Metamath Proof Explorer


Theorem topdifinfeq

Description: Two different ways of defining the collection from Exercise 3 of Munkres p. 83. (Contributed by ML, 18-Jul-2020)

Ref Expression
Assertion topdifinfeq ⊢ x ∈ 𝒫 A | ¬ A ∖ x ∈ Fin ∨ A ∖ x = ∅ ∨ A ∖ x = A = x ∈ 𝒫 A | ¬ A ∖ x ∈ Fin ∨ x = ∅ ∨ x = A

Proof

Step Hyp Ref Expression
1 velpw ⊢ x ∈ 𝒫 A ↔ x ⊆ A
2 sseqin2 ⊢ x ⊆ A ↔ A ∩ x = x
3 1 2 bitri ⊢ x ∈ 𝒫 A ↔ A ∩ x = x
4 eqeq1 ⊢ A ∩ x = x → A ∩ x = ∅ ↔ x = ∅
5 3 4 sylbi ⊢ x ∈ 𝒫 A → A ∩ x = ∅ ↔ x = ∅
6 disj3 ⊢ A ∩ x = ∅ ↔ A = A ∖ x
7 eqcom ⊢ A = A ∖ x ↔ A ∖ x = A
8 6 7 bitri ⊢ A ∩ x = ∅ ↔ A ∖ x = A
9 5 8 bitr3di ⊢ x ∈ 𝒫 A → x = ∅ ↔ A ∖ x = A
10 eqss ⊢ x = A ↔ x ⊆ A ∧ A ⊆ x
11 ssdif0 ⊢ A ⊆ x ↔ A ∖ x = ∅
12 11 bicomi ⊢ A ∖ x = ∅ ↔ A ⊆ x
13 1 12 anbi12i ⊢ x ∈ 𝒫 A ∧ A ∖ x = ∅ ↔ x ⊆ A ∧ A ⊆ x
14 10 13 bitr4i ⊢ x = A ↔ x ∈ 𝒫 A ∧ A ∖ x = ∅
15 14 baib ⊢ x ∈ 𝒫 A → x = A ↔ A ∖ x = ∅
16 9 15 orbi12d ⊢ x ∈ 𝒫 A → x = ∅ ∨ x = A ↔ A ∖ x = A ∨ A ∖ x = ∅
17 orcom ⊢ A ∖ x = A ∨ A ∖ x = ∅ ↔ A ∖ x = ∅ ∨ A ∖ x = A
18 16 17 bitrdi ⊢ x ∈ 𝒫 A → x = ∅ ∨ x = A ↔ A ∖ x = ∅ ∨ A ∖ x = A
19 18 orbi2d ⊢ x ∈ 𝒫 A → ¬ A ∖ x ∈ Fin ∨ x = ∅ ∨ x = A ↔ ¬ A ∖ x ∈ Fin ∨ A ∖ x = ∅ ∨ A ∖ x = A
20 19 bicomd ⊢ x ∈ 𝒫 A → ¬ A ∖ x ∈ Fin ∨ A ∖ x = ∅ ∨ A ∖ x = A ↔ ¬ A ∖ x ∈ Fin ∨ x = ∅ ∨ x = A
21 20 rabbiia ⊢ x ∈ 𝒫 A | ¬ A ∖ x ∈ Fin ∨ A ∖ x = ∅ ∨ A ∖ x = A = x ∈ 𝒫 A | ¬ A ∖ x ∈ Fin ∨ x = ∅ ∨ x = A