Talkin' with the Rooster - coq/coq GitHub Wiki

My goal is ..., how can I prove it?

My goal is a conjunction, how can I prove it?

Use some theorem or assumption or use the split tactic:

Coq < Goal forall A B:Prop, A->B-> A/\B.

1 subgoal

  ============================
  forall A B : Prop, A -> B -> A /\ B

Unnamed_thm < intros.

1 subgoal
  
  A : Prop
  B : Prop
  H : A
  H0 : B
  ============================
  A /\ B

Unnamed_thm < split.

2 subgoals
  
  A : Prop
  B : Prop
  H : A
  H0 : B
  ============================
  A
subgoal 2 is:
 B

Unnamed_thm < assumption.

1 subgoal
  
  A : Prop
  B : Prop
  H : A
  H0 : B
  ============================
  B

Unnamed_thm < assumption.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal contains a conjunction as a hypothesis, how can I use it?

If you want to decompose your hypothesis into several hypothesess you can use the destruct tactic:

Coq < Goal forall A B : Prop, A /\ B -> B.

1 subgoal
  
  ============================
  forall A B : Prop, A /\ B -> B

Unnamed_thm < intros.

1 subgoal
  
  A : Prop
  B : Prop
  H : A /\ B
  ============================
  B

Unnamed_thm < destruct H as [H1 H2].

1 subgoal
  
  A : Prop
  B : Prop
  H : A /\ B
  H0 : A
  H1 : B
  ============================
  B

Unnamed_thm < assumption.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

You can also perform the destruction at the time of introduction:

Coq < Goal forall A B : Prop, A /\ B -> B.

1 subgoal

  ============================
  forall A B : Prop, A /\ B -> B

Unnamed_thm < intros A B [H1 H2].

1 subgoal

  A, B : Prop
  H1 : A
  H2 : B
  ============================
  B

Unnamed_thm < assumption.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is a disjunction, how can I prove it?

You can prove the left part or the right part of the disjunction using the left or right tactics. If you want to do a classical reasoning step, use the classic axiom to prove the right part with the assumption that the left part of the disjunction is false.

Coq < Goal forall A B : Prop, A -> A \/ B.

1 subgoal
  
  ============================
  forall A B : Prop, A -> A \/ B

Unnamed_thm < intros.

1 subgoal
  
  A : Prop
  B : Prop
  H : A
  ============================
  A \/ B

Unnamed_thm < left.

1 subgoal
  
  A : Prop
  B : Prop
  H : A
  ============================
  A

Unnamed_thm < assumption.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

An example using classical reasoning:

Coq < Require Import Classical.

Coq < Ltac classical_right := 
      match goal with 
      | _:_ |-?X1 \/ _ => (elim (classic X1);intro;[left;trivial|right])
      end.
classical_right is defined

Coq < Ltac classical_left := 
      match goal with 
      | _:_ |- _ \/?X1 => (elim (classic X1);intro;[right;trivial|left])
      end.
classical_left is defined

Coq < Goal forall A B:Prop, (~A -> B) -> A \/ B.

1 subgoal
  
  ============================
  forall A B : Prop, (~ A -> B) -> A \/ B

Unnamed_thm < intros.

1 subgoal
  
  A : Prop
  B : Prop
  H : ~ A -> B
  ============================
  A \/ B

Unnamed_thm < classical_right.

1 subgoal
  
  A : Prop
  B : Prop
  H : ~ A -> B
  H0 : ~ A
  ============================
  B

Unnamed_thm < auto.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is an universally quantified statement, how can I prove it?

Use some theorem or assumption or introduce the quantified variable in the context using the intro tactic. If there are several variables you can use the intros tactic. A good habit is to provide names for these variables: Coq will do it anyway, but such automatic naming decreases legibility and robustness.

My goal contains an universally quantified statement, how can I use it?

If the universally quantified assumption matches the goal you can use the apply tactic. If it is an equation you can use the rewrite tactic. Otherwise you can use the specialize tactic to instantiate the quantified variables with terms. The variant assert(Ht := H t) makes a copy of assumption H before instantiating it.

My goal is an existential, how can I prove it?

Use some theorem or assumption or exhibit the witness using the exists tactic.

Coq < Goal exists x:nat, forall y, x + y = y.

1 subgoal
  
  ============================
  exists x : nat, forall y : nat, x + y = y

Unnamed_thm < exists 0.

1 subgoal
  
  ============================
  forall y : nat, 0 + y = y

Unnamed_thm < intros.

1 subgoal
  
  y : nat
  ============================
  0 + y = y

Unnamed_thm < auto.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is solvable by some lemma, how can I prove it?

Just use the apply tactic.

Coq < Lemma mylemma : forall x, x + 0 = x.

1 subgoal
  
  ============================
  forall x : nat, x + 0 = x

mylemma < auto.
No more subgoals.

mylemma < Qed.
mylemma is defined

Coq < Goal 3 + 0 = 3.
1 subgoal
  
  ============================
  3 + 0 = 3

Unnamed_thm < apply mylemma.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal contains `False` as a hypothesis, how can I prove it?

You can use the contradiction or intuition tactics.

My goal is an equality of two convertible terms, how can I prove it?

Just use the reflexivity tactic.

Coq < Goal forall x, 0 + x = x.

1 subgoal
  
  ============================
  forall x : nat, 0 + x = x

Unnamed_thm < intros.

1 subgoal
  
  x : nat
  ============================
  0 + x = x

Unnamed_thm < reflexivity.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is `a let x := a in ...` , how can I prove it?

Just use the intro tactic.

My goal is `a let (a, ..., b) := c in`, how can I prove it?

Just use the destruct c as (a,...,b) tactic.

My goal contains some existential hypotheses, how can I use it?

As with conjunctive hypotheses, you can use the destruct tactic or the intros tactic to decompose them into several hypotheses.

Coq < Require Import Arith.

Coq < Goal forall x, (exists y, x * y = 1) -> x = 1.

1 subgoal
  ============================
  forall x : nat, (exists y : nat, x * y = 1) -> x = 1

Unnamed_thm < intros x [y H].

1 subgoal

  x, y : nat
  H : x * y = 1
  ============================
  x = 1

Unnamed_thm < apply mult_is_one in H.

1 subgoal

  x, y : nat
  H : x = 1 /\ y = 1
  ============================
  x = 1

Unnamed_thm < easy.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is an equality, how can I swap the left and right hand terms?

Just use the symmetry tactic.

Coq < Goal forall x y : nat, x = y -> y = x.

1 subgoal
  
  ============================
  forall x y : nat, x = y -> y = x

Unnamed_thm < intros.

1 subgoal
  
  x : nat
  y : nat
  H : x = y
  ============================
  y = x

Unnamed_thm < symmetry.

1 subgoal
  
  x : nat
  y : nat
  H : x = y
  ============================
  x = y

Unnamed_thm < assumption.
No more subgoals.

Unnamed_thm  < Qed.
Unnamed_thm is defined

My hypothesis is an equality, how can I swap the left and right hand terms?

Just use the symmetry in tactic.

Coq < Goal forall x y : nat, x=y -> y=x.

1 subgoal
  
  ============================
  forall x y : nat, x = y -> y = x

Unnamed_thm < intros.

1 subgoal
  
  x : nat
  y : nat
  H : x = y
  ============================
  y = x

Unnamed_thm < symmetry in H.

1 subgoal
  
  x : nat
  y : nat
  H : y = x
  ============================
  y = x

Unnamed_thm < assumption.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal would be solvable using `apply;assumption` if it would not create meta-variables, how can I prove it?

You can use eapply yourtheorem;eauto but it won't work in all cases! (for example if more than one hypothesis matches one of the subgoals generated by eapply) so you should rather use try solve [eapply yourtheorem;eauto], otherwise some metavariables may be incorrectly instantiated.

Coq < Lemma trans : forall x y z : nat, x=y -> y=z -> x=z.

1 subgoal
  
  ============================
  forall x y z : nat, x = y -> y = z -> x = z

trans < intros.

1 subgoal
  
  x, y, z : nat
  H : x = y
  H0 : y = z
  ============================
  x = z

trans < transitivity y;assumption.
No more subgoals.

trans < Qed.
trans is defined

Coq < Goal forall x y z : nat, x=y -> y=z -> x=z.

1 subgoal
  
  ============================
  forall x y z : nat, x = y -> y = z -> x = z

Unnamed_thm < intros.

1 subgoal
  
  x, y, z : nat
  H : x = y
  H0 : y = z
  ============================
  x = z

Unnamed_thm < eapply trans;eauto.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

Coq < Goal forall x y z t : nat, x=y -> x=t -> y=z -> x=z.

1 subgoal
  
  ============================
  forall x y z t : nat, x = y -> x = t -> y = z -> x = z

Unnamed_thm0 < intros.

1 subgoal
  
  x, y, z, t : nat
  H : x = y
  H0 : x = t
  H1 : y = z
  ============================
  x = z

Unnamed_thm0 < eapply trans;eauto.

1 subgoal
  
  x, y, z, t : nat
  H : x = y
  H0 : x = t
  H1 : y = z
  ============================
  t = z

Unnamed_thm0 < Undo.

1 subgoal
  
  x, y, z, t : nat
  H : x = y
  H0 : x = t
  H1 : y = z
  ============================
  x = z

Unnamed_thm0 < eapply trans.

2 subgoals
  
  x, y, z, t : nat
  H : x = y
  H0 : x = t
  H1 : y = z
  ============================
   x = ?y

subgoal 2 is:
 ?y = z

Unnamed_thm0 < apply H.

1 subgoal
  
  x, y, z, t : nat
  H : x = y
  H0 : x = t
  H1 : y = z
  ============================
  y = z

Unnamed_thm0 < auto.
No more subgoals.

Unnamed_thm0 < Qed.
Unnamed_thm0 is defined

Coq < Goal forall x y z t : nat, x=y -> x=t -> y=z -> x=z.

1 subgoal
  
  ============================
  forall x y z t : nat, x = y -> x = t -> y = z -> x = z

Unnamed_thm1 < intros.

1 subgoal
  
  x, y, z, t : nat
  H : x = y
  H0 : x = t
  H1 : y = z
  ============================
  x = z

Unnamed_thm1 < eapply trans;eauto.

1 subgoal
  
  x, y, z, t : nat
  H : x = y
  H0 : x = t
  H1 : y = z
  ============================
  t = z

Unnamed_thm1 < Undo.

1 subgoal
  
  x, y, z, t : nat
  H : x = y
  H0 : x = t
  H1 : y = z
  ============================
  x = z

Unnamed_thm1 < try solve [eapply trans;eauto].

1 subgoal
  
  x, y, z, t : nat
  H : x = y
  H0 : x = t
  H1 : y = z
  ============================
  x = z

Unnamed_thm1 < eapply trans.

2 subgoals
  
  x, y, z, t : nat
  H : x = y
  H0 : x = t
  H1 : y = z
  ============================
  x = ?y

subgoal 2 is:
 ?y = z

Unnamed_thm1 < apply H.
1 subgoal
  
  x , y, z, t : nat
  H : x = y
  H0 : x = t
  H1 : y = z
  ============================
  y = z

Unnamed_thm1 < auto.
No more subgoals.

Unnamed_thm1 < Qed.
Unnamed_thm1 is defined

My goal is solvable by some lemma within a set of lemmas and I don't want to remember which one, how can I prove it?

You can use what is called a "hints' base".

Coq < Require Import ZArith.
Coq < Require Ring.
Coq < Local Open Scope Z_scope.

Coq < Lemma toto1 : 1 + 1 = 2.

1 subgoal
  
  ============================
  1 + 1 = 2

toto1 < ring.
No more subgoals.

toto1 < Qed.
toto1 is defined

Coq < Lemma toto2 : 2 + 2 = 4.

1 subgoal
  
  ============================
  2 + 2 = 4

toto2 < ring.
No more subgoals.

toto2 < Qed.
toto2 is defined

Coq < Lemma toto3 : 2 + 1 = 3.

1 subgoal
  
  ============================
  2 + 1 = 3

toto3 < ring.
No more subgoals.

toto3 < Qed.
toto3 is defined

Coq < Hint Resolve toto1 toto2 toto3 : mybase.
Coq < Goal 2+(1+1)=4. 
1 subgoal
  
  ============================
   2 + (1 + 1) = 4

Unnamed_thm < auto with mybase.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is one of the hypotheses, how can I prove it?

Use the assumption tactic.

Coq < Goal 1 = 1 -> 1 = 1.

1 subgoal
  
  ============================
  1 = 1 -> 1 = 1

Unnamed_thm < intro.

1 subgoal
  
  H : 1 = 1
  ============================
  1 = 1

Unnamed_thm < assumption.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal appears twice in the hypotheses and I want to choose which one is used, how can I do it?

Use the exact tactic.

Coq < Goal 1 = 1 -> 1 = 1 -> 1 = 1.

1 subgoal
  
  ============================
  1 = 1 -> 1 = 1 -> 1 = 1

Unnamed_thm < intros.

1 subgoal
  
  H : 1 = 1
  H0 : 1 = 1
  ============================
  1 = 1

Unnamed_thm < exact H0.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

What is the difference between applying one hypothesis or another in the context of the last question?

From a proof point of view it is equivalent but if you want to extract a program from your proof, the two hypotheses can lead to different programs.

My goal is a propositional tautology, how can I prove it?

Just use the tauto tactic.

Coq < Goal forall A B:Prop, A -> (A \/ B) /\ A.

1 subgoal
  
  ============================
  forall A B : Prop, A -> (A \/ B) /\ A

Unnamed_thm < intros.

1 subgoal
  
  A : Prop
  B : Prop
  H : A
  ============================
  (A \/ B) /\ A

Unnamed_thm < tauto.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is a first order formula, how can I prove it?

Just use the semi-decision tactic firstorder.

My goal is solvable by a sequence of rewrites, how can I prove it?

Just use the congruence tactic.

Coq < Goal forall a b c d e, a = d -> b = e -> c + b = d -> c + e = a.

1 subgoal
  
  ============================
  forall a b c d e : Z, a = d -> b = e -> c + b = d -> c + e = a

Unnamed_thm < intros.

1 subgoal
  
  a, b, c, d, e : Z
  H : a = d
  H0 : b = e
  H1 : c + b = d
  ============================
  c + e = a

Unnamed_thm < congruence.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is a disequality solvable by a sequence of rewrites, how can I prove it?

Just use the congruence tactic.

Coq < Goal forall a b c d, a <> d -> b = a -> d = c + b -> b <> c + b.

1 subgoal
  
  ============================
  forall a b c d : Z, a <> d -> b = a -> d = c + b -> b <> c + b

Unnamed_thm < intros.

1 subgoal
  
  a, b, c, d : Z
  H : a <> d
  H0 : b = a
  H1 : d = c + b
  ============================
  b <> c + b

Unnamed_thm < congruence.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is an equality on some ring (e.g. natural numbers), how can I prove it?

Just use the ring tactic.

Coq < Require Import ZArith.
Coq < Require Ring.
Coq < Local Open Scope Z_scope.

Coq < Goal forall a b : Z, (a + b) * (a + b) = a * a + 2 * a * b + b * b. 

1 subgoal
  
  ============================
  forall a b : Z, (a + b) * (a + b) = a * a + 2 * a * b + b * b

Unnamed_thm < intros.

1 subgoal
  
  a, b : Z
  ============================
  (a + b) * (a + b) = a * a + 2 * a * b + b * b

Unnamed_thm < ring.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is an equality on some field (e.g. real numbers), how can I prove it?

Just use the field tactic.

Coq < Require Import Reals.
Coq < Require Ring.
Coq < Local Open Scope R_scope.

Coq < Goal forall a b : R, b * a <> 0 -> (a / b) * (b / a) = 1.

1 subgoal
  
  ============================
  forall a b : R, b * a <> 0 -> a / b * (b / a) = 1

Unnamed_thm < intros.

1 subgoal
  
  a, b : R
  H : b * a <> 0
  ============================
  a / b * (b / a) = 1

Unnamed_thm < field.

1 subgoal
  
  a, b : R
  H : b * a <> 0
  ============================
  a <> 0 /\ b <> 0

Unnamed_thm < split ; auto with real.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is an inequality on integers in Presburger's arithmetic (an expression built from `+`, `-`, constants and variables), how can I prove it?

Coq < Require Import ZArith.
Coq < Require Omega.
Coq < Local Open Scope Z_scope.

Coq < Goal forall a : Z, a > 0 -> a + a > a. 

1 subgoal
  
  ============================
  forall a : Z, a > 0 -> a + a > a

Unnamed_thm < intros.

1 subgoal
  
  a : Z
  H : a > 0
  ============================
  a + a > a

Unnamed_thm < omega.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

My goal is an equation solvable using equational hypothesis on some ring (e.g. natural numbers), how can I prove it?

You need the gb tactic (see Loïc Pottier's homepage).

Tactics usage

I want to state a fact that I will use later as a hypothesis, how can I do it?

If you want to use forward reasoning (first proving the fact and then using it) you just need to use the assert tactic. If you want to use backward reasoning (proving your goal using an assumption and then proving the assumption) use the cut tactic.

Coq < Goal forall A B C D : Prop, (A -> B) -> (B -> C) -> A -> C.

1 subgoal
  
  ============================
  forall A B C : Prop, Prop -> (A -> B) -> (B -> C) -> A -> C

Unnamed_thm < intros.

1 subgoal
  
  A, B, C, D : Prop
  H : A -> B
  H0 : B -> C
  H1 : A
  ============================
  C

Unnamed_thm < assert (A -> C).

2 subgoals
  
  A, B, C, D : Prop
  H : A -> B
  H0 : B -> C
  H1 : A
  ============================
  A -> C

subgoal 2 is:
 C

Unnamed_thm < intro;apply H0;apply H;assumption.

1 subgoal
  
  A, B, C, D : Prop
  H : A -> B
  H0 : B -> C
  H1 : A
  H2 : A -> C
  ============================
  C

Unnamed_thm < apply H2.

1 subgoal
  
  A, B, C, D : Prop
  H : A -> B
  H0 : B -> C
  H1 : A
  H2 : A -> C
  ============================
  A

Unnamed_thm < assumption.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

Coq < Goal forall A B C D : Prop, (A -> B) -> (B -> C) -> A -> C.

1 subgoal
  
  ============================
  forall A B C : Prop, Prop -> (A -> B) -> (B -> C) -> A -> C

Unnamed_thm0 < intros.

1 subgoal
  
  A, B, C, D : Prop
  H : A -> B
  H0 : B -> C
  H1 : A
  ============================
  C

Unnamed_thm0 < cut (A -> C).

2 subgoals
  
  A, B, C, D : Prop
  H : A -> B
  H0 : B -> C
  H1 : A
  ============================
  (A -> C) -> C

subgoal 2 is:
 A -> C

Unnamed_thm0 < intro.

2 subgoals
  
  A, B, C, D : Prop
  H : A -> B
  H0 : B -> C
  H1 : A
  H2 : A -> C
  ============================
   C

subgoal 2 is:
 A -> C

Unnamed_thm0 < apply H2;assumption.

1 subgoal
  
  A, B, C, D : Prop
  H : A -> B
  H0 : B -> C
  H1 : A
  ============================
  A -> C

Unnamed_thm0 < intro;apply H0;apply H;assumption.
No more subgoals.

Unnamed_thm0 < Qed.
Unnamed_thm0 is defined

I want to state a fact that I will use later as an hypothesis and prove it later, how can I do it?

You can use cut followed by intro or you can use the following Ltac command:

Ltac assert_later t := cut t;[intro|idtac].

What is the difference between `Qed` and `Defined`?

These two commands perform type checking, but when Defined is used the new definition is set as transparent, otherwise it is defined as opaque (see this section of the Glossary).

How can I know what an automation tactic does in my example?

You can use its info variant: info_auto, info_trivial, info_eauto.

Why doesn't `auto` work? How can I fix it?

You can increase the depth of the proof search or add some lemmas in the base of hints. Perhaps you may need to use eauto.

What is `eauto`?

This is the same tactic as auto, but it relies on eapply instead of apply.

How can I speed up `auto`?

You can use info auto to replace auto by the tactics it generates. You can split your hint bases into smaller ones.

What is the equivalent of `tauto` for classical logic?

Currently there are no equivalent tactic for classical logic. You can use Gödel's "not not" translation.

I want to replace some term with another in the goal, how can I do it?

If one of your hypothesis (say H) states that the terms are equal you can use the rewrite tactic. Otherwise you can use the replace with tactic.

I want to replace some term with another in an hypothesis, how can I do it?

You can use the rewrite in tactic.

I want to replace some symbol with its definition, how can I do it?

You can use the unfold tactic.

How can I reduce some term?

You can use the simpl tactic.

How can I declare a shortcut for some term?

You can use the set or pose tactics.

How can I perform case analysis?

You can use the case or destruct tactics.

How can I prevent the `case` tactic from losing information?

You may want to use the (now standard) case_eq tactic. See page 159 of the book Coq’Art.

Why should I name my intros?

When you use the intro tactic you don't have to give a name to your hypothesis. If you do so the name will be generated by Coq but your scripts may be less robust. If you add some hypothesis to your theorem (or change their order), you will have to change your proof to adapt to the new names.

How can I automatize the naming?

You can use the Show Intro or Show Intros commands to generate the names and use your editor to generate a fully named intro tactic. This can be automatized within xemacs.

Coq < Goal forall A B C : Prop, A -> B -> C -> A /\ B /\ C.

1 subgoal
  
  ============================
  forall A B C : Prop, A -> B -> C -> A /\ B /\ C

Unnamed_thm < Show Intros.
A B C H H0 H1

Unnamed_thm < (* A B C H H0 H1 *)
              intros A B C H H0 H1.

1 subgoal
  
  A, B, C : Prop
  H : A
  H0 : B
  H1 : C
  ============================
  A /\ B /\ C

Unnamed_thm < repeat split;assumption.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

I want to automatize the use of some tactic, how can I do it?

You need to use the proof with T command and add ... at the end of your sentences. For instance:

Coq < Goal forall A B C : Prop, A -> B /\ C -> A /\ B /\ C.

1 subgoal
  
  ============================
  forall A B C : Prop, A -> B /\ C -> A /\ B /\ C

Unnamed_thm < Proof with assumption.

1 subgoal

  ============================
  forall A B C : Prop, A -> B /\ C -> A /\ B /\ C

Unnamed_thm < intros.

1 subgoal
  
  A, B, C : Prop
  H : A
  H0 : B /\ C
  ============================
  A /\ B /\ C

Unnamed_thm < split...
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

I want to execute the `proof with` tactic only if it solves the goal, how can I do it?

You need to use the try and solve tactics. For instance:

Coq < Require Import ZArith.
Coq < Require Ring.
Coq < Local Open Scope Z_scope.

Coq < Goal forall a b c : Z, a + b = b + a.

1 subgoal
  
  ============================
  forall a b : Z, Z -> a + b = b + a

Unnamed_thm < Proof with try solve [ring].

1 subgoal

  ============================
  forall a b : Z, Z -> a + b = b + a

Unnamed_thm < intros...
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

How can I do the opposite of the `intro` tactic?

You can use the generalize tactic.

Coq < Goal forall A B : Prop, A -> B -> A /\ B.

1 subgoal
  
  ============================
  forall A B : Prop, A -> B -> A /\ B

Unnamed_thm < intros.

1 subgoal
  
  A, B : Prop
  H : A
  H0 : B
  ============================
  A /\ B

Unnamed_thm < generalize H.

1 subgoal
  
  A, B : Prop
  H : A
  H0 : B
  ============================
  A -> A /\ B

Unnamed_thm < intro.

1 subgoal
  
  A, B : Prop
  H : A
  H0 : B
  H1 : A
  ============================
  A /\ B

Unnamed_thm < auto.
No more subgoals.

Unnamed_thm < Qed.
Unnamed_thm is defined

One of the hypotheses is an equality between a variable and some term. I want to get rid of this variable, how can I do it?

You can use the subst tactic. This will rewrite the equality everywhere and clear the assumption.

What can I do if I get the message "generated subgoal term has metavariables in it"?

You should use the eapply tactic, this will generate some goals containing metavariables.

How can I instantiate some metavariable?

Just use the instantiate tactic.

What is the use of the `pattern` tactic?

The pattern tactic transforms the current goal, performing beta-expansion on all the applications featuring this tactic's argument. For instance, if the current goal includes a subterm phi(t), then pattern t transforms the subterm into (fun x:A => phi(x)) t. This can be useful when apply fails on matching, to abstract the appropriate terms.

What is the difference between `assert`, `cut`, and `generalize`?

For people that are interested in proof rendering, assert, pose, and cut are not rendered the same as generalize (see the HELM experimental rendering tool at http://helm.cs.unibo.it, link HELM, link COQ Online). Indeed, generalize builds a beta-expanded term while assert, pose and cut use a let-in.

So this:

(* Goal is T *)
generalize (H1 H2).
(* Goal is A->T *)
... a proof of A->T ...

is rendered into something like:

(h) ... the proof of A->T ...
    we proved A->T
(h0) by (H1 H2) we proved A
by (h h0) we proved T

while

(* Goal is T *)
assert q := (H1 H2).
(* Goal is A *)
... a proof of A ...
(* Goal is A |- T *)
... a proof of T ...

is rendered into something like:

(q) ... the proof of A ...
    we proved A
... the proof of T ...
we proved T

Otherwise said, generalize is not rendered in a forward-reasoning way, while assert is.

What can I do if Coqcan does not infer some implicit argument?

You can state explicitly what this implicit argument is. See below.

How can I make explicit some implicit argument?

Just use A:=term where A is the argument. For instance, if you want to use the existence of "nil" on nat*nat lists:

exists (nil (A:=(nat*nat))).

Proof management

How can I change the order of the subgoals?

You can use the Focus command to concentrate on some goal. When the goal is proved you will see the remaining goals.

How can I change the order of the hypotheses?

You can use the Move ... after command.

How can I change the name of a hypothesis?

You can use the Rename ... into command.

How can I delete hypotheses?

You can use the clear command.

How can I use a proof which is not finished?

You can use the Admitted command to state your current proof as an axiom. You can use the give_up tactic to omit a portion of a proof.

How can I state a conjecture?

You can use the Admitted command to state your current proof as an axiom.

What is the difference between a lemma, a fact and a theorem?

From Coq's point of view there is no difference. But some tools can behave differently when you use a lemma rather than a theorem. For instance, coqdoc will not generate documentation for the lemmas within your development.

How can I organize my proofs?

You can organize your proofs using the section mechanism of Coq. Have a look at the manual for further information.