使用Coq證明有限集的冪集是有限的

Question

在試圖證明某些事情時，我遇到了一個無辜的看起來聲稱，我沒有在Coq證明。 聲稱對於給定的有限集合，powerset也是有限的。 該聲明在下面的Coq代碼中給出。

我查看了關於有限集和有限集的有限集和事實的Coq文檔，但是我找不到將powerset解構為子集聯合的東西（這樣可以使用Union_is_finite構造函數）。 另一種方法可能是顯示powerset的基數為2 ^ | S | 但在這里我當然不知道如何處理證據。

From Coq Require Export Sets.Ensembles.
From Coq Require Export Sets.Powerset.
From Coq Require Export Sets.Finite_sets.

Lemma powerset_finite {T} (S : Ensemble T) :
  Finite T S -> Finite (Ensemble T) (Power_set T S).
Proof.
  (* I don't know how to proceed. *)
Admitted.

Answer 1

我沒有完全解決它，因為我自己在這個證據上苦苦掙扎。 我只是沿着你的思路轉移它。 現在問題的症結在於，證明一組n個元素的冪集的基數是2 ^ n。

From Coq Require Export Sets.Ensembles.
From Coq Require Export Sets.Powerset.
From Coq Require Export Sets.Finite_sets.
From Coq Require Export Sets.Finite_sets_facts.

Fixpoint exp (n m : nat) : nat :=
  match m with
    | 0 => 1
    | S m' => n * (exp n m')
  end.

Theorem power_set_empty :
  forall (U : Type), Power_set _ (Empty_set U) = Singleton _ (Empty_set _).
Proof with auto with sets.
  intros U.
  apply Extensionality_Ensembles.
  unfold Same_set. split. 
  + unfold Included. intros x Hin.
    inversion Hin; subst. 
    apply Singleton_intro.
    symmetry. apply less_than_empty; auto.

  + unfold Included. intros x Hin.
    constructor. inversion Hin; subst.
    unfold Included; intros; assumption.
Qed.

Lemma cardinality_power_set :
  forall (U : Type) (A : Ensemble U) (n : nat),
    cardinal U A n -> cardinal _ (Power_set _ A) (exp 2 n).
Proof.
  intros U A n. revert A.
  induction n; cbn; intros He Hc.
  + inversion Hc; subst. rewrite power_set_empty.
    Search Singleton.
    rewrite <- Empty_set_zero'.
    constructor; repeat auto with sets. 
  + inversion Hc; subst; clear Hc.
Admitted.



Lemma powerset_finite {T} (S : Ensemble T) :
  Finite T S -> Finite (Ensemble T) (Power_set T S).
Proof.
  intros Hf.
  destruct (finite_cardinal _ S Hf) as [n Hc].
  eapply cardinal_finite with (n := exp 2 n).
  apply cardinality_power_set; auto.
Qed.