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(** In this example we show how to use the [firstorder] tactic to
prove statements in first-order logic. We consider the Frobenius
rule, which states that existential quantifiers and conjunctions
Theorem frobenius (A : Set) (p : A -> Prop) (q : Prop) :
(exists x : A, q /\ p x) <-> (q /\ exists x : A, p x).
(** We stated a theorem with _parameters_ [A], [p] and [q]. An
alternative would be to quantify universally over [A], [p] and [q].
Both forms are acceptable. Coq 8.3 will automatically perform the [intros]
rule on parameters. *)
(** Coq's [firstorder] tactic is used for intuitionistic first-order reasoning.
Let us try it. *)
(** And we are done. *)
(** According to the Curry-Howard isomorphism there is a
correspondence between propositions and sets/types. Thus there is
also a set-theoretic version of Frobenius rule. *)
Theorem frobenius_set1 (A : Set) (p : A -> Set) (q : Set) :
({x : A & q * p x} -> q * {x : A & p x})%type.
(** The above notation requires some explanation. First of all, the enclosing
[(...)%type] is there to tell Coq that the [*] symbol should be interpreted
in the context of types, as opposed to natural numbers. Without it, Coq thinks
that [q * p x] is multiplication of natural numbers and complains accordingly.
The notation [{x : A & ...}] means "the set of all [x]'s from [A] that satisfy
the condition ...". Also, equivalence [<->] is only used for propositions, so
have to split the theorem into two function spaces (implications) by hand. *)
Theorem frobenius_set2 (A : Set) (p : A -> Set) (q : Set) :
(q * {x : A & p x} -> {x : A & q * p x})%type.
(** What proofs did [firstoder] actually find? *)
Print frobenius_set2.
(** Uhhhh. *)
(* Here are some exercise. *)
(* Exercise: in [frobenius], replace conjunctions with disjunctions
and see what [firstorder] reduces the statement to. You will discover
a missing condition. Add the condition to the original statement and
rerun [firstorder]. *)
(* Exercise: the dual Frobenius rule says that universal quantifiers
commute with disjunctions. State the theorem and prove it using
the Law of Excluded Middle. *)
(* Exercise (hard): prove that the dual Frobenius rule is actually equivalent
to the Law of Excluded Middle. *)