If I'm in Coq and I find myself in a situation with a goal like so:
==================
x = y -> y = x
Is there a tactic that can can take care of this in one swoop? As it is, I'm writing
intros H. rewrite -> H. reflexivity.
But it's a bit clunky.
To "flip" an equality H: x = y you can use symmetry in H. If you want to flip the goal, simply use symmetry.
If you're looking for a single tactic, then the easy tactic handles this one immediately:
Coq < Parameter x y : nat.
x is assumed
y is assumed
Coq < Lemma sym : x = y -> y = x.
1 subgoal
============================
x = y -> y = x
sym < easy.
No more subgoals.
If you take a look at the proof that the easy tactic found, the key part is an application of eq_sym:
sym < Show Proof.
(fun H : x = y => eq_sym H)
The heavier-weight auto tactic will also handle this goal in a single step. For a slightly lower-level proof that produces exactly the same proof term, you can use the symmetry tactic (which also automatically does the necessary intro for you):
sym < Restart.
1 subgoal
============================
x = y -> y = x
sym < symmetry.
1 subgoal
H : x = y
============================
x = y
sym < assumption.
No more subgoals.
sym < Show Proof.
(fun H : x = y => eq_sym H)
Related
In my list of hypothesis, I have:
X : Type
l' : list X
n' : nat
H : S (length l') = S n'
My goal is length l' = n'.
So I tried f_equal in H. But I get the following error:
Syntax error: [tactic:ltac_use_default] expected after [tactic:tactic] (in [vernac:tactic_command]).
Am I wrong in thinking I should be able to apply f_equal to H in order to remove the S on both sides?
f_equal is about congruence of equality. It can be used to prove f x = f y from x = y. However, it cannot be used to deduce x = y from f x = f y because that is not true in general, only when f is injective.
Here it is a particular case as S is a constructor of an inductive type, and constructors are indeed injective. You could for instance use tactics like inversion H to obtain the desired equality.
Another solution involving f_equal would be to apply a function that removes the S like
Definition removeS n :=
match n with
| S m => m
| 0 => 0
end.
and then use
apply (f_equal removeS) in H.
f_equal tells you that if x = y, then f x = f y. In other words, when you have x = y and need f x = f y, you can use f_equal.
Your situation is the reverse. You have f x = f y and you need x = y, so you can't use f_equal.
If you think about your conclusion, it is only true when S is an injection. You need a different tactic.
I am interested in how does dependent-typed theorem provers do substitution on the context. I find a thing called "intro ->" in Coq, as described in:
https://coq.inria.fr/refman/proof-engine/tactics.html#intropattern-rarrow-ex
In the example, it says the goal:
x, y, z : nat
H : x = y
y = z -> x = z
will become
x, z : nat
H : x = z
x = z
after an application of "intros ->".
I am not sure how is the step of substituting H:x = y into H:x = z be done, what is the rule of the logic of Coq is this step according to? It seems like a step of rewriting, and according to my current knowledge most of these rewriting are according to equalities, so the replacement from H:x = y into H:x = z should come from an equality like (H:x = y) = (H:x= z) ($\diamond$), but such an equality is ill-formed because, in principle, we require the LHS and RHS of an equality to have the same type, so the well-formedness of such equality ($\diamond$) depends on the assumption that y = z.
Could someone please help explaining how does "intro ->" work here? Do we need to get "equalities" like ($\diamond$) involved?
In Coq, rewriting on the context is a derived notion. The tactic intros -> in this case is roughly equivalent to the following tactics:
intros H'. rewrite H' in H. clear y H'.
The tactic rewrite H' in H., in turn, is roughly equivalent to
revert H. rewrite H'. intros H.
In words, to rewrite on the context, Coq first transfers all the relevant hypotheses to the goal, rewrites the goal and then brings the hypotheses back to the context.
Rewriting the goal uses the following elimination principle for the equality type:
forall [A : Type] [a : A] (P : A -> Prop), P a -> forall b : A, b = a -> P b
For rewriting H' above, we would have:
A := T
a := z
P := fun c => x = c -> x = z
b := y
so,
P b == x = y -> x = z
P a == x = z -> x = z
Indeed, ignoring the move steps to and from the context, Coq went from P b to P a when you performed the rewrite.
What effect does the following tactic have on the goal and the assumptions?
I know what induction on variables and named hypothesis do, but am unclear about induction on a number.
Induction 1
From the Coq Reference Manual: https://coq.inria.fr/distrib/current/refman/proof-engine/tactics.html#coq:tacn.induction
(...) induction num behaves as intros until num followed by induction applied to the last introduced hypothesis.
And for intros until num: https://coq.inria.fr/distrib/current/refman/proof-engine/tactics.html#coq:tacv.intros
intros until num: This repeats intro until the num-th non-dependent product.
Example
On the subgoal forall x y : nat, x = y -> y = x the tactic intros until 1 is equivalent to intros x y H, as x = y -> y = x is the first non-dependent product.
On the subgoal forall x y z : nat, x = y -> y = x the tactic intros until 1 is equivalent to intros x y z as the product on z can be rewritten as a non-dependent product: forall x y : nat, nat -> x = y -> y = x.
For reference, there is an index of standard tactics in the Manual where those can easily be looked up: https://coq.inria.fr/distrib/current/refman/coq-tacindex.html
(There are other indices in there that you may find interesting as well.)
From this example:
Example foo : forall (X : Type) (x y z : X) (l j : list X),
x :: y :: l = z :: j ->
y :: l = x :: j ->
x = y.
It can be solved just doing inversion on the second hypothesis:
Proof.
intros X x y z l j eq1 eq2. inversion eq2. reflexivity. Qed.
However, doing inversion also in the first hypothesis, yields apparently contradictory hypothesis:
Proof.
intros X x y z l j eq1 eq2. inversion eq2. inversion eq1. reflexivity. Qed.
Because, in this last proof, the generated hypothesis are:
H0 : y = x
H1 : l = j
H2 : x = z
H3 : y :: l = j
But, if I'm not missing something obvious, it is impossible for both H1 and H3 to be true at the same time.
Can someone explain me what is going on? Is it just that the example is "bad designed" (both hypothesis are contradictory) and that Coq inversion tactic just swallows them? Is it a principle of explosion based on two hypothesis considered together? If so, is it then possible to prove the example just by deriving anything from falsehood? How?
Your example is assuming contradictory hypotheses: they imply that length l + 2 is equal to length l + 1.
Require Import Coq.Lists.List.
Require Import Omega.
Example foo : forall (X : Type) (x y z : X) (l j : list X),
x :: y :: l = z :: j ->
y :: l = x :: j ->
x = y.
Proof.
intros X x y z l j eq1 eq2.
apply (f_equal (#length _)) in eq1.
apply (f_equal (#length _)) in eq2.
simpl in *.
omega.
Qed.
By the principle of explosion, it is not surprising that Coq is able to derive a contradictory context.
Besides this small oddity, there is nothing wrong with the fact that the generated hypotheses are contradictory: such contexts can arise even if the original hypotheses are consistent. Consider the following (admittedly contrived) proof:
Goal forall b c : bool, b = c -> c = b.
Proof.
intros b c e.
destruct b, c.
- reflexivity.
- discriminate.
- discriminate.
- reflexivity.
Qed.
The second and third branches have contradictory hypotheses (true = false and false = true), even if the original hypothesis, b = c, is innocuous. This example is a bit different from the original one, because the contradiction was not obtained by combining hypotheses. Instead, when we call destruct, we promise Coq to prove the conclusion by considering a few subgoals obtained by case analyses. If some of the subgoals happen to be contradictory, even better: there won't be any work to do there.
I'm new to Coq and am doing some exercises to get more familiar with it.
My understanding is that proving a proposition in Coq "really" is writing down a type in Gallina and then showing that it's inhabited using tactics to combine terms together in deterministic ways.
I'm wondering if there's a way to get a pretty-printed representation of the actual term, with all the tactics removed.
In the example below, an anonymous term of type plus_comm (x y : N) : plus x y = plus y x is ultimately produced... I think. What should I do if I want to look at it? In a certain sense, I'm curious what the tactics "desugar" to.
Here's the code in question, lifted essentially verbatim from a tutorial on YouTube https://www.youtube.com/watch?v=OaIn7g8BAIc.
Inductive N : Type :=
| O : N
| S : N -> N
.
Fixpoint plus (x y : N) : N :=
match x with
| O => y
| S x' => S (plus x' y)
end.
Lemma plus_0 (x : N) : plus x O = x.
Proof.
induction x.
- simpl. reflexivity.
- simpl. rewrite IHx. reflexivity.
Qed.
Lemma plus_S (x y : N) : plus x (S y) = S(plus x y).
Proof.
induction x.
- simpl. reflexivity.
- simpl. rewrite IHx. reflexivity.
Qed.
Lemma plus_comm (x y : N) : plus x y = plus y x.
Proof.
induction x.
- simpl. rewrite plus_0. reflexivity.
- simpl. rewrite IHx. rewrite plus_S. reflexivity.
Qed.
First of all, plus_comm is not a part of the type. You get a term named plus_comm of type forall x y : N, plus x y = plus y x. You can check it using the following command
Check plus_comm.
So, an alternative way of defining the plus_comm lemma is
Lemma plus_comm : forall x y : N, plus x y = plus y x.
As a side note: in this case you'll need to add intros x y. (or just intros.) after the Proof. part.
Tactics (and the means to glue them together) are a metalanguage called Ltac, because they are used to produce terms of another language, called Gallina, which is the specification language of Coq.
For example, forall x y : N, plus x y = plus y x is an instance of Gallina sentence as well as the body of the plus function. To obtain the term attached to plus_comm use the Print command:
Print plus_comm.
plus_comm =
fun x y : N =>
N_ind (fun x0 : N => plus x0 y = plus y x0)
(eq_ind_r (fun n : N => y = n) eq_refl (plus_0 y))
(fun (x0 : N) (IHx : plus x0 y = plus y x0) =>
eq_ind_r (fun n : N => S n = plus y (S x0))
(eq_ind_r (fun n : N => S (plus y x0) = n) eq_refl (plus_S y x0))
IHx) x
: forall x y : N, plus x y = plus y x
It is not an easy read, but with some experience you'll be able to understand it.
Incidentally, here is how we could have proved the lemma not using tactics:
Definition plus_comm : forall x y : N, plus x y = plus y x :=
fix IH (x y : N) :=
match x return plus x y = plus y x with
| O => eq_sym (plus_0 y)
| S x => eq_ind _ (fun p => S p = plus y (S x)) (eq_sym (plus_S y x)) _ (eq_sym (IH x y))
end.
To explain a few things: fix is the means of defining recursive functions, eq_sym is used to change x = y into y = x, and eq_ind corresponds to the rewrite tactic.