# UseAutoTheory and Practice of Automation in Coq Proofs

(* Chapter written and maintained by Arthur Chargueraud *)

In a machine-checked proof, every single detail has to be
justified. This can result in huge proof scripts. Fortunately,
Coq comes with a proof-search mechanism and with several decision
procedures that enable the system to automatically synthesize
simple pieces of proof. Automation is very powerful when set up
appropriately. The purpose of this chapter is to explain the
basics of how automation works in Coq.
The chapter is organized in two parts. The first part focuses on a
general mechanism called "proof search." In short, proof search
consists in naively trying to apply lemmas and assumptions in all
possible ways. The second part describes "decision procedures",
which are tactics that are very good at solving proof obligations
that fall in some particular fragments of the logic of Coq.
Many of the examples used in this chapter consist of small lemmas
that have been made up to illustrate particular aspects of automation.
These examples are completely independent from the rest of the Software
Foundations course. This chapter also contains some bigger examples
which are used to explain how to use automation in realistic proofs.
These examples are taken from other chapters of the course (mostly
from STLC), and the proofs that we present make use of the tactics
from the library LibTactics.v, which is presented in the chapter
UseTactics.

From Coq Require Import Arith.Arith.

From PLF Require Import Maps.

From PLF Require Import Smallstep.

From PLF Require Import LibTactics.

From PLF Require Stlc.

From PLF Require Imp.

From Coq Require Import Lists.List.

From PLF Require Import Maps.

From PLF Require Import Smallstep.

From PLF Require Import LibTactics.

From PLF Require Stlc.

From PLF Require Imp.

From Coq Require Import Lists.List.

# Basic Features of Proof Search

## Strength of Proof Search

## Basics

The second example illustrates a proof where a sequence of
two calls to apply are needed. The goal is to prove that
if Q n implies P n for any n and if Q n holds for any n,
then P 2 holds.

If we are interested to see which proof auto came up with,
one possibility is to look at the generated proof-term,
using the command:
Print solving_by_apply.
The proof term is:
fun (P Q : nat → Prop) (H : ∀ n : nat, Q n → P n) (H
This essentially means that auto applied the hypothesis H
(the first one), and then applied the hypothesis H
The tactic auto can invoke apply but not eapply. So, auto
cannot exploit lemmas whose instantiation cannot be directly
deduced from the proof goal. To exploit such lemmas, one needs to
invoke the tactic eauto, which is able to call eapply.
In the following example, the first hypothesis asserts that P n
is true when Q m is true for some m, and the goal is to prove
that Q 1 implies P 2. This implication follows directly from
the hypothesis by instantiating m as the value 1. The
following proof script shows that eauto successfully solves the
goal, whereas auto is not able to do so.

_{0}: ∀ n : nat, Q n) ⇒ H 2 (H_{0}2)_{0}(the second one).
Lemma solving_by_eapply : ∀ (P Q : nat→Prop),

(∀ n m, Q m → P n) →

Q 1 →

P 2.

Proof. auto. eauto. Qed.

(∀ n m, Q m → P n) →

Q 1 →

P 2.

Proof. auto. eauto. Qed.

## Conjunctions

However, when an assumption is a conjunction, auto and eauto
are not able to exploit this conjunction. It can be quite
surprising at first that eauto can prove very complex goals but
that it fails to prove that F ∧ F' implies F. The tactics
iauto and jauto are able to decompose conjunctions from the context.
Here is an example.

Lemma solving_conj_hyp : ∀ (F F' : Prop),

F ∧ F' →

F.

Proof. auto. eauto. jauto. (* or iauto *) Qed.

F ∧ F' →

F.

Proof. auto. eauto. jauto. (* or iauto *) Qed.

The tactic jauto is implemented by first calling a
pre-processing tactic called jauto_set, and then calling
eauto. So, to understand how jauto works, one can directly
call the tactic jauto_set.

Next is a more involved goal that can be solved by iauto and
jauto.

Lemma solving_conj_more : ∀ (P Q R : nat→Prop) (F : Prop),

(F ∧ (∀ n m, (Q m ∧ R n) → P n)) →

(F → R 2) →

Q 1 →

P 2 ∧ F.

Proof. jauto. (* or iauto *) Qed.

(F ∧ (∀ n m, (Q m ∧ R n) → P n)) →

(F → R 2) →

Q 1 →

P 2 ∧ F.

Proof. jauto. (* or iauto *) Qed.

The strategy of iauto and jauto is to run a global analysis of
the top-level conjunctions, and then call eauto. For this
reason, those tactics are not good at dealing with conjunctions
that occur as the conclusion of some universally quantified
hypothesis. The following example illustrates a general weakness
of Coq proof search mechanisms.

Lemma solving_conj_hyp_forall : ∀ (P Q : nat→Prop),

(∀ n, P n ∧ Q n) →

P 2.

Proof.

auto. eauto. iauto. jauto.

(* Nothing works, so we have to do some of the work by hand *)

intros. destruct (H 2). auto.

Qed.

(∀ n, P n ∧ Q n) →

P 2.

Proof.

auto. eauto. iauto. jauto.

(* Nothing works, so we have to do some of the work by hand *)

intros. destruct (H 2). auto.

Qed.

This situation is slightly disappointing, since automation is
able to prove the following goal, which is very similar. The
only difference is that the universal quantification has been
distributed over the conjunction.

Lemma solved_by_jauto : ∀ (P Q : nat→Prop) (F : Prop),

(∀ n, P n) ∧ (∀ n, Q n) →

P 2.

Proof. jauto. (* or iauto *) Qed.

(∀ n, P n) ∧ (∀ n, Q n) →

P 2.

Proof. jauto. (* or iauto *) Qed.

However, only iauto is able to automate reasoning on the
disjunctions that appear in the context. For example, iauto can
prove that F ∨ F' entails F' ∨ F.

More generally, iauto can deal with complex combinations of
conjunctions, disjunctions, and negations. Here is an example.

Lemma solving_tauto : ∀ (F

((¬F

(F

(F

¬F

Proof. iauto. Qed.

_{1}F_{2}F_{3}: Prop),((¬F

_{1}∧ F_{3}) ∨ (F_{2}∧ ¬F_{3})) →(F

_{2}→ F_{1}) →(F

_{2}→ F_{3}) →¬F

_{2}.Proof. iauto. Qed.

However, the ability of iauto to automatically perform a case
analysis on disjunctions comes with a downside: iauto may be
very slow. If the context involves several hypotheses with
disjunctions, iauto typically generates an exponential number of
subgoals on which eauto is called. One major advantage of jauto
compared with iauto is that it never spends time performing this
kind of case analyses.

## Existentials

Lemma solving_exists_goal : ∀ (f : nat→Prop),

f 2 →

∃ x, f x.

Proof.

auto. (* observe that auto does not deal with existentials, *)

eauto. (* whereas eauto, iauto and jauto solve the goal *)

Qed.

f 2 →

∃ x, f x.

Proof.

auto. (* observe that auto does not deal with existentials, *)

eauto. (* whereas eauto, iauto and jauto solve the goal *)

Qed.

A major strength of jauto over the other proof search tactics is
that it is able to exploit the existentially-quantified
hypotheses, i.e., those of the form ∃ x, P.

Lemma solving_exists_hyp : ∀ (f g : nat→Prop),

(∀ x, f x → g x) →

(∃ a, f a) →

(∃ a, g a).

Proof.

auto. eauto. iauto. (* All of these tactics fail, *)

jauto. (* whereas jauto succeeds. *)

(* For the details, run intros. jauto_set. eauto *)

Qed.

(∀ x, f x → g x) →

(∃ a, f a) →

(∃ a, g a).

Proof.

auto. eauto. iauto. (* All of these tactics fail, *)

jauto. (* whereas jauto succeeds. *)

(* For the details, run intros. jauto_set. eauto *)

Qed.

## Negation

Lemma negation_study_1 : ∀ (P : nat→Prop),

P 0 →

(∀ x, ¬ P x) →

False.

Proof.

intros P H

eauto. (* It fails to see that HX applies *)

unfold not in ×. eauto.

Qed.

P 0 →

(∀ x, ¬ P x) →

False.

Proof.

intros P H

_{0}HX.eauto. (* It fails to see that HX applies *)

unfold not in ×. eauto.

Qed.

For this reason, the tactics iauto and jauto systematically
invoke unfold not in × as part of their pre-processing. So,
they are able to solve the previous goal right away.

Lemma negation_study_2 : ∀ (P : nat→Prop),

P 0 →

(∀ x, ¬ P x) →

False.

Proof. jauto. (* or iauto *) Qed.

P 0 →

(∀ x, ¬ P x) →

False.

Proof. jauto. (* or iauto *) Qed.

We will come back later on to the behavior of proof search with
respect to the unfolding of definitions.

## Equalities

To automate more advanced reasoning on equalities, one should
rather try to use the tactic congruence, which is presented at
the end of this chapter in the "Decision Procedures" section.

## Search Depth

The reason auto fails to solve the goal is because there are
too many conjunctions. If there had been only five of them, auto
would have successfully solved the proof, but six is too many.
The tactic auto limits the number of lemmas and hypotheses
that can be applied in a proof, so as to ensure that the proof
search eventually terminates. By default, the maximal number
of steps is five. One can specify a different bound, writing
for example auto 6 to search for a proof involving at most
six steps. For example, auto 6 would solve the previous lemma.
(Similarly, one can invoke eauto 6 or intuition eauto 6.)
The argument n of auto n is called the "search depth."
The tactic auto is simply defined as a shorthand for auto 5.
The behavior of auto n can be summarized as follows. It first
tries to solve the goal using reflexivity and assumption. If
this fails, it tries to apply a hypothesis (or a lemma that has
been registered in the hint database), and this application
produces a number of sugoals. The tactic auto (n-1) is then
called on each of those subgoals. If all the subgoals are solved,
the job is completed, otherwise auto n tries to apply a
different hypothesis.
During the process, auto n calls auto (n-1), which in turn
might call auto (n-2), and so on. The tactic auto 0 only
tries reflexivity and assumption, and does not try to apply
any lemma. Overall, this means that when the maximal number of
steps allowed has been exceeded, the auto tactic stops searching
and backtracks to try and investigate other paths.
The following lemma admits a unique proof that involves exactly
three steps. So, auto n proves this goal iff n is greater than
three.

Lemma search_depth_1 : ∀ (P : nat→Prop),

P 0 →

(P 0 → P 1) →

(P 1 → P 2) →

(P 2).

Proof.

auto 0. (* does not find the proof *)

auto 1. (* does not find the proof *)

auto 2. (* does not find the proof *)

auto 3. (* finds the proof *)

(* more generally, auto n solves the goal if n ≥ 3 *)

Qed.

P 0 →

(P 0 → P 1) →

(P 1 → P 2) →

(P 2).

Proof.

auto 0. (* does not find the proof *)

auto 1. (* does not find the proof *)

auto 2. (* does not find the proof *)

auto 3. (* finds the proof *)

(* more generally, auto n solves the goal if n ≥ 3 *)

Qed.

We can generalize the example by introducing an assumption
asserting that P k is derivable from P (k-1) for all k,
and keep the assumption P 0. The tactic auto, which is the
same as auto 5, is able to derive P k for all values of k
less than 5. For example, it can prove P 4.

Lemma search_depth_3 : ∀ (P : nat→Prop),

(* Hypothesis H

(* Hypothesis H

(* Goal: *) (P 4).

Proof. auto. Qed.

(* Hypothesis H

_{1}: *) (P 0) →(* Hypothesis H

_{2}: *) (∀ k, P (k - 1) → P k) →(* Goal: *) (P 4).

Proof. auto. Qed.

However, to prove P 5, one needs to call at least auto 6.

Lemma search_depth_4 : ∀ (P : nat→Prop),

(* Hypothesis H

(* Hypothesis H

(* Goal: *) (P 5).

Proof. auto. auto 6. Qed.

(* Hypothesis H

_{1}: *) (P 0) →(* Hypothesis H

_{2}: *) (∀ k, P (k - 1) → P k) →(* Goal: *) (P 5).

Proof. auto. auto 6. Qed.

Because auto looks for proofs at a limited depth, there are
cases where auto can prove a goal F and can prove a goal
F' but cannot prove F ∧ F'. In the following example,
auto can prove P 4 but it is not able to prove P 4 ∧ P 4,
because the splitting of the conjunction consumes one proof step.
To prove the conjunction, one needs to increase the search depth,
using at least auto 6.

Lemma search_depth_5 : ∀ (P : nat→Prop),

(* Hypothesis H

(* Hypothesis H

(* Goal: *) (P 4 ∧ P 4).

Proof. auto. auto 6. Qed.

(* Hypothesis H

_{1}: *) (P 0) →(* Hypothesis H

_{2}: *) (∀ k, P (k - 1) → P k) →(* Goal: *) (P 4 ∧ P 4).

Proof. auto. auto 6. Qed.

## Backtracking

Lemma working_of_auto_1 : ∀ (P : nat→Prop),

(* Hypothesis H

(* Hypothesis H

(* Hypothesis H

(* Goal: *) (P 2).

(* Uncomment "debug" in the following line to see the debug trace: *)

Proof. intros P H

(* Hypothesis H

_{1}: *) (P 0) →(* Hypothesis H

_{2}: *) (∀ k, P (k - 1) → P k) →(* Hypothesis H

_{3}: *) (∀ k, P (k + 1) → P k) →(* Goal: *) (P 2).

(* Uncomment "debug" in the following line to see the debug trace: *)

Proof. intros P H

_{1}H_{2}H_{3}. (* debug *) eauto. Qed.
The output message produced by debug eauto is as follows.
It seems that eauto was quite lucky there, as it never even
tried to use the hypothesis H

1 depth=5 1.1 depth=4 simple apply HThe depth indicates the value of n with which eauto n is called. The tactics shown in the message indicate that the first thing that eauto has tried to do is to apply H_{2}1.1.1 depth=3 simple apply H_{2}1.1.1.1 depth=3 exact H_{1}

_{2}. The effect of applying H_{2}is to replace the goal P 2 with the goal P 1. Then, again, H_{2}has been applied, changing the goal P 1 into P 0. At that point, the goal was exactly the hypothesis H_{1}._{3}at any time. The reason is that auto always tried to use the H_{2}first. So, let's permute the hypotheses H_{2}and H_{3}and see what happens.
Lemma working_of_auto_2 : ∀ (P : nat→Prop),

(* Hypothesis H

(* Hypothesis H

(* Hypothesis H

(* Goal: *) (P 2).

Proof. intros P H

(* Hypothesis H

_{1}: *) (P 0) →(* Hypothesis H

_{3}: *) (∀ k, P (k + 1) → P k) →(* Hypothesis H

_{2}: *) (∀ k, P (k - 1) → P k) →(* Goal: *) (P 2).

Proof. intros P H

_{1}H_{3}H_{2}. (* debug *) eauto. Qed.
This time, the output message suggests that the proof search
investigates many possibilities. If we print the proof term:
Print working_of_auto_2.
we observe that the proof term refers to H
In turns out that the proof goes through the proof obligation P 3,
even though it is not required to do so. The following tree drawing
describes all the goals that eauto has been going through.
The process goes on and on, until backtracking to P 3 and trying
to apply H

_{3}. Thus the proof is not the simplest one, since only H_{2}and H_{1}are needed.|5||4||3||2||1||0| -- below, tabulation indicates the depth [P 2] -> [P 3] -> [P 4] -> [P 5] -> [P 6] -> [P 7] -> [P 5] -> [P 4] -> [P 5] -> [P 3] --> [P 3] -> [P 4] -> [P 5] -> [P 3] -> [P 2] -> [P 3] -> [P 1] -> [P 2] -> [P 3] -> [P 4] -> [P 5] -> [P 3] -> [P 2] -> [P 3] -> [P 1] -> [P 1] -> [P 2] -> [P 3] -> [P 1] -> [P 0] -> !! Done !!The first few lines read as follows. To prove P 2, eauto 5 has first tried to apply H

_{3}, producing the subgoal P 3. To solve it, eauto 4 has tried again to apply H_{3}, producing the goal P 4. Similarly, the search goes through P 5, P 6 and P 7. When reaching P 7, the tactic eauto 0 is called but as it is not allowed to try and apply any lemma, it fails. So, we come back to the goal P 6, and try this time to apply hypothesis H_{2}, producing the subgoal P 5. Here again, eauto 0 fails to solve this goal._{3}three times in a row, going through P 2 and P 1 and P 0. This search tree explains why eauto came up with a proof term starting with an application of H_{3}.## Adding Hints

Lemma nat_le_refl : ∀ (x:nat), x ≤ x.

Proof. apply le_n. Qed.

Local Hint Resolve nat_le_refl : core.

Proof. apply le_n. Qed.

Local Hint Resolve nat_le_refl : core.

A convenient shorthand for adding all the constructors of an
inductive datatype as hints is the command Hint Constructors
mydatatype.
Warning: some lemmas, such as transitivity results, should
not be added as hints as they would very badly affect the
performance of proof search. The description of this problem
and the presentation of a general work-around for transitivity
lemmas appear further on.

## Integration of Automation in Tactics

Ltac auto_star ::= a_new_definition. Observe the use of ::= instead of :=, which indicates that the tactic is being rebound to a new definition. So, the default definition is as follows.

Ltac auto_star ::= try solve [ jauto ].

Nearly all standard Coq tactics and all the tactics from
"LibTactics" can be called with a star symbol. For example, one
can invoke subst×, destruct× H, inverts× H, lets× I: H x,
specializes× H x, and so on... There are two notable exceptions.
The tactic auto× is just another name for the tactic
auto_star. And the tactic apply× H calls eapply H (or the
more powerful applys H if needed), and then calls auto_star.
Note that there is no eapply× H tactic, use apply× H
instead.
In large developments, it can be convenient to use two degrees of
automation. Typically, one would use a fast tactic, like auto,
and a slower but more powerful tactic, like jauto. To allow for
a smooth coexistence of the two form of automation, LibTactics.v
also defines a "tilde" version of tactics, like apply¬ H,
destruct¬ H, subst¬, auto¬ and so on. The meaning of the
tilde symbol is described by the auto_tilde tactic, whose
default implementation is auto.

Ltac auto_tilde ::= auto.

In the examples that follow, only auto_star is needed.
An alternative, possibly more efficient version of auto_star is the
following":
Ltac auto_star ::= try solve eassumption | auto | jauto .
With the above definition, auto_star first tries to solve the
goal using the assumptions; if it fails, it tries using auto,
and if this still fails, then it calls jauto. Even though
jauto is strictly stronger than eassumption and auto, it
makes sense to call these tactics first, because, when the
succeed, they save a lot of time, and when they fail to prove
the goal, they fail very quickly.".

# Example Proofs using Automation

Recall the original proof of the determinism lemma for the IMP
language, shown below.

Theorem ceval_deterministic: ∀ c st st

st =[ c ]=> st

st =[ c ]=> st

st

Proof.

intros c st st

generalize dependent st

induction E

- (* E_Skip *) reflexivity.

- (* E_Asgn *) reflexivity.

- (* E_Seq *)

assert (st' = st'0) as EQ

{ (* Proof of assertion *) apply IHE1_1; assumption. }

subst st'0.

apply IHE1_2. assumption.

(* E_IfTrue *)

- (* b

apply IHE1. assumption.

- (* b

rewrite H in H

(* E_IfFalse *)

- (* b

rewrite H in H

- (* b

apply IHE1. assumption.

(* E_WhileFalse *)

- (* b

reflexivity.

- (* b

rewrite H in H

(* E_WhileTrue *)

- (* b

rewrite H in H

- (* b

assert (st' = st'0) as EQ

{ (* Proof of assertion *) apply IHE1_1; assumption. }

subst st'0.

apply IHE1_2. assumption.

Qed.

_{1}st_{2},st =[ c ]=> st

_{1}→st =[ c ]=> st

_{2}→st

_{1}= st_{2}.Proof.

intros c st st

_{1}st_{2}E_{1}E_{2}.generalize dependent st

_{2}.induction E

_{1}; intros st_{2}E_{2}; inversion E_{2}; subst.- (* E_Skip *) reflexivity.

- (* E_Asgn *) reflexivity.

- (* E_Seq *)

assert (st' = st'0) as EQ

_{1}.{ (* Proof of assertion *) apply IHE1_1; assumption. }

subst st'0.

apply IHE1_2. assumption.

(* E_IfTrue *)

- (* b

_{1}reduces to true *)apply IHE1. assumption.

- (* b

_{1}reduces to false (contradiction) *)rewrite H in H

_{5}. inversion H_{5}.(* E_IfFalse *)

- (* b

_{1}reduces to true (contradiction) *)rewrite H in H

_{5}. inversion H_{5}.- (* b

_{1}reduces to false *)apply IHE1. assumption.

(* E_WhileFalse *)

- (* b

_{1}reduces to true *)reflexivity.

- (* b

_{1}reduces to false (contradiction) *)rewrite H in H

_{2}. inversion H_{2}.(* E_WhileTrue *)

- (* b

_{1}reduces to true (contradiction) *)rewrite H in H

_{4}. inversion H_{4}.- (* b

_{1}reduces to false *)assert (st' = st'0) as EQ

_{1}.{ (* Proof of assertion *) apply IHE1_1; assumption. }

subst st'0.

apply IHE1_2. assumption.

Qed.

Exercise: rewrite this proof using auto whenever possible.
(The solution uses auto 9 times.)

Theorem ceval_deterministic': ∀ c st st

st =[ c ]=> st

st =[ c ]=> st

st

Proof.

(* FILL IN HERE *) admit.

Admitted.

_{1}st_{2},st =[ c ]=> st

_{1}→st =[ c ]=> st

_{2}→st

_{1}= st_{2}.Proof.

(* FILL IN HERE *) admit.

Admitted.

In fact, using automation is not just a matter of calling auto
in place of one or two other tactics. Using automation is about
rethinking the organization of sequences of tactics so as to
minimize the effort involved in writing and maintaining the proof.
This process is eased by the use of the tactics from
LibTactics.v. So, before trying to optimize the way automation
is used, let's first rewrite the proof of determinism:

- use introv H instead of intros x H,
- use gen x instead of generalize dependent x,
- use inverts H instead of inversion H; subst,
- use tryfalse to handle contradictions, and get rid of
the cases where beval st b
_{1}= true and beval st b_{1}= false both appear in the context.

Theorem ceval_deterministic'': ∀ c st st

st =[ c ]=> st

st =[ c ]=> st

st

Proof.

introv E

induction E

- auto.

- auto.

- assert (st' = st'0). auto. subst. auto.

- auto.

- auto.

- auto.

- assert (st' = st'0). auto. subst. auto.

Qed.

_{1}st_{2},st =[ c ]=> st

_{1}→st =[ c ]=> st

_{2}→st

_{1}= st_{2}.Proof.

introv E

_{1}E_{2}. gen st_{2}.induction E

_{1}; intros; inverts E_{2}; tryfalse.- auto.

- auto.

- assert (st' = st'0). auto. subst. auto.

- auto.

- auto.

- auto.

- assert (st' = st'0). auto. subst. auto.

Qed.

To obtain a nice clean proof script, we have to remove the calls
assert (st' = st'0). Such a tactic call is not nice
because it refers to some variables whose name has been
automatically generated. This kind of tactics tend to be very
brittle. The tactic assert (st' = st'0) is used to assert the
conclusion that we want to derive from the induction
hypothesis. So, rather than stating this conclusion explicitly, we
are going to ask Coq to instantiate the induction hypothesis,
using automation to figure out how to instantiate it. The tactic
forwards, described in LibTactics.v precisely helps with
instantiating a fact. So, let's see how it works out on our
example.

Theorem ceval_deterministic''': ∀ c st st

st =[ c ]=> st

st =[ c ]=> st

st

Proof.

(* Let's replay the proof up to the assert tactic. *)

introv E

induction E

- auto.

- auto.

(* We duplicate the goal for comparing different proofs. *)

- dup 4.

(* The old proof: *)

+ assert (st' = st'0). apply IHE1_1. apply H

(* produces H: st' = st'0. *) skip.

(* The new proof, without automation: *)

+ forwards: IHE1_1. apply H

(* produces H: st' = st'0. *) skip.

(* The new proof, with automation: *)

+ forwards: IHE1_1. eauto.

(* produces H: st' = st'0. *) skip.

(* The new proof, with integrated automation: *)

+ forwards*: IHE1_1.

(* produces H: st' = st'0. *) skip.

Abort.

_{1}st_{2},st =[ c ]=> st

_{1}→st =[ c ]=> st

_{2}→st

_{1}= st_{2}.Proof.

(* Let's replay the proof up to the assert tactic. *)

introv E

_{1}E_{2}. gen st_{2}.induction E

_{1}; intros; inverts E_{2}; tryfalse.- auto.

- auto.

(* We duplicate the goal for comparing different proofs. *)

- dup 4.

(* The old proof: *)

+ assert (st' = st'0). apply IHE1_1. apply H

_{1}.(* produces H: st' = st'0. *) skip.

(* The new proof, without automation: *)

+ forwards: IHE1_1. apply H

_{1}.(* produces H: st' = st'0. *) skip.

(* The new proof, with automation: *)

+ forwards: IHE1_1. eauto.

(* produces H: st' = st'0. *) skip.

(* The new proof, with integrated automation: *)

+ forwards*: IHE1_1.

(* produces H: st' = st'0. *) skip.

Abort.

To polish the proof script, it remains to factorize the calls
to auto, using the star symbol. The proof of determinism can then
be rewritten in just 4 lines, including no more than 10 tactics.

Theorem ceval_deterministic'''': ∀ c st st

st =[ c ]=> st

st =[ c ]=> st

st

Proof.

introv E

induction E

- forwards*: IHE1_1. subst×.

- forwards*: IHE1_1. subst×.

Qed.

End DeterministicImp.

_{1}st_{2},st =[ c ]=> st

_{1}→st =[ c ]=> st

_{2}→st

_{1}= st_{2}.Proof.

introv E

_{1}E_{2}. gen st_{2}.induction E

_{1}; intros; inverts× E_{2}; tryfalse.- forwards*: IHE1_1. subst×.

- forwards*: IHE1_1. subst×.

Qed.

End DeterministicImp.

## Preservation for STLC

Set Warnings "-notation-overridden,-parsing,-deprecated-hint-without-locality".

From PLF Require StlcProp.

Module PreservationProgressStlc.

Import Stlc.

Import StlcProp.

Import STLC.

Import STLCProp.

From PLF Require StlcProp.

Module PreservationProgressStlc.

Import Stlc.

Import StlcProp.

Import STLC.

Import STLCProp.

Consider the proof of perservation of STLC, shown below.
This proof already uses eauto through the triple-dot
mechanism.

Theorem preservation : ∀ t t' T,

empty ⊢ t \in T →

t --> t' →

empty ⊢ t' \in T.

Proof with eauto.

intros t t' T HT. generalize dependent t'.

remember empty as Gamma.

induction HT;

intros t' HE; subst;

try solve [inversion HE; subst; auto].

- (* T_App *)

inversion HE; subst...

(* Most of the cases are immediate by induction,

and eauto takes care of them *)

+ (* ST_AppAbs *)

apply substitution_preserves_typing with T

inversion HT

Qed.

empty ⊢ t \in T →

t --> t' →

empty ⊢ t' \in T.

Proof with eauto.

intros t t' T HT. generalize dependent t'.

remember empty as Gamma.

induction HT;

intros t' HE; subst;

try solve [inversion HE; subst; auto].

- (* T_App *)

inversion HE; subst...

(* Most of the cases are immediate by induction,

and eauto takes care of them *)

+ (* ST_AppAbs *)

apply substitution_preserves_typing with T

_{2}...inversion HT

_{1}...Qed.

Exercise: rewrite this proof using tactics from LibTactics
and calling automation using the star symbol rather than the
triple-dot notation. More precisely, make use of the tactics
inverts× and applys× to call auto× after a call to
inverts or to applys.

Theorem preservation' : ∀ t t' T,

empty ⊢ t \in T →

t --> t' →

empty ⊢ t' \in T.

Proof.

(* FILL IN HERE *) admit.

Admitted.

empty ⊢ t \in T →

t --> t' →

empty ⊢ t' \in T.

Proof.

(* FILL IN HERE *) admit.

Admitted.

Theorem progress : ∀ t T,

empty ⊢ t \in T →

value t ∨ ∃ t', t --> t'.

Proof with eauto.

intros t T Ht.

remember empty as Gamma.

induction Ht; subst Gamma; auto.

(* auto solves all three cases in which t is a value *)

- (* T_Var *)

(* contradictory: variables cannot be typed in an

empty context *)

discriminate H.

- (* T_App *)

(* t = t

value or steps... *)

right. destruct IHHt1...

+ (* t

destruct IHHt2...

× (* t

eapply canonical_forms_fun in Ht

destruct Ht

∃ (<{ [x:=t

× (* t

destruct H

+ (* t

destruct H as [t

- (* T_If *)

right. destruct IHHt1...

+ (* t

destruct (canonical_forms_bool t

+ (* t

destruct H as [t

Qed.

empty ⊢ t \in T →

value t ∨ ∃ t', t --> t'.

Proof with eauto.

intros t T Ht.

remember empty as Gamma.

induction Ht; subst Gamma; auto.

(* auto solves all three cases in which t is a value *)

- (* T_Var *)

(* contradictory: variables cannot be typed in an

empty context *)

discriminate H.

- (* T_App *)

(* t = t

_{1}t_{2}. Proceed by cases on whether t_{1}is avalue or steps... *)

right. destruct IHHt1...

+ (* t

_{1}is a value *)destruct IHHt2...

× (* t

_{2}is also a value *)eapply canonical_forms_fun in Ht

_{1}; [|assumption].destruct Ht

_{1}as [x [t_{0}H_{1}]]. subst.∃ (<{ [x:=t

_{2}]t_{0}}>)...× (* t

_{2}steps *)destruct H

_{0}as [t_{2}' Hstp]. ∃ (<{t_{1}t_{2}'}>)...+ (* t

_{1}steps *)destruct H as [t

_{1}' Hstp]. ∃ (<{t_{1}' t_{2}}>)...- (* T_If *)

right. destruct IHHt1...

+ (* t

_{1}is a value *)destruct (canonical_forms_bool t

_{1}); subst; eauto.+ (* t

_{1}also steps *)destruct H as [t

_{1}' Hstp]. ∃ <{if t_{1}' then t_{2}else t_{3}}>...Qed.

Exercise: optimize the above proof.
Hint: make use of destruct× and inverts×.
The solution fits on 10 short lines.

Theorem progress' : ∀ t T,

empty ⊢ t \in T →

value t ∨ ∃ t', t --> t'.

Proof.

(* FILL IN HERE *) admit.

Admitted.

End PreservationProgressStlc.

empty ⊢ t \in T →

value t ∨ ∃ t', t --> t'.

Proof.

(* FILL IN HERE *) admit.

Admitted.

End PreservationProgressStlc.

Consider the proof relating a small-step reduction judgment
to a big-step reduction judgment.

Theorem multistep__eval : ∀ t v,

normal_form_of t v → ∃ n, v = C n ∧ t ==> n.

Proof.

intros t v Hnorm.

unfold normal_form_of in Hnorm.

inversion Hnorm as [Hs Hnf]; clear Hnorm.

rewrite nf_same_as_value in Hnf. inversion Hnf. clear Hnf.

∃ n. split. reflexivity.

induction Hs; subst.

- (* multi_refl *)

apply E_Const.

- (* multi_step *)

eapply step__eval. eassumption. apply IHHs. reflexivity.

Qed.

normal_form_of t v → ∃ n, v = C n ∧ t ==> n.

Proof.

intros t v Hnorm.

unfold normal_form_of in Hnorm.

inversion Hnorm as [Hs Hnf]; clear Hnorm.

rewrite nf_same_as_value in Hnf. inversion Hnf. clear Hnf.

∃ n. split. reflexivity.

induction Hs; subst.

- (* multi_refl *)

apply E_Const.

- (* multi_step *)

eapply step__eval. eassumption. apply IHHs. reflexivity.

Qed.

Our goal is to optimize the above proof. It is generally
easier to isolate inductions into separate lemmas. So,
we are going to first prove an intermediate result
that consists of the judgment over which the induction
is being performed.
Exercise: prove the following result, using tactics
introv, induction and subst, and apply×.
The solution fits on 3 short lines.

Theorem multistep_eval_ind : ∀ t v,

t -->* v → ∀ n, C n = v → t ==> n.

Proof.

(* FILL IN HERE *) admit.

Admitted.

t -->* v → ∀ n, C n = v → t ==> n.

Proof.

(* FILL IN HERE *) admit.

Admitted.

Exercise: using the lemma above, simplify the proof of
the result multistep__eval. You should use the tactics
introv, inverts, split× and apply×.
The solution fits on 2 lines.

Theorem multistep__eval' : ∀ t v,

normal_form_of t v → ∃ n, v = C n ∧ t ==> n.

Proof.

(* FILL IN HERE *) admit.

Admitted.

normal_form_of t v → ∃ n, v = C n ∧ t ==> n.

Proof.

(* FILL IN HERE *) admit.

Admitted.

If we try to combine the two proofs into a single one,
we will likely fail, because of a limitation of the
induction tactic. Indeed, this tactic loses
information when applied to a property whose arguments
are not reduced to variables, such as t -->* (C n).
You will thus need to use the more powerful tactic called
dependent induction. (This tactic is available only after
importing the Program library, as we did above.)
Exercise: prove the lemma multistep__eval without invoking
the lemma multistep_eval_ind, that is, by inlining the proof
by induction involved in multistep_eval_ind, using the
tactic dependent induction instead of induction.
The solution fits on 6 lines.

Theorem multistep__eval'' : ∀ t v,

normal_form_of t v → ∃ n, v = C n ∧ t ==> n.

Proof.

(* FILL IN HERE *) admit.

Admitted.

End Semantics.

normal_form_of t v → ∃ n, v = C n ∧ t ==> n.

Proof.

(* FILL IN HERE *) admit.

Admitted.

End Semantics.

From Coq Require Import Lia.

From PLF Require Import References.

Require Import Program.

Module PreservationProgressReferences.

Import STLCRef.

Hint Resolve store_weakening extends_refl : core.

From PLF Require Import References.

Require Import Program.

Module PreservationProgressReferences.

Import STLCRef.

Hint Resolve store_weakening extends_refl : core.

The proof of preservation for STLCRef can be found in chapter
References. The optimized proof script is more than twice
shorter. The following material explains how to build the
optimized proof script. The resulting optimized proof script for
the preservation theorem appears afterwards.

Theorem preservation : ∀ ST t t' T st st',

empty ; ST ⊢ t \in T →

store_well_typed ST st →

t / st --> t' / st' →

∃ ST',

extends ST' ST ∧

empty ; ST' ⊢ t' \in T ∧

store_well_typed ST' st'.

Proof.

(* old: Proof with eauto using store_weakening, extends_refl.

new: Proof., and the two lemmas are registered as hints

before the proof of the lemma, possibly inside a section in

order to restrict the scope of the hints. *)

remember empty as Gamma.

introv Ht. gen t'.

induction Ht; introv HST Hstep;

(* old: subst; try solve_by_invert; inversion Hstep; subst; try (eauto using store_weakening, extends_refl)

new: subst Gamma; inverts Hstep; eauto.

We want to be more precise on what exactly we substitute,

and we do not want to call try solve_by_invert which

is way too slow. *)

subst Gamma; inverts Hstep; eauto.

(* T_App *)

- (* ST_AppAbs *)

(* old:

exists ST.

inversion Ht

split; try split... eapply substitution_preserves_typing... *)

(* new: we use inverts in place of inversion and splits to

split the conjunction, and applys× in place of eapply... *)

∃ ST. inverts Ht

- (* ST_App1 *)

(* old:

eapply IHHt1 in H

destruct H

exists ST'... *)

(* new: The tactic eapply IHHt1 in H

But H

there eauto can figure this out on its own. The tactic

forwards is used to instantiate all the arguments of IHHt1,

producing existential variables and subgoals when needed. *)

forwards: IHHt1. eauto. eauto. eauto.

(* At this point, we need to decompose the hypothesis H that has

just been created by forwards. This is done by the first part

of the preprocessing phase of jauto. *)

jauto_set_hyps; intros.

(* It remains to decompose the goal, which is done by the second

part of the preprocessing phase of jauto. *)

jauto_set_goal; intros.

(* All the subgoals produced can then be solved by eauto. *)

eauto. eauto. eauto.

- (* ST_App2 *)

(* old:

eapply IHHt2 in H

destruct H

exists ST'... *)

(* new: this time, we need to call forwards on IHHt2,

and we call jauto right away, by writing forwards×,

proving the goal in a single tactic! *)

forwards*: IHHt2.

(* The same trick works for many of the other subgoals. *)

- forwards*: IHHt.

- forwards*: IHHt.

- forwards*: IHHt1.

- forwards*: IHHt2.

- forwards*: IHHt1.

- (* T_Ref *)

+ (* ST_RefValue *)

(* old:

exists (ST ++ T

inversion HST; subst.

split.

{ apply extends_app. }

split.

{ replace <{ Ref T

with <{ Ref {store_Tlookup (length st) (ST ++ T

{ apply T_Loc.

rewrite <- H. rewrite app_length, add_comm. simpl. lia. }

unfold store_Tlookup. rewrite <- H. rewrite nth_eq_last.

reflexivity. }

apply store_well_typed_app; assumption. *)

(* new: In this proof case, we need to perform an inversion

without removing the hypothesis. The tactic inverts keep

serves exactly this purpose. *)

∃ (ST ++ T

(* The proof of the first subgoal needs no change *)

apply extends_app.

(* For the second subgoal, we use the tactic applys_eq to avoid

a manual replace before T_loc can be applied. *)

applys_eq T_Loc.

(* The next proof case is hard to polish because it relies on the

lemma app_nth2 whose statement is not automation-friendly.

We'll come back to this proof case further on. *)

unfold store_Tlookup. rewrite <- H. rewrite× app_nth2.

(* Last, we replace apply ..; assumption with apply× .. *)

rewrite minus_diag. simpl. reflexivity.

(* To justify the inequality, there is no need to call rewrite <- H,

because the tactic lia is able to exploit H on its own.

So, only the rewriting of app_length and the call to the

tactic lia remain, with a call to simpl to unfold the

definition of app. *)

rewrite app_length. simpl. lia.

apply× store_well_typed_app.

- forwards*: IHHt.

- (* T_Deref *)

+ (* ST_DerefLoc *)

(* old:

exists ST. split; try split...

destruct HST as _ Hsty.

replace T

apply Hsty...

inversion Ht; subst... *)

(* new: we start by calling ∃ ST and splits×. *)

∃ ST. splits×.

(* new: we replace destruct HST as [_ Hsty] by the following *)

lets [_ Hsty]: HST.

(* new: then we use the tactic applys_eq to avoid the need to

perform a manual replace before applying Hsty. *)

applys_eq× Hsty.

(* new: we then can call inverts in place of inversion;subst *)

inverts× Ht.

- forwards*: IHHt.

- (* T_Assign *)

+ (* ST_Assign *)

(* old:

exists ST. split; try split...

eapply assign_pres_store_typing...

inversion Ht

(* new: simply using nicer tactics *)

∃ ST. splits×. applys× assign_pres_store_typing. inverts× Ht

- forwards*: IHHt1.

- forwards*: IHHt2.

Qed.

empty ; ST ⊢ t \in T →

store_well_typed ST st →

t / st --> t' / st' →

∃ ST',

extends ST' ST ∧

empty ; ST' ⊢ t' \in T ∧

store_well_typed ST' st'.

Proof.

(* old: Proof with eauto using store_weakening, extends_refl.

new: Proof., and the two lemmas are registered as hints

before the proof of the lemma, possibly inside a section in

order to restrict the scope of the hints. *)

remember empty as Gamma.

introv Ht. gen t'.

induction Ht; introv HST Hstep;

(* old: subst; try solve_by_invert; inversion Hstep; subst; try (eauto using store_weakening, extends_refl)

new: subst Gamma; inverts Hstep; eauto.

We want to be more precise on what exactly we substitute,

and we do not want to call try solve_by_invert which

is way too slow. *)

subst Gamma; inverts Hstep; eauto.

(* T_App *)

- (* ST_AppAbs *)

(* old:

exists ST.

inversion Ht

_{1}; subst.split; try split... eapply substitution_preserves_typing... *)

(* new: we use inverts in place of inversion and splits to

split the conjunction, and applys× in place of eapply... *)

∃ ST. inverts Ht

_{1}. splits×. applys× substitution_preserves_typing.- (* ST_App1 *)

(* old:

eapply IHHt1 in H

_{0}...destruct H

_{0}as ST' [Hext [Hty Hsty]].exists ST'... *)

(* new: The tactic eapply IHHt1 in H

_{0}... applies IHHt1 to H_{0}.But H

_{0}is only thing that IHHt1 could be applied to, sothere eauto can figure this out on its own. The tactic

forwards is used to instantiate all the arguments of IHHt1,

producing existential variables and subgoals when needed. *)

forwards: IHHt1. eauto. eauto. eauto.

(* At this point, we need to decompose the hypothesis H that has

just been created by forwards. This is done by the first part

of the preprocessing phase of jauto. *)

jauto_set_hyps; intros.

(* It remains to decompose the goal, which is done by the second

part of the preprocessing phase of jauto. *)

jauto_set_goal; intros.

(* All the subgoals produced can then be solved by eauto. *)

eauto. eauto. eauto.

- (* ST_App2 *)

(* old:

eapply IHHt2 in H

_{5}...destruct H

_{5}as ST' [Hext [Hty Hsty]].exists ST'... *)

(* new: this time, we need to call forwards on IHHt2,

and we call jauto right away, by writing forwards×,

proving the goal in a single tactic! *)

forwards*: IHHt2.

(* The same trick works for many of the other subgoals. *)

- forwards*: IHHt.

- forwards*: IHHt.

- forwards*: IHHt1.

- forwards*: IHHt2.

- forwards*: IHHt1.

- (* T_Ref *)

+ (* ST_RefValue *)

(* old:

exists (ST ++ T

_{1}::nil).inversion HST; subst.

split.

{ apply extends_app. }

split.

{ replace <{ Ref T

_{1}}>with <{ Ref {store_Tlookup (length st) (ST ++ T

_{1}::nil)} }>.{ apply T_Loc.

rewrite <- H. rewrite app_length, add_comm. simpl. lia. }

unfold store_Tlookup. rewrite <- H. rewrite nth_eq_last.

reflexivity. }

apply store_well_typed_app; assumption. *)

(* new: In this proof case, we need to perform an inversion

without removing the hypothesis. The tactic inverts keep

serves exactly this purpose. *)

∃ (ST ++ T

_{1}::nil). inverts keep HST. splits.(* The proof of the first subgoal needs no change *)

apply extends_app.

(* For the second subgoal, we use the tactic applys_eq to avoid

a manual replace before T_loc can be applied. *)

applys_eq T_Loc.

(* The next proof case is hard to polish because it relies on the

lemma app_nth2 whose statement is not automation-friendly.

We'll come back to this proof case further on. *)

unfold store_Tlookup. rewrite <- H. rewrite× app_nth2.

(* Last, we replace apply ..; assumption with apply× .. *)

rewrite minus_diag. simpl. reflexivity.

(* To justify the inequality, there is no need to call rewrite <- H,

because the tactic lia is able to exploit H on its own.

So, only the rewriting of app_length and the call to the

tactic lia remain, with a call to simpl to unfold the

definition of app. *)

rewrite app_length. simpl. lia.

apply× store_well_typed_app.

- forwards*: IHHt.

- (* T_Deref *)

+ (* ST_DerefLoc *)

(* old:

exists ST. split; try split...

destruct HST as _ Hsty.

replace T

_{1}with (store_Tlookup l ST).apply Hsty...

inversion Ht; subst... *)

(* new: we start by calling ∃ ST and splits×. *)

∃ ST. splits×.

(* new: we replace destruct HST as [_ Hsty] by the following *)

lets [_ Hsty]: HST.

(* new: then we use the tactic applys_eq to avoid the need to

perform a manual replace before applying Hsty. *)

applys_eq× Hsty.

(* new: we then can call inverts in place of inversion;subst *)

inverts× Ht.

- forwards*: IHHt.

- (* T_Assign *)

+ (* ST_Assign *)

(* old:

exists ST. split; try split...

eapply assign_pres_store_typing...

inversion Ht

_{1}; subst... *)(* new: simply using nicer tactics *)

∃ ST. splits×. applys× assign_pres_store_typing. inverts× Ht

_{1}.- forwards*: IHHt1.

- forwards*: IHHt2.

Qed.

Let's come back to the proof case that was hard to optimize.
The difficulty comes from the statement of nth_eq_last, which
takes the form nth (length l) (l ++ x::nil) d = x. This lemma is
hard to exploit because its first argument, length l, mentions
a list l that has to be exactly the same as the l occuring in
l ++ x::nil. In practice, the first argument is often a natural
number n that is provably equal to length l yet that is not
syntactically equal to length l. There is a simple fix for
making nth_eq_last easy to apply: introduce the intermediate
variable n explicitly, so that the goal becomes
nth n (l ++ x::nil) d = x, with a premise asserting n = length l.

Lemma nth_eq_last' : ∀ (A : Type) (l : list A) (x d : A) (n : nat),

n = length l → nth n (l ++ x::nil) d = x.

Proof. intros. subst. apply nth_eq_last. Qed.

n = length l → nth n (l ++ x::nil) d = x.

Proof. intros. subst. apply nth_eq_last. Qed.

The proof case for ref from the preservation theorem then
becomes much easier to prove, because rewrite nth_eq_last'
now succeeds.

Lemma preservation_ref : ∀ (st:store) (ST : store_ty) T

length ST = length st →

<{ Ref T

Proof.

intros. dup.

(* A first proof, with an explicit unfold *)

unfold store_Tlookup. rewrite× nth_eq_last'.

(* A second proof, with a call to fequal *)

fequal. symmetry. apply× nth_eq_last'.

Qed.

_{1},length ST = length st →

<{ Ref T

_{1}}> = <{ Ref {store_Tlookup (length st) (ST ++ T_{1}::nil)} }>.Proof.

intros. dup.

(* A first proof, with an explicit unfold *)

unfold store_Tlookup. rewrite× nth_eq_last'.

(* A second proof, with a call to fequal *)

fequal. symmetry. apply× nth_eq_last'.

Qed.

The optimized proof of preservation is summarized next.

Theorem preservation' : ∀ ST t t' T st st',

empty ; ST ⊢ t \in T →

store_well_typed ST st →

t / st --> t' / st' →

∃ ST',

extends ST' ST ∧

empty ; ST' ⊢ t' \in T ∧

store_well_typed ST' st'.

Proof.

remember empty as Gamma.

introv Ht. gen t'.

induction Ht; introv HST Hstep; subst Gamma; inverts Hstep; eauto.

- ∃ ST. inverts Ht

- forwards*: IHHt1.

- forwards*: IHHt2.

- forwards*: IHHt.

- forwards*: IHHt.

- forwards*: IHHt1.

- forwards*: IHHt2.

- forwards*: IHHt1.

- ∃ (ST ++ T

apply extends_app.

applys_eq T_Loc.

unfold store_Tlookup. rewrite× nth_eq_last'.

rewrite app_length. simpl. lia.

apply× store_well_typed_app.

- forwards*: IHHt.

- ∃ ST. splits×. lets [_ Hsty]: HST.

applys_eq× Hsty. inverts× Ht.

- forwards*: IHHt.

- ∃ ST. splits×. applys× assign_pres_store_typing. inverts× Ht

- forwards*: IHHt1.

- forwards*: IHHt2.

Qed.

empty ; ST ⊢ t \in T →

store_well_typed ST st →

t / st --> t' / st' →

∃ ST',

extends ST' ST ∧

empty ; ST' ⊢ t' \in T ∧

store_well_typed ST' st'.

Proof.

remember empty as Gamma.

introv Ht. gen t'.

induction Ht; introv HST Hstep; subst Gamma; inverts Hstep; eauto.

- ∃ ST. inverts Ht

_{1}. splits×. applys× substitution_preserves_typing.- forwards*: IHHt1.

- forwards*: IHHt2.

- forwards*: IHHt.

- forwards*: IHHt.

- forwards*: IHHt1.

- forwards*: IHHt2.

- forwards*: IHHt1.

- ∃ (ST ++ T

_{1}::nil). inverts keep HST. splits.apply extends_app.

applys_eq T_Loc.

unfold store_Tlookup. rewrite× nth_eq_last'.

rewrite app_length. simpl. lia.

apply× store_well_typed_app.

- forwards*: IHHt.

- ∃ ST. splits×. lets [_ Hsty]: HST.

applys_eq× Hsty. inverts× Ht.

- forwards*: IHHt.

- ∃ ST. splits×. applys× assign_pres_store_typing. inverts× Ht

_{1}.- forwards*: IHHt1.

- forwards*: IHHt2.

Qed.

## Progress for STLCRef

Theorem progress : ∀ ST t T st,

empty ; ST ⊢ t \in T →

store_well_typed ST st →

(value t ∨ ∃ t' st', t / st --> t' / st').

Proof.

introv Ht HST. remember empty as Gamma.

induction Ht; subst Gamma; tryfalse; try solve [left*].

- right. destruct× IHHt1 as [K|].

inverts K; inverts Ht

destruct× IHHt2.

- right. destruct× IHHt as [K|].

inverts K; try solve [inverts Ht]. eauto.

- right. destruct× IHHt as [K|].

inverts K; try solve [inverts Ht]. eauto.

- right. destruct× IHHt1 as [K|].

inverts K; try solve [inverts Ht

destruct× IHHt2 as [M|].

inverts M; try solve [inverts Ht

- right. destruct× IHHt1 as [K|].

inverts K; try solve [inverts Ht

- right. destruct× IHHt.

- right. destruct× IHHt as [K|].

inverts K; inverts Ht as M.

inverts HST as N. rewrite× N in M.

- right. destruct× IHHt1 as [K|].

destruct× IHHt2.

inverts K; inverts Ht

inverts HST as N. rewrite× N in M.

Qed.

End PreservationProgressReferences.

empty ; ST ⊢ t \in T →

store_well_typed ST st →

(value t ∨ ∃ t' st', t / st --> t' / st').

Proof.

introv Ht HST. remember empty as Gamma.

induction Ht; subst Gamma; tryfalse; try solve [left*].

- right. destruct× IHHt1 as [K|].

inverts K; inverts Ht

_{1}.destruct× IHHt2.

- right. destruct× IHHt as [K|].

inverts K; try solve [inverts Ht]. eauto.

- right. destruct× IHHt as [K|].

inverts K; try solve [inverts Ht]. eauto.

- right. destruct× IHHt1 as [K|].

inverts K; try solve [inverts Ht

_{1}].destruct× IHHt2 as [M|].

inverts M; try solve [inverts Ht

_{2}]. eauto.- right. destruct× IHHt1 as [K|].

inverts K; try solve [inverts Ht

_{1}]. destruct× n.- right. destruct× IHHt.

- right. destruct× IHHt as [K|].

inverts K; inverts Ht as M.

inverts HST as N. rewrite× N in M.

- right. destruct× IHHt1 as [K|].

destruct× IHHt2.

inverts K; inverts Ht

_{1}as M.inverts HST as N. rewrite× N in M.

Qed.

End PreservationProgressReferences.

Consider the inversion lemma for typing judgment
of abstractions in a type system with subtyping.

Lemma abs_arrow : ∀ x S

empty ⊢ \x:S

T

∧ (x ⊢> S

_{1}s_{2}T_{1}T_{2},empty ⊢ \x:S

_{1},s_{2}\in (T_{1}→T_{2}) →T

_{1}<: S_{1}∧ (x ⊢> S

_{1}; empty) ⊢ s_{2}\in T_{2}.
Proof with eauto.

intros x S

apply typing_inversion_abs in Hty.

destruct Hty as [S

apply sub_inversion_arrow in Hsub.

destruct Hsub as [U

injection Heq as Heq; subst...

Qed.

intros x S

_{1}s_{2}T_{1}T_{2}Hty.apply typing_inversion_abs in Hty.

destruct Hty as [S

_{2}[Hsub Hty1]].apply sub_inversion_arrow in Hsub.

destruct Hsub as [U

_{1}[U_{2}[Heq [Hsub1 Hsub2]]]].injection Heq as Heq; subst...

Qed.

Exercise: optimize the proof script, using
introv, lets and inverts×. In particular,
you will find it useful to replace the pattern
apply K in H. destruct H as I with lets I: K H.
The solution fits on 3 lines.

Lemma abs_arrow' : ∀ x S

empty ⊢ \x:S

T

∧ (x ⊢> S

Proof.

(* FILL IN HERE *) admit.

Admitted.

End SubtypingInversion.

_{1}s_{2}T_{1}T_{2},empty ⊢ \x:S

_{1},s_{2}\in (T_{1}→T_{2}) →T

_{1}<: S_{1}∧ (x ⊢> S

_{1}; empty) ⊢ s_{2}\in T_{2}.Proof.

(* FILL IN HERE *) admit.

Admitted.

End SubtypingInversion.

## Stating Lemmas in the Right Way

Lemma order_matters_1 : ∀ (P : nat→Prop),

(∀ n m, P m → m ≠ 0 → P n) →

P 2 →

P 1.

Proof.

eauto. (* Success *)

(* The proof: intros P H K. eapply H. apply K. auto. *)

Qed.

Lemma order_matters_2 : ∀ (P : nat→Prop),

(∀ n m, m ≠ 0 → P m → P n) →

P 5 →

P 1.

Proof.

eauto. (* Failure *)

(* To understand why, let us replay the previous proof *)

intros P H K.

eapply H.

(* The application of eapply has left two subgoals,

?X ≠ 0 and P ?X, where ?X is an existential variable. *)

(* Solving the first subgoal is easy for eauto: it suffices

to instantiate ?X as the value 1, which is the simplest

value that satisfies ?X ≠ 0. *)

eauto.

(* But then the second goal becomes P 1, which is where we

started from. So, eauto gets stuck at this point. *)

Abort.

(∀ n m, P m → m ≠ 0 → P n) →

P 2 →

P 1.

Proof.

eauto. (* Success *)

(* The proof: intros P H K. eapply H. apply K. auto. *)

Qed.

Lemma order_matters_2 : ∀ (P : nat→Prop),

(∀ n m, m ≠ 0 → P m → P n) →

P 5 →

P 1.

Proof.

eauto. (* Failure *)

(* To understand why, let us replay the previous proof *)

intros P H K.

eapply H.

(* The application of eapply has left two subgoals,

?X ≠ 0 and P ?X, where ?X is an existential variable. *)

(* Solving the first subgoal is easy for eauto: it suffices

to instantiate ?X as the value 1, which is the simplest

value that satisfies ?X ≠ 0. *)

eauto.

(* But then the second goal becomes P 1, which is where we

started from. So, eauto gets stuck at this point. *)

Abort.

It is very important to understand that the hypothesis ∀ n
m, P m → m ≠ 0 → P n is eauto-friendly, whereas ∀ n m, m
≠ 0 → P m → P n really isn't. Guessing a value of m for
which P m holds and then checking that m ≠ 0 holds works well
because there are few values of m for which P m holds. So, it
is likely that eauto comes up with the right one. On the other
hand, guessing a value of m for which m ≠ 0 and then checking
that P m holds does not work well, because there are many values
of m that satisfy m ≠ 0 but not P m.

## Unfolding of Definitions During Proof-Search

Proving that myFact under the assumption that P x holds for
any x should be trivial. Yet, auto fails to prove it unless we
unfold the definition of myFact explicitly.

Lemma demo_hint_unfold_goal_1 :

(∀ x, P x) →

myFact.

Proof.

auto. (* Proof search doesn't know what to do, *)

unfold myFact. auto. (* unless we unfold the definition. *)

Qed.

(∀ x, P x) →

myFact.

Proof.

auto. (* Proof search doesn't know what to do, *)

unfold myFact. auto. (* unless we unfold the definition. *)

Qed.

To automate the unfolding of definitions that appear as proof
obligation, one can use the command Hint Unfold myFact to tell
Coq that it should always try to unfold myFact when myFact
appears in the goal.

This time, automation is able to see through the definition
of myFact.

However, the Hint Unfold mechanism only works for unfolding
definitions that appear in the goal. In general, proof search does
not unfold definitions from the context. For example, assume we
want to prove that P 3 holds under the assumption that True →
myFact.

Lemma demo_hint_unfold_context_1 :

(True → myFact) →

P 3.

Proof.

intros.

auto. (* fails *)

unfold myFact in ×. auto. (* succeeds *)

Qed.

(True → myFact) →

P 3.

Proof.

intros.

auto. (* fails *)

unfold myFact in ×. auto. (* succeeds *)

Qed.

There is actually one exception to the previous rule: a constant
occuring in an hypothesis is automatically unfolded if the
hypothesis can be directly applied to the current goal. For example,
auto can prove myFact → P 3, as illustrated below.

## Automation for Proving Absurd Goals

Equivalently, one could state that a number greater than three is
not less than or equal to 3.

In fact, both statements are equivalent to a third one stating
that x ≤ 3 and x > 3 are contradictory, in the sense that
they imply False.

The following investigation aim at figuring out which of the three
statments is the most convenient with respect to proof
automation. The following material is enclosed inside a Section,
so as to restrict the scope of the hints that we are adding. In
other words, after the end of the section, the hints added within
the section will no longer be active.

Let's try to add the first lemma, le_not_gt, as hint,
and see whether we can prove that the proposition
∃ x, x ≤ 3 ∧ x > 3 is absurd.

Hint Resolve le_not_gt : core.

Lemma demo_auto_absurd_1 :

(∃ x, x ≤ 3 ∧ x > 3) →

False.

Proof.

intros. jauto_set. (* decomposes the assumption *)

(* debug *) eauto. (* does not see that le_not_gt could apply *)

eapply le_not_gt. eauto. eauto.

Qed.

Lemma demo_auto_absurd_1 :

(∃ x, x ≤ 3 ∧ x > 3) →

False.

Proof.

intros. jauto_set. (* decomposes the assumption *)

(* debug *) eauto. (* does not see that le_not_gt could apply *)

eapply le_not_gt. eauto. eauto.

Qed.

The lemma gt_not_le is symmetric to le_not_gt, so it will not
be any better. The third lemma, le_gt_false, is a more useful
hint, because it concludes on False, so proof search will try to
apply it when the current goal is False.

Hint Resolve le_gt_false : core.

Lemma demo_auto_absurd_2 :

(∃ x, x ≤ 3 ∧ x > 3) →

False.

Proof.

dup.

(* detailed version: *)

intros. jauto_set. (* debug *) eauto.

(* short version: *)

jauto.

Qed.

Lemma demo_auto_absurd_2 :

(∃ x, x ≤ 3 ∧ x > 3) →

False.

Proof.

dup.

(* detailed version: *)

intros. jauto_set. (* debug *) eauto.

(* short version: *)

jauto.

Qed.

In summary, a lemma of the form H
That said, one should be careful with adding lemmas whose
conclusion is False as hint. The reason is that whenever
reaching the goal False, the proof search mechanism will
potentially try to apply all the hints whose conclusion is False
before applying the appropriate one.

_{1}→ H_{2}→ False is a much more effective hint than H_{1}→ ¬ H_{2}, even though the two statments are equivalent up to the definition of the negation symbol ¬.
Adding lemmas whose conclusion is False as hint can be, locally,
a very effective solution. However, this approach does not scale
up for global hints. For most practical applications, it is
reasonable to give the name of the lemmas to be exploited for
deriving a contradiction. The tactic false H, provided by
LibTactics serves that purpose: false H replaces the goal
with False and calls eapply H. Its behavior is described next.
Observe that any of the three statements le_not_gt, gt_not_le
or le_gt_false can be used.

Lemma demo_false : ∀ x,

(x ≤ 3) →

(x > 3) →

4 = 5.

Proof.

intros. dup 4.

(* A failed proof: *)

- false. eapply le_gt_false.

+ auto. (* here, auto does not prove ?x ≤ 3 by using H but

by using the lemma le_refl : ∀ x, x ≤ x. *)

(* The second subgoal becomes 3 > 3, which is not provable. *)

+ skip.

(* A correct proof: *)

- false. eapply le_gt_false.

+ eauto. (* here, eauto uses H, as expected, to prove ?x ≤ 3 *)

+ eauto. (* so the second subgoal becomes x > 3 *)

(* The same proof using false: *)

- false le_gt_false. eauto. eauto.

(* The lemmas le_not_gt and gt_not_le work as well *)

- false le_not_gt. eauto. eauto.

Abort.

(x ≤ 3) →

(x > 3) →

4 = 5.

Proof.

intros. dup 4.

(* A failed proof: *)

- false. eapply le_gt_false.

+ auto. (* here, auto does not prove ?x ≤ 3 by using H but

by using the lemma le_refl : ∀ x, x ≤ x. *)

(* The second subgoal becomes 3 > 3, which is not provable. *)

+ skip.

(* A correct proof: *)

- false. eapply le_gt_false.

+ eauto. (* here, eauto uses H, as expected, to prove ?x ≤ 3 *)

+ eauto. (* so the second subgoal becomes x > 3 *)

(* The same proof using false: *)

- false le_gt_false. eauto. eauto.

(* The lemmas le_not_gt and gt_not_le work as well *)

- false le_not_gt. eauto. eauto.

Abort.

In the above example, false le_gt_false; eauto proves the goal,
but false le_gt_false; auto does not, because auto does not
correctly instantiate the existential variable. Note that false×
le_gt_false would not work either, because the star symbol tries
to call auto first. So, there are two possibilities for
completing the proof: either call false le_gt_false; eauto, or
call false× (le_gt_false 3).

## Automation for Transitivity Lemmas

Parameter typ : Type.

Parameter subtype : typ → typ → Prop.

Parameter subtype_refl : ∀ T,

subtype T T.

Parameter subtype_trans : ∀ S T U,

subtype S T → subtype T U → subtype S U.

Parameter subtype : typ → typ → Prop.

Parameter subtype_refl : ∀ T,

subtype T T.

Parameter subtype_trans : ∀ S T U,

subtype S T → subtype T U → subtype S U.

Adding reflexivity as hint is generally a good idea,
so let's add reflexivity of subtyping as hint.

Hint Resolve subtype_refl : core.

Adding transitivity as hint is generally a bad idea. To
understand why, let's add it as hint and see what happens.
Because we cannot remove hints once we've added them, we are going
to open a "Section," so as to restrict the scope of the
transitivity hint to that section.

Now, consider the goal ∀ S T, subtype S T, which clearly has
no hope of being solved. Let's call eauto on this goal.

Lemma transitivity_bad_hint_1 : ∀ S T,

subtype S T.

Proof.

intros. (* debug *) eauto. (* Investigates 106 applications... *)

Abort.

subtype S T.

Proof.

intros. (* debug *) eauto. (* Investigates 106 applications... *)

Abort.

Note that after closing the section, the hint subtype_trans
is no longer active.

In the previous example, the proof search has spent a lot of time
trying to apply transitivity and reflexivity in every possible
way. Its process can be summarized as follows. The first goal is
subtype S T. Since reflexivity does not apply, eauto invokes
transitivity, which produces two subgoals, subtype S ?X and
subtype ?X T. Solving the first subgoal, subtype S ?X, is
straightforward, it suffices to apply reflexivity. This unifies
?X with S. So, the second sugoal, subtype ?X T,
becomes subtype S T, which is exactly what we started from...
The problem with the transitivity lemma is that it is applicable
to any goal concluding on a subtyping relation. Because of this,
eauto keeps trying to apply it even though it most often doesn't
help to solve the goal. So, one should never add a transitivity
lemma as a hint for proof search.
There is a general workaround for having automation to exploit
transitivity lemmas without giving up on efficiency. This workaround
relies on a powerful mechanism called "external hint." This
mechanism allows to manually describe the condition under which
a particular lemma should be tried out during proof search.
For the case of transitivity of subtyping, we are going to tell
Coq to try and apply the transitivity lemma on a goal of the form
subtype S U only when the proof context already contains an
assumption either of the form subtype S T or of the form
subtype T U. In other words, we only apply the transitivity
lemma when there is some evidence that this application might
help. To set up this "external hint," one has to write the
following.

Hint Extern 1 (subtype ?S ?U) ⇒

match goal with

| H: subtype S ?T ⊢ _ ⇒ apply (@subtype_trans S T U)

| H: subtype ?T U ⊢ _ ⇒ apply (@subtype_trans S T U)

end : core.

match goal with

| H: subtype S ?T ⊢ _ ⇒ apply (@subtype_trans S T U)

| H: subtype ?T U ⊢ _ ⇒ apply (@subtype_trans S T U)

end : core.

This hint declaration can be understood as follows.
Note: the same external hint can be reused for any other transitive
relation, simply by renaming subtype into the name of that relation.
Let us see an example illustrating how the hint works.

- "Hint Extern" introduces the hint.
- The number "1" corresponds to a priority for proof search. It doesn't matter so much what priority is used in practice.
- The pattern subtype ?S ?U describes the kind of goal on which the pattern should apply. The question marks are used to indicate that the variables ?S and ?U should be bound to some value in the rest of the hint description.
- The construction match goal with ... end tries to recognize patterns in the goal, or in the proof context, or both.
- The first pattern is H: subtype S ?T ⊢ _. It indices that the context should contain an hypothesis H of type subtype S ?T, where S has to be the same as in the goal, and where ?T can have any value.
- The symbol ⊢ _ at the end of H: subtype S ?T ⊢ _ indicates that we do not impose further condition on how the proof obligation has to look like.
- The branch ⇒ apply (@subtype_trans S T U) that follows indicates that if the goal has the form subtype S U and if there exists an hypothesis of the form subtype S T, then we should try and apply transitivity lemma instantiated on the arguments S, T and U. (Note: the symbol @ in front of subtype_trans is only actually needed when the "Implicit Arguments" feature is activated.)
- The other branch, which corresponds to an hypothesis of the form H: subtype ?T U is symmetrical.

Lemma transitivity_workaround_1 : ∀ T

subtype T

subtype T

subtype T

subtype T

Proof.

intros. (* debug *) eauto. (* The trace shows the external hint being used *)

Qed.

_{1}T_{2}T_{3}T_{4},subtype T

_{1}T_{2}→subtype T

_{2}T_{3}→subtype T

_{3}T_{4}→subtype T

_{1}T_{4}.Proof.

intros. (* debug *) eauto. (* The trace shows the external hint being used *)

Qed.

We may also check that the new external hint does not suffer from the
complexity blow up.

Lemma transitivity_workaround_2 : ∀ S T,

subtype S T.

Proof.

intros. (* debug *) eauto. (* Investigates 0 applications *)

Abort.

subtype S T.

Proof.

intros. (* debug *) eauto. (* Investigates 0 applications *)

Abort.

# Decision Procedures

## Lia

Here is an example. Let x and y be two natural numbers
(they cannot be negative). Assume y is less than 4, assume
x+x+1 is less than y, and assume x is not zero. Then,
it must be the case that x is equal to one.

Lemma lia_demo_1 : ∀ (x y : nat),

(y ≤ 4) →

(x + x + 1 ≤ y) →

(x ≠ 0) →

(x = 1).

Proof. intros. lia. Qed.

(y ≤ 4) →

(x + x + 1 ≤ y) →

(x ≠ 0) →

(x = 1).

Proof. intros. lia. Qed.

Another example: if z is the mean of x and y, and if the
difference between x and y is at most 4, then the difference
between x and z is at most 2.

Lemma lia_demo_2 : ∀ (x y z : nat),

(x + y = z + z) →

(x - y ≤ 4) →

(x - z ≤ 2).

Proof. intros. lia. Qed.

(x + y = z + z) →

(x - y ≤ 4) →

(x - z ≤ 2).

Proof. intros. lia. Qed.

One can proof False using omega if the mathematical facts
from the context are contradictory. In the following example,
the constraints on the values x and y cannot be all
satisfied in the same time.

Note: omega can prove a goal by contradiction only if its
conclusion reduces to False. The tactic omega always fails
when the conclusion is an arbitrary proposition P, even though
False implies any proposition P (by ex_falso_quodlibet).

Lemma lia_demo_4 : ∀ (x y : nat) (P : Prop),

(x + 5 ≤ y) →

(y - x < 3) →

P.

Proof.

intros.

(* The lia tactic can solve non arithmetic goals

provided there is a contradiction that it can derive

from arithmetic hypotheses *)

lia.

Qed.

(x + 5 ≤ y) →

(y - x < 3) →

P.

Proof.

intros.

(* The lia tactic can solve non arithmetic goals

provided there is a contradiction that it can derive

from arithmetic hypotheses *)

lia.

Qed.

## Ring

Require Import ZArith.

Module RingDemo.

Open Scope Z_scope.

(* Arithmetic symbols are now interpreted in Z *)

Lemma ring_demo : ∀ (x y z : Z),

x × (y + z) - z × 3 × x

= x × y - 2 × x × z.

Proof. intros. ring. Qed.

End RingDemo.

Module RingDemo.

Open Scope Z_scope.

(* Arithmetic symbols are now interpreted in Z *)

Lemma ring_demo : ∀ (x y z : Z),

x × (y + z) - z × 3 × x

= x × y - 2 × x × z.

Proof. intros. ring. Qed.

End RingDemo.

## Congruence

Lemma congruence_demo_1 :

∀ (f : nat→nat→nat) (g h : nat→nat) (x y z : nat),

f (g x) (g y) = z →

2 = g x →

g y = h z →

f 2 (h z) = z.

Proof. intros. congruence. Qed.

∀ (f : nat→nat→nat) (g h : nat→nat) (x y z : nat),

f (g x) (g y) = z →

2 = g x →

g y = h z →

f 2 (h z) = z.

Proof. intros. congruence. Qed.

Moreover, congruence is able to exploit universally quantified
equalities, for example ∀ a, g a = h a.

Lemma congruence_demo_2 :

∀ (f : nat→nat→nat) (g h : nat→nat) (x y z : nat),

(∀ a, g a = h a) →

f (g x) (g y) = z →

g x = 2 →

f 2 (h y) = z.

Proof. congruence. Qed.

∀ (f : nat→nat→nat) (g h : nat→nat) (x y z : nat),

(∀ a, g a = h a) →

f (g x) (g y) = z →

g x = 2 →

f 2 (h y) = z.

Proof. congruence. Qed.

Next is an example where congruence is very useful.

Lemma congruence_demo_4 : ∀ (f g : nat→nat),

(∀ a, f a = g a) →

f (g (g 2)) = g (f (f 2)).

Proof. congruence. Qed.

(∀ a, f a = g a) →

f (g (g 2)) = g (f (f 2)).

Proof. congruence. Qed.

The tactic congruence is able to prove a contradiction if the
goal entails an equality that contradicts an inequality available
in the proof context.

Lemma congruence_demo_3 :

∀ (f g h : nat→nat) (x : nat),

(∀ a, f a = h a) →

g x = f x →

g x ≠ h x →

False.

Proof. congruence. Qed.

∀ (f g h : nat→nat) (x : nat),

(∀ a, f a = h a) →

g x = f x →

g x ≠ h x →

False.

Proof. congruence. Qed.

One of the strengths of congruence is that it is a very fast
tactic. So, one should not hesitate to invoke it wherever it might
help.

# Summary

- auto automatically applies reflexivity, assumption, and apply.
- eauto moreover tries eapply, and in particular can instantiate
existentials in the conclusion.
- iauto extends eauto with support for negation, conjunctions, and
disjunctions. However, its support for disjunction can make it
exponentially slow.
- jauto extends eauto with support for negation, conjunctions, and
existential at the head of hypothesis.
- congruence helps reasoning about equalities and inequalities.
- lia proves arithmetic goals with equalities and inequalities,
but it does not support multiplication.
- ring proves arithmetic goals with multiplications, but does not support inequalities.

- automation is all about balance: not enough automation makes proofs
not very robust on change, whereas too much automation makes proofs
very hard to fix when they break.
- if a lemma is not goal directed (i.e., some of its variables do not
occur in its conclusion), then the premises need to be ordered in
such a way that proving the first premises maximizes the chances of
correctly instantiating the variables that do not occur in the conclusion.
- a lemma whose conclusion is False should only be added as a local
hint, i.e., as a hint within the current section.
- a transitivity lemma should never be considered as hint; if automation
of transitivity reasoning is really necessary, an Extern Hint needs
to be set up.
- a definition usually needs to be accompanied with a Hint Unfold.

(* 2021-05-26 09:58 *)