may find it easiest to start with the induction principle and work backwards to the inductive definition.)

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However, unlike False, which we'll use extensively, True is basically a theoretical curiosity: since it is trivial to prove as a goal, it carries no useful information as a hypothesis.

Negation

The logical complement of a proposition A is written not A or, for shorthand, ~A:

The intuition is that, if A is not true, then anything at all (even False) should follow from assuming A.

It takes a little practice to get used to working with negation in Coq. Even though you can see perfectly well why something is true, it can be a little hard at first to figure out how to get things into the right configuration so that Coq can see it! Here are proofs of a view familiar facts about negation to get you warmed up.

Exercise: 2 stars (double_neg_inf)

As an exercise, write an informal proof of double_neg:

Theorem: For any prop A, A -> ~~A

Proof: (* FILL IN HERE *)
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Exercise: 2 stars

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Exercise: 1 star

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Exercise: 1 star

Theorem five_not_even in Logic.v confirms the unsurprising fact that that five is not an even number. Prove this more interesting fact:

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Exercise: 4 stars, optional (classical_axioms)

For those who like a challenge, here is an exercise taken from the Coq'Art book (p. 123). The following five statements are often considered as characterizations of classical logic (as opposed to constructive logic, which is what is "built in" to Coq). We can't prove them in Coq, but we can consistently add any one of them as an unproven axiom if we wish to work in classical logic. Prove that these five propositions are equivalent.

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Inequality

Saying x <> y is just the same as saying ~(x = y).

Since inequality involves a negation, it again requires a little practice to be able to work with it fluently. Here is one very useful trick. If you are trying to prove a goal that is nonsensical (e.g., the goal state is false = true), apply the lemma ex_falso_quodlibet to change the goal to False. This makes it easier to use assumptions of the form ~P that are available in the context -- in particular, assumptions of the form x<>y.

Exercise: 2 stars

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Exercise: 4 stars, optional

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Existential quantification

Another extremely important logical connective is existential quantification.

The intuition behind this definition is that, in order to give evidence for the assertion "there exists an x for which the proposition P holds" we must actually name a witness -- a specific value x -- and then give evidence for P x.

We can use Coq's notation definition facility to introduce more standard notation for writing existentially quantified propositions, exactly parallel to the built-in syntax for universally quantified propositions. Instead of writing (ex nat (fun x => ev x)) to express the proposition that there exists some number that is even, for example, we can write exists x:nat, ev x. (It is not necessary to understand the details of how the definition works.)

We can use the same tactics as always for manipulate existentials. For example, if to prove an existential, we apply the constructor ex_intro. Since the premise of ex_intro involves a variable (witness) that does not appear in its conclusion, we need to explicitly give its value when we use apply.

Note that we have to explicitly give the witness.

Or, instead of writing apply ex_intro with e, we can write the convenient shorthand exists e.

Conversely, if we have an existential hypothesis in the context, we can eliminate it with inversion. Note the use of the as... pattern to name the variable that Coq introduces to name the witness value and get evidence that the hypothesis holds for the witness. (If we don't explicitly choose one, Coq will just call it witness, which makes proofs confusing.)

Exercise: 1 star

Prove that "P holds for all x" and "there is no x for which P does not hold" are equivalent assertions.

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Exercise: 2 stars

Prove that existential quantification distributes over disjunction.

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Equality.

Even Coq's equality relation is not actually built in. It has the following inductive definition.

(We enclose this definition in a module to avoid confusion with the standard library equality (which we have used extensively already).

This definition is a bit subtle. The way to think about it is that, given a set A, it defines a family of propositions "x is equal to y," indexed by pairs of values (x and y) from A. There is just one way of constructing evidence for members of this family: by applying the constructor refl_equal to two identical arguments.

Here is an alternate definition, from the Coq standard library.

Exercise: 3 stars, optional

Verify that the two definitions of equality are equivalent.

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The advantage of the second definition is that the induction principle that Coq derives for it is precisely the familiar principle of Leibniz equality: what we mean when we say "x and y are equal" is that every property on A that is true of x is also true of y.

Inversion, again

We've seen inversion used with both equality hypotheses and hypotheses asserting inductively defined propositions. Now that we've seen that these are actually the same thing, let's take a closer look at how inversion behaves...

In general, the inversion tactic

• takes a hypothesis H whose type is some inductively defined proposition P
  
  
• for each constructor C in P's definition
  
  
  • generates a new subgoal in which we assume H was built with C
    
    
  • adds the arguments (premises) of C to the context of the subgoal as extra hypotheses
    
    
  • matches the conclusion (result type) of C against the current goal and calculates a set of equalities that must hold in order for C to be applicable
    
    
  • adds these equalities to the context of the subgoal
    
    
  • if the equalities are not satisfiable (e.g., they involve things like S n = O, immediately solves the subgoal.

EXAMPLE: If we invert a hypothesis built with or, there are two constructors, so two subgoals get generated. The conclusion (result type) of the constructor -- A \/ B -- doesn't place any restrictions on the form of A or B, so we don't get any extra equalities in the context of the subgoal.

EXAMPLE: If we invert a hypothesis built with and, there is only one constructor, so only one subgoal gets generated. Again, the conclusion (result type) of the constructor -- A /\ B -- doesn't place any restrictions on the form of A or B, so we don't get any extra equalities in the context of the subgoal. The constructor does have two arguments, though, and these can be seen in the context in the subgoal.

EXAMPLE: If we invert a hypothesis built with eq, there is again only one constructor, so only one subgoal gets generated. Now, though, the form of the refl_equal constructor does give us some extra information: it tells us that the two arguments to eq must be equal! The inversion tactic adds this to the context.

Relations as propositions

A proposition parameterized over a number (like ev) can be thought of as a predicate -- i.e., it defines a subset of nat, namely those numbers for which the proposition is provable. In the same way, a two-argument proposition can be thought of as a relation -- i.e., it defines a set of pairs for which the proposition is provable. In the next few sections, we explore the consequences of this observation.

We've already seen an inductive definition of one fundamental relation: equality. Another useful one is the "less than or equal to" relation on numbers:

This definition should be fairly intuitive. It says that there are two ways to give evidence that one number is less than or equal to another: either observe that they are the same number, or give evidence that the first is less than or equal to the predecessor of the second.

This is a fine definition of the <= relation, but we can streamline it a little by observing that the left-hand argument n is the same everywhere in the definition, so we can actually make it a "general parameter" to the whole definition, rather than an argument to each constructor.

Why is the second one better (even though it looks less symmetric)? Because it gives us a simpler induction principle.

By contrast, the induction principle that Coq calculates for the first definition has a lot of extra quantifiers, which makes it messier to work with when proving things by induction. Here is the induction principle for the first le:

Proofs of facts about <= using the constructors le_n and le_S follow the same patterns as proofs about predicates, like ev in the previous chapter. We can apply the constructors to prove <= goals (e.g., to show that 3<=3 or 3<=6), and we can use tactics like inversion to extract information from <= hypotheses in the context (e.g., to prove that ~(2 <= 1).)

Here are some sanity checks on the definition. (Notice that, although these are the same kind of simple "unit tests" as we gave for the testing functions we wrote in the first few lectures, we must construct their proofs explicitly -- simpl and reflexivity don't do the job, because the proofs aren't just a matter of simplifying computations.

The "strictly less than" relation n < m can now be defined in terms of le.

A few more simple relations on numbers

Exercise: 2 stars

Define an inductive relation total_relation that holds betfween every pair of natural numbers.

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Exercise: 2 stars

Define an inductive relation empty_relation (on numbers) that never holds.

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Exercise: 3 stars (R_provability)

We can define three-place relations, four-place relations, etc., in just the same way as binary relations. For example, consider the following three-place relation on numbers:

• Which of the following propositions are provable?
  • R 1 1 2
  • R 2 2 6
    
    
• If we dropped constructor c5 from the definition of R, would the set of provable propositions change? Briefly (1 sentence) explain your answer.
  
  
• If we dropped constructor c4 from the definition of R, would the set of provable propositions change? Briefly (1 sentence) explain your answer.

(* FILL IN HERE *)
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Exercise: 4 stars, challenge problem (filter_challenge)

One of the main purposes of Coq is to prove that programs match their specifications. To this end, let's prove that our definition of filter matches a specification. Here is the specification, written out informally in English.

Suppose we have a set X, a function test: X->bool, and a list l of type list X. Suppose further that l is an in-order merge of two lists, l1 and l2, such that every item in l1 satisfies test and no item in l2 satisfies test. Then filter test l = l1.

Your job is to translate this specification into a Coq theorem and prove it. (Hint: You'll need to begin by defining what it means for one list to be a merge of two others. Do this with an inductive relation, not a Fixpoint.)

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Relations, in General

We've now defined a few particular relations. As you probably remember from your undergraduate discrete math course, there are a lot of ways of discussing and describing relations in general -- ways of classifying relations (are they reflexive, transitive, etc.), theorems that can be proved generically about classes of relations, constructions that build one relation from another, etc. Let us pause here to review a few of these that will be useful in what follows.

A relation on a set X is a proposition parameterized by two Xs -- i.e., it is a logical assertion involving two values from the set X. (This definition is just giving a name to the idea that we've been using for the past few minutes.)

A relation R on a set X is a partial function if, for every x, there is at most one y such that R x y -- or, to put it differently, if R x y1 and R x y2 together imply y1 = y2.

Now, we can prove that the next_nat relation is a partial function...

...but <= is not a partial function.

Exercise: 2 stars

Show that the total_relation you defined above is not a partial function.

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Exercise: 2 stars

Show that the empty_relation you defined above is a partial function.

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Exercise: 2 stars

We can also prove lt_trans more laboriously by induction, without using le_trans. Do this.

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Exercise: 2 stars

Prove the same thing again by induction on o

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The transitivity of le, in turn, can be used to prove some facts that will be useful later (e.g., for the proof of antisymmetry in the next section)...

Exercise: 1 star

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Exercise: 2 stars (le_Sn_n_inf)

Provide an informal proof of the following theorem:

Theorem: For every n, ~(S n <= n)

A formal proof of this is an optional exercise below, but try the informal proof without doing the formal proof first

Proof: (* FILL IN HERE *)
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Exercise: 1 star

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Exercise: 2 stars

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Exercise: 2 stars

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A relation is called a "partial order" when it's reflexive, antisymmetric, and transitive. In the Coq standard library it's just "order" for short.

A preorder is almost like a partial order, but doesn't have to be antisymmetric.

More facts about <= and <

Let's pause briefly to record several facts about the <= and < relations that we are going to need later in the course. The proofs make good practice exercises.

Exercise: 2 stars, optional (le_exercises)

Hint: Do the right induction!

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