In traditional mathematics, the concept of equality is usually taken as a given. On the other hand, in type theory, equality is a very contentious subject. There are at least three different notions of equality that are important, and researchers are actively investigating new definitions of what it means for two terms to be equal. Even once we fix a notion of equality, there are inevitably tricky issues that arise in proving properties of programs that manipulate equality proofs explicitly. In this chapter, we will focus on design patterns for circumventing these tricky issues, and we will introduce the different notions of equality as they are germane.

The Definitional Equality

We have seen many examples so far where proof goals follow "by computation." That is, we apply computational reduction rules to reduce the goal to a normal form, at which point it follows trivially. Exactly when this works and when it does not depends on the details of Coq's definitional equality. This is an untyped binary relation appearing in the formal metatheory of CIC. CIC contains a typing rule allowing the conclusion from the premise and a proof that and are definitionally equal.

The cbv tactic will help us illustrate the rules of Coq's definitional equality. We redefine the natural number predecessor function in a somewhat convoluted way and construct a manual proof that it returns 0 when applied to 1.

CIC follows the traditions of lambda calculus in associating reduction rules with Greek letters. Coq can certainly be said to support the familiar alpha reduction rule, which allows capture-avoiding renaming of bound variables, but we never need to apply alpha explicitly, since Coq uses a de Bruijn representation that encodes terms canonically.

The delta rule is for unfolding global definitions. We can use it here to unfold the definition of pred'. We do this with the cbv tactic, which takes a list of reduction rules and makes as many call-by-value reduction steps as possible, using only those rules. There is an analogous tactic lazy for call-by-need reduction.

  ============================
   (fun x : nat => match x with
                   | 0 => 0
                   | S n' => let y := n' in y
                   end) 1 = 0
 

At this point, we want to apply the famous beta reduction of lambda calculus, to simplify the application of a known function abstraction.

  ============================
   match 1 with
   | 0 => 0
   | S n' => let y := n' in y
   end = 0
 

Next on the list is the iota reduction, which simplifies a single match term by determining which pattern matches.

  ============================
   (fun n' : nat => let y := n' in y) 0 = 0
 

Now we need another beta reduction.

  ============================
   (let y := 0 in y) = 0
 



The final reduction rule is zeta, which replaces a let expression by its body with the appropriate term subsituted.

  ============================
   0 = 0
 

The standard eq relation is critically dependent on the definitional equality. eq is often called a propositional equality, because it reifies definitional equality as a proposition that may or may not hold. Standard axiomatizations of an equality predicate in first-order logic define equality in terms of properties it has, like reflexivity, symmetry, and transitivity. In contrast, for eq in Coq, those properties are implicit in the properties of the definitional equality, which are built into CIC's metatheory and the implementation of Gallina. We could add new rules to the definitional equality, and eq would keep its definition and methods of use.

This all may make it sound like the choice of eq's definition is unimportant. To the contrary, in this chapter, we will see examples where alternate definitions may simplify proofs. Before that point, we will introduce effective proof methods for goals that use proofs of the standard propositional equality "as data."

Heterogeneous Lists Revisited

One of our example dependent data structures from the last chapter was heterogeneous lists and their associated "cursor" type. The recursive version poses some special challenges related to equality proofs, since it uses such proofs in its definition of member types.

We can define a map-like function for fhlists.

For the inductive versions of the ilist definitions, we proved a lemma about the interaction of get and imap. It was a strategic choice not to attempt such a proof for the definitions that we just gave, because that sets us on a collision course with the problems that are the subject of this chapter.

Part of our single remaining subgoal is:


  a0 : a = elm
  ============================
   match a0 in (_ = a2) return (C a2) with
   | refl_equal => f a1
   end = f match a0 in (_ = a2) return (B a2) with
           | refl_equal => a1
           end
 

This seems like a trivial enough obligation. The equality proof a0 must be refl_equal, since that is the only constructor of eq. Therefore, both the matches reduce to the point where the conclusion follows by reflexivity.


    destruct a0.

User error: Cannot solve a second-order unification problem
 

This is one of Coq's standard error messages for informing us that its heuristics for attempting an instance of an undecidable problem about dependent typing have failed. We might try to nudge things in the right direction by stating the lemma that we believe makes the conclusion trivial.


    assert (a0 = refl_equal _).

The term "refl_equal ?98" has type "?98 = ?98"
 while it is expected to have type "a = elm"
 

In retrospect, the problem is not so hard to see. Reflexivity proofs only show x = x for particular values of x, whereas here we are thinking in terms of a proof of a = elm, where the two sides of the equality are not equal syntactically. Thus, the essential lemma we need does not even type-check!

Is it time to throw in the towel? Luckily, the answer is "no." In this chapter, we will see several useful patterns for proving obligations like this.

For this particular example, the solution is surprisingly straightforward. destruct has a simpler sibling case which should behave identically for any inductive type with one constructor of no arguments.

  ============================
   f a1 = f a1
 

It seems that destruct was trying to be too smart for its own good.

It will be helpful to examine the proof terms generated by this sort of strategy. A simpler example illustrates what is going on.

simple destruct pf is a convenient form for applying case. It runs intro to bring into scope all quantified variables up to its argument.

lemma1 =
fun (x : A) (pf : x = elm) =>
match pf as e in (_ = y) return (0 = match e with
                                     | refl_equal => 0
                                     end) with
| refl_equal => refl_equal 0
end
     : forall (x : A) (pf : x = elm), 0 = match pf with
                                          | refl_equal => 0
                                          end
 

Using what we know about shorthands for match annotations, we can write this proof in shorter form manually.

Surprisingly, what seems at first like a simpler lemma is harder to prove.

    simple destruct pf.

User error: Cannot solve a second-order unification problem
 

Nonetheless, we can adapt the last manual proof to handle this theorem.

We can try to prove a lemma that would simplify proofs of many facts like lemma2:

    simple destruct pf.

User error: Cannot solve a second-order unification problem

This time, even our manual attempt fails.


  Definition lemma3' :=
    fun (x : A) (pf : x = x) =>
      match pf as pf' in (_ = x') return (pf' = refl_equal x') with
        | refl_equal => refl_equal _
      end.

The term "refl_equal x'" has type "x' = x'" while it is expected to have type
 "x = x'"
 

The type error comes from our return annotation. In that annotation, the as-bound variable pf' has type x = x', refering to the in-bound variable x'. To do a dependent match, we must choose a fresh name for the second argument of eq. We are just as constrained to use the "real" value x for the first argument. Thus, within the return clause, the proof we are matching on must equate two non-matching terms, which makes it impossible to equate that proof with reflexivity.

Nonetheless, it turns out that, with one catch, we can prove this lemma.

UIP_refl
     : forall (U : Type) (x : U) (p : x = x), p = refl_equal x
 

UIP_refl comes from the Eqdep module of the standard library. Do the Coq authors know of some clever trick for building such proofs that we have not seen yet? If they do, they did not use it for this proof. Rather, the proof is based on an axiom.

eq_rect_eq =
fun U : Type => Eq_rect_eq.eq_rect_eq U
     : forall (U : Type) (p : U) (Q : U -> Type) (x : Q p) (h : p = p),
       x = eq_rect p Q x p h
 

eq_rect_eq states a "fact" that seems like common sense, once the notation is deciphered. eq_rect is the automatically-generated recursion principle for eq. Calling eq_rect is another way of matching on an equality proof. The proof we match on is the argument h, and x is the body of the match. eq_rect_eq just says that matches on proofs of p = p, for any p, are superfluous and may be removed.

Perhaps surprisingly, we cannot prove eq_rect_eq from within Coq. This proposition is introduced as an axiom; that is, a proposition asserted as true without proof. We cannot assert just any statement without proof. Adding False as an axiom would allow us to prove any proposition, for instance, defeating the point of using a proof assistant. In general, we need to be sure that we never assert inconsistent sets of axioms. A set of axioms is inconsistent if its conjunction implies False. For the case of eq_rect_eq, consistency has been verified outside of Coq via "informal" metatheory.

This axiom is equivalent to another that is more commonly known and mentioned in type theory circles.

Streicher_K =
fun U : Type => UIP_refl__Streicher_K U (UIP_refl U)
     : forall (U : Type) (x : U) (P : x = x -> Prop),
       P (refl_equal x) -> forall p : x = x, P p
 

This is the unfortunately-named "Streicher's axiom K," which says that a predicate on properly-typed equality proofs holds of all such proofs if it holds of reflexivity.

Type-Casts in Theorem Statements

Sometimes we need to use tricks with equality just to state the theorems that we care about. To illustrate, we start by defining a concatenation function for fhlists.

We might like to prove that fhapp is associative.


  Theorem fhapp_ass : forall ls1 ls2 ls3
    (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
    fhapp hls1 (fhapp hls2 hls3) = fhapp (fhapp hls1 hls2) hls3.

The term
 "fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3) (fhapp (ls1:=ls1) (ls2:=ls2) hls1 hls2)
    hls3" has type "fhlist B ((ls1 ++ ls2) ++ ls3)"
 while it is expected to have type "fhlist B (ls1 ++ ls2 ++ ls3)"
 

This first cut at the theorem statement does not even type-check. We know that the two fhlist types appearing in the error message are always equal, by associativity of normal list append, but this fact is not apparent to the type checker. This stems from the fact that Coq's equality is intensional, in the sense that type equality theorems can never be applied after the fact to get a term to type-check. Instead, we need to make use of equality explicitly in the theorem statement.

The first remaining subgoal looks trivial enough:


  ============================
   fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
   match pf in (_ = ls) return (fhlist B ls) with
   | refl_equal => fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
   end
 

We can try what worked in previous examples.


    case pf.

User error: Cannot solve a second-order unification problem
 

It seems we have reached another case where it is unclear how to use a dependent match to implement case analysis on our proof. The UIP_refl theorem can come to our rescue again.

  ============================
   fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
   fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
 

Our second subgoal is trickier.


  pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
  ============================
   (a0,
   fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
     (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
   match pf in (_ = ls) return (fhlist B ls) with
   | refl_equal =>
       (a0,
       fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
         (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
   end

    rewrite (UIP_refl _ _ pf).

The term "pf" has type "a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3"
 while it is expected to have type "?556 = ?556"
 

We can only apply UIP_refl on proofs of equality with syntactically equal operands, which is not the case of pf here. We will need to manipulate the form of this subgoal to get us to a point where we may use UIP_refl. A first step is obtaining a proof suitable to use in applying the induction hypothesis. Inversion on the structure of pf is sufficient for that.

  pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
  pf' : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3
  ============================
   (a0,
   fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
     (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
   match pf in (_ = ls) return (fhlist B ls) with
   | refl_equal =>
       (a0,
       fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
         (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
   end
 

Now we can rewrite using the inductive hypothesis.

  ============================
   (a0,
   match pf' in (_ = ls) return (fhlist B ls) with
   | refl_equal =>
       fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
         (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3
   end) =
   match pf in (_ = ls) return (fhlist B ls) with
   | refl_equal =>
       (a0,
       fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
         (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
   end
 

We have made an important bit of progress, as now only a single call to fhapp appears in the conclusion. Trying case analysis on our proofs still will not work, but there is a move we can make to enable it. Not only does just one call to fhapp matter to us now, but it also does not matter what the result of the call is. In other words, the subgoal should remain true if we replace this fhapp call with a fresh variable. The generalize tactic helps us do exactly that.

   forall f : fhlist B ((ls1 ++ ls2) ++ ls3),
   (a0,
   match pf' in (_ = ls) return (fhlist B ls) with
   | refl_equal => f
   end) =
   match pf in (_ = ls) return (fhlist B ls) with
   | refl_equal => (a0, f)
   end
 

The conclusion has gotten markedly simpler. It seems counterintuitive that we can have an easier time of proving a more general theorem, but that is exactly the case here and for many other proofs that use dependent types heavily. Speaking informally, the reason why this kind of activity helps is that match annotations only support variables in certain positions. By reducing more elements of a goal to variables, built-in tactics can have more success building match terms under the hood.

In this case, it is helpful to generalize over our two proofs as well.

   forall (pf0 : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
     (pf'0 : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3)
     (f : fhlist B ((ls1 ++ ls2) ++ ls3)),
   (a0,
   match pf'0 in (_ = ls) return (fhlist B ls) with
   | refl_equal => f
   end) =
   match pf0 in (_ = ls) return (fhlist B ls) with
   | refl_equal => (a0, f)
   end
 

To an experienced dependent types hacker, the appearance of this goal term calls for a celebration. The formula has a critical property that indicates that our problems are over. To get our proofs into the right form to apply UIP_refl, we need to use associativity of list append to rewrite their types. We could not do that before because other parts of the goal require the proofs to retain their original types. In particular, the call to fhapp that we generalized must have type (ls1 ++ ls2) ++ ls3, for some values of the list variables. If we rewrite the type of the proof used to type-cast this value to something like ls1 ++ ls2 ++ ls3 = ls1 ++ ls2 ++ ls3, then the lefthand side of the equality would no longer match the type of the term we are trying to cast.

However, now that we have generalized over the fhapp call, the type of the term being type-cast appears explicitly in the goal and may be rewritten as well. In particular, the final masterstroke is rewriting everywhere in our goal using associativity of list append.

  ============================
   forall (pf0 : a :: ls1 ++ ls2 ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
     (pf'0 : ls1 ++ ls2 ++ ls3 = ls1 ++ ls2 ++ ls3)
     (f : fhlist B (ls1 ++ ls2 ++ ls3)),
   (a0,
   match pf'0 in (_ = ls) return (fhlist B ls) with
   | refl_equal => f
   end) =
   match pf0 in (_ = ls) return (fhlist B ls) with
   | refl_equal => (a0, f)
   end
 

We can see that we have achieved the crucial property: the type of each generalized equality proof has syntactically equal operands. This makes it easy to finish the proof with UIP_refl.

Heterogeneous Equality

There is another equality predicate, defined in the JMeq module of the standard library, implementing heterogeneous equality.

Inductive JMeq (A : Type) (x : A) : forall B : Type, B -> Prop :=
    JMeq_refl : JMeq x x
 

JMeq stands for "John Major equality," a name coined by Conor McBride as a sort of pun about British politics. JMeq starts out looking a lot like eq. The crucial difference is that we may use JMeq on arguments of different types. For instance, a lemma that we failed to establish before is trivial with JMeq. It makes for prettier theorem statements to define some syntactic shorthand first.

There is no quick way to write such a proof by tactics, but the underlying proof term that we want is trivial.

Suppose that we want to use UIP_refl' to establish another lemma of the kind of we have run into several times so far.

All in all, refreshingly straightforward, but there really is no such thing as a free lunch. The use of rewrite is implemented in terms of an axiom:

JMeq_eq
     : forall (A : Type) (x y : A), x == y -> x = y
 

It may be surprising that we cannot prove that heterogeneous equality implies normal equality. The difficulties are the same kind we have seen so far, based on limitations of match annotations.

We can redo our fhapp associativity proof based around JMeq.

This time, the naive theorem statement type-checks.

Even better, crush discharges the first subgoal automatically. The second subgoal is:


  ============================
   (a0,
   fhapp (B:=B) (ls1:=ls1) (ls2:=ls2 ++ ls3) b
     (fhapp (B:=B) (ls1:=ls2) (ls2:=ls3) hls2 hls3)) ==
   (a0,
   fhapp (B:=B) (ls1:=ls1 ++ ls2) (ls2:=ls3)
     (fhapp (B:=B) (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
 

It looks like one rewrite with the inductive hypothesis should be enough to make the goal trivial.


    rewrite IHls1.

Error: Impossible to unify "fhlist B ((ls1 ++ ?1572) ++ ?1573)" with
 "fhlist B (ls1 ++ ?1572 ++ ?1573)"
 

We see that JMeq is not a silver bullet. We can use it to simplify the statements of equality facts, but the Coq type-checker uses non-trivial heterogeneous equality facts no more readily than it uses standard equality facts. Here, the problem is that the form (e1, e2) is syntactic sugar for an explicit application of a constructor of an inductive type. That application mentions the type of each tuple element explicitly, and our rewrite tries to change one of those elements without updating the corresponding type argument.

We can get around this problem by another multiple use of generalize. We want to bring into the goal the proper instance of the inductive hypothesis, and we also want to generalize the two relevant uses of fhapp.

  ============================
   forall (f : fhlist B (ls1 ++ ls2 ++ ls3))
     (f0 : fhlist B ((ls1 ++ ls2) ++ ls3)), f == f0 -> (a0, f) == (a0, f0)
 

Now we can rewrite with append associativity, as before.

  ============================
   forall f f0 : fhlist B (ls1 ++ ls2 ++ ls3), f == f0 -> (a0, f) == (a0, f0)
 

From this point, the goal is trivial.

Equivalence of Equality Axioms

Assuming axioms (like axiom K and JMeq_eq) is a hazardous business. The due diligence associated with it is necessarily global in scope, since two axioms may be consistent alone but inconsistent together. It turns out that all of the major axioms proposed for reasoning about equality in Coq are logically equivalent, so that we only need to pick one to assert without proof. In this section, we demonstrate this by showing how each the previous two sections' approaches reduces to the other logically.

To show that JMeq and its axiom let us prove UIP_refl, we start from the lemma UIP_refl' from the previous section. The rest of the proof is trivial.

The other direction is perhaps more interesting. Assume that we only have the axiom of the Eqdep module available. We can define JMeq in a way that satisfies the same interface as the combination of the JMeq module's inductive definition and axiom.

We say that, by definition, x and y are equal if and only if there exists a proof pf that their types are equal, such that x equals the result of casting y with pf. This statement can look strange from the standpoint of classical math, where we almost never mention proofs explicitly with quantifiers in formulas, but it is perfectly legal Coq code.

We can easily prove a theorem with the same type as that of the JMeq_refl constructor of JMeq.

The proof of an analogue to JMeq_eq is a little more interesting, but most of the action is in appealing to UIP_refl.

  H : exists pf : A = A,
        x = match pf in (_ = T) return T with
            | refl_equal => y
            end
  ============================
   x = y
 

  x0 : A = A
  H : x = match x0 in (_ = T) return T with
          | refl_equal => y
          end
  ============================
   x = y
 

  x0 : A = A
  ============================
   match x0 in (_ = T) return T with
   | refl_equal => y
   end = y
 

We see that, in a very formal sense, we are free to switch back and forth between the two styles of proofs about equality proofs. One style may be more convenient than the other for some proofs, but we can always intercovert between our results. The style that does not use heterogeneous equality may be preferable in cases where many results do not require the tricks of this chapter, since then the use of axioms is avoided altogether for the simple cases, and a wider audience will be able to follow those "simple" proofs. On the other hand, heterogeneous equality often makes for shorter and more readable theorem statements.

It is worth remarking that it is possible to avoid axioms altogether for equalities on types with decidable equality. The Eqdep_dec module of the standard library contains a parametric proof of UIP_refl for such cases.

Equality of Functions

The following seems like a reasonable theorem to want to hold, and it does hold in set theory.
   Theorem S_eta : S = (fun n => S n).
 

Unfortunately, this theorem is not provable in CIC without additional axioms. None of the definitional equality rules force function equality to be extensional. That is, the fact that two functions return equal results on equal inputs does not imply that the functions are equal. We can assert function extensionality as an axiom.

This axiom has been verified metatheoretically to be consistent with CIC and the two equality axioms we considered previously. With it, the proof of S_eta is trivial.

The same axiom can help us prove equality of types, where we need to "reason under quantifiers."

There are no immediate opportunities to apply ext_eq, but we can use change to fix that.

2 subgoals
  
  ============================
   (nat -> True) = (nat -> True)

subgoal 2 is:
 forall x : nat, match x with
                 | 0 => True
                 | S _ => True
                 end = True

Exercises

1. Implement and prove correct a substitution function for simply-typed lambda calculus. In particular:
   1. Define a datatype type of lambda types, including just booleans and function types.
   2. Define a type family exp : list type -> type -> Type of lambda expressions, including boolean constants, variables, and function application and abstraction.
   3. Implement a definitional interpreter for exps, by way of a recursive function over expressions and substitutions for free variables, like in the related example from the last chapter.
   4. Implement a function subst : forall t' ts t, exp (t' :: ts) t -> exp ts t' -> exp ts t. The type of the first expression indicates that its most recently bound free variable has type t'. The second expression also has type t', and the job of subst is to substitute the second expression for every occurrence of the "first" variable of the first expression.
   5. Prove that subst preserves program meanings. That is, prove
      forall t' ts t (e : exp (t' :: ts) t) (e' : exp ts t') (s : hlist typeDenote ts),
        expDenote (subst e e') s = expDenote e (expDenote e' s ::: s)
       
      
      where ::: is an infix operator for heterogeneous "cons" that is defined in the book's DepList module.
   The material presented up to this point should be sufficient to enable a good solution of this exercise, with enough ingenuity. If you get stuck, it may be helpful to use the following structure. None of these elements need to appear in your solution, but we can at least guarantee that there is a reasonable solution based on them.
   1. The DepList module will be useful. You can get the standard dependent list definitions there, instead of copying-and-pasting from the last chapter. It is worth reading the source for that module over, since it defines some new helpful functions and notations that we did not use last chapter.
   2. Define a recursive function liftVar : forall ts1 ts2 t t', member t (ts1 ++ ts2) -> member t (ts1 ++ t' :: ts2). This function should "lift" a de Bruijn variable so that its type refers to a new variable inserted somewhere in the index list.
   3. Define a recursive function lift' : forall ts t (e : exp ts t) ts1 ts2 t', ts = ts1 ++ ts2 -> exp (ts1 ++ t' :: ts2) t which performs a similar lifting on an exp. The convoluted type is to get around restrictions on match annotations. We delay "realizing" that the first index of e is built with list concatenation until after a dependent match, and the new explicit proof argument must be used to cast some terms that come up in the match body.
   4. Define a function lift : forall ts t t', exp ts t -> exp (t' :: ts) t, which handles simpler top-level lifts. This should be an easy one-liner based on lift'.
   5. Define a recursive function substVar : forall ts1 ts2 t t', member t (ts1 ++ t' :: ts2) -> (t' = t) + member t (ts1 ++ ts2). This function is the workhorse behind substitution applied to a variable. It returns inl to indicate that the variable we pass to it is the variable that we are substituting for, and it returns inr to indicate that the variable we are examining is not the one we are substituting for. In the first case, we get a proof that the necessary typing relationship holds, and, in the second case, we get the original variable modified to reflect the removal of the substitutee from the typing context.
   6. Define a recursive function subst' : forall ts t (e : exp ts t) ts1 t' ts2, ts = ts1 ++ t' :: ts2 -> exp (ts1 ++ ts2) t' -> exp (ts1 ++ ts2) t. This is the workhorse of substitution in expressions, employing the same proof-passing trick as for lift'. You will probably want to use lift somewhere in the definition of subst'.
   7. Now subst should be a one-liner, defined in terms of subst'.
   8. Prove a correctness theorem for each auxiliary function, leading up to the proof of subst correctness.
   9. All of the reasoning about equality proofs in these theorems follows a regular pattern. If you have an equality proof that you want to replace with refl_equal somehow, run generalize on that proof variable. Your goal is to get to the point where you can rewrite with the original proof to change the type of the generalized version. To avoid type errors (the infamous "second-order unification" failure messages), it will be helpful to run generalize on other pieces of the proof context that mention the equality's lefthand side. You might also want to use generalize dependent, which generalizes not just one variable but also all variables whose types depend on it. generalize dependent has the sometimes-helpful property of removing from the context all variables that it generalizes. Once you do manage the mind-bending trick of using the equality proof to rewrite its own type, you will be able to rewrite with UIP_refl.
   10. A variant of the ext_eq axiom from the end of this chapter is available in the book module Axioms, and you will probably want to use it in the lift' and subst' correctness proofs.
   11. The change tactic should come in handy in the proofs about lift and subst, where you want to introduce "extraneous" list concatenations with nil to match the forms of earlier theorems.
   12. Be careful about destructing a term "too early." You can use generalize on proof terms to bring into the proof context any important propositions about the term. Then, when you destruct the term, it is updated in the extra propositions, too. The case_eq tactic is another alternative to this approach, based on saving an equality between the original term and its new form.