The apply Tactic

We can use apply when some hypothesis or an earlier lemma exactly matches the goal:

Here, we could finish with "rewrite → eq₂. reflexivity." as we have done several times before. We can finish this proof in a single step by using the apply tactic instead:

apply also works with conditional hypotheses:

The apply with Tactic

The following silly example uses two rewrites in a row to get from [a;b] to [e;f].

Since this is a common pattern, we might like to pull it out as a lemma recording, once and for all, the fact that equality is transitive.

Applying this lemma to the example above requires a slight variation on apply:

Doing apply trans_eq doesn't work! But...

does.

The injection and discriminate Tactics

In Coq, the constructors of inductive types are injective and disjoint. E.g., for nat...

• if S n = S m then it must hold that n = m (that is, S is injective aka one-to-one)
• O is not equal to S n for any n (that is, O and S are disjoint)

For example, we can prove the injectivity of S by using the pred function defined in Basics.v.

As a convenience, the injection tactic allows us to exploit injectivity of any constructor.

Here's a more interesting example that shows how injection can derive multiple equations at once.

The "as" variant of injection permits us to choose names for the equations and immediately introduce them as hypotheses, rather than as premises. The original equality, in this case H, is also removed from the hypotheses with this variant.

So much for injectivity of constructors. What about disjointness? The principle of disjointness says that two terms beginning with different constructors (like O and S, or true and false) can never be equal. This means that, any time we find ourselves working in a context where we've assumed that two such terms are equal, we are justified in concluding anything we want to (because the assumption is nonsensical). The discriminate tactic embodies this principle: It is used on a hypothesis involving an equality between different constructors (e.g., S n = O), and it solves the current goal immediately. For example:

This is an instance of a logical principle known as the principle of explosion, which asserts that a contradictory hypothesis entails anything, even false things!

The injectivity of constructors allows us to reason that ∀ (n m : nat), S n = S m → n = m. The converse of this implication is an instance of a more general fact about both constructors and functions, which we will find convenient in a few places below:

Or use tactic f_equal.

Using Tactics on Hypotheses

Many tactics come with "... in ..." variants that work on hypotheses instead of goals.

The ordinary apply tactic is a form of "backward reasoning": it says "We're trying to prove X and we know Y → X, so if we can prove Y we'll be done." By contrast, the variant apply... in... is "forward reasoning": it says "We know Y and we know Y → X, so we also know X."

Varying the Induction Hypothesis

Recall this function for doubling a natural number (from the Induction chapter):

Suppose we want to show that double is injective — i.e., that it maps different arguments to different results. The way we start this proof is a little bit delicate:

At this point, the induction hypothesis, IHn', does not give us n' = m' — there is an extra S in the way — so the goal is not provable.

The successful proof of double_injective leaves m in the goal statement at the point where the induction tactic is invoked on n:

The strategy of doing fewer intros before an induction to obtain a more general IH doesn't always work by itself; sometimes some rearrangement of quantified variables is needed. Suppose, for example, that we wanted to prove double_injective by induction on m instead of n.

The problem is that, to do induction on m, we must first introduce n. (If we simply say induction m without introducing anything first, Coq will automatically introduce n for us!)

What we can do instead is to first introduce all the quantified variables and then re-generalize one or more of them, selectively taking variables out of the context and putting them back at the beginning of the goal. The generalize dependent tactic does this.

Unfolding Definitions

It sometimes happens that we need to manually unfold a name that has been introduced by a Definition so that we can manipulate its right-hand side. For example, if we define...

... and try to prove a simple fact about square...

... we appear to be stuck: simpl doesn't simplify anything at this point, and since we haven't proved any other facts about square, there is nothing we can apply or rewrite with. To make progress, we can manually unfold the definition of square:

Now we have plenty to work with: both sides of the equality are expressions involving multiplication, and we have lots of facts about multiplication at our disposal. In particular, we know that it is commutative and associative, and from these it is not hard to finish the proof.

At this point, some discussion of unfolding and simplification is in order. You may already have observed that tactics like simpl, reflexivity, and apply will often unfold the definitions of functions automatically when this allows them to make progress. For example, if we define foo m to be the constant 5...

.... then the simpl in the following proof (or the reflexivity, if we omit the simpl) will unfold foo m to (fun x ⇒ 5) m and then further simplify this expression to just 5.

However, this automatic unfolding is somewhat conservative. For example, if we define a slightly more complicated function involving a pattern match...

...then the analogous proof will get stuck:

The reason that simpl doesn't make progress here is that it notices that, after tentatively unfolding bar m, it is left with a match whose scrutinee, m, is a variable, so the match cannot be simplified further. It is not smart enough to notice that the two branches of the match are identical, so it gives up on unfolding bar m and leaves it alone. Similarly, tentatively unfolding bar (m+1) leaves a match whose scrutinee is a function application (that cannot itself be simplified, even after unfolding the definition of +), so simpl leaves it alone.

At this point, there are two ways to make progress. One is to use destruct m to break the proof into two cases, each focusing on a more concrete choice of m (O vs S _). In each case, the match inside of bar can now make progress, and the proof is easy to complete.

This approach works, but it depends on our recognizing that the match hidden inside bar is what was preventing us from making progress.

A more straightforward way to make progress is to explicitly tell Coq to unfold bar.

Now it is apparent that we are stuck on the match expressions on both sides of the =, and we can use destruct to finish the proof without thinking too hard.

Using destruct on Compound Expressions

The destruct tactic can be used on expressions as well as variables:

The eqn: part of the destruct tactic is optional: We've chosen to include it most of the time, just for the sake of documentation, but many Coq proofs omit it. When destructing compound expressions, however, the information recorded by the eqn: can actually be critical: if we leave it out, then destruct can sometimes erase information we need to complete a proof.

Adding the eqn: qualifier saves this information so we can use it.

Micro Sermon

Mindless proof-hacking is a terrible temptation... Try to resist!