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 ∣ ( ¬ ( 𝐴𝑥 ) ∈ Fin ∨ ( ( 𝐴𝑥 ) = ∅ ∨ ( 𝐴𝑥 ) = 𝐴 ) ) } = { 𝑥 ∈ 𝒫 𝐴 ∣ ( ¬ ( 𝐴𝑥 ) ∈ Fin ∨ ( 𝑥 = ∅ ∨ 𝑥 = 𝐴 ) ) }

Proof

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