Metamath Proof Explorer


Theorem pwwf

Description: A power set is well-founded iff the base set is. (Contributed by Mario Carneiro, 8-Jun-2013) (Revised by Mario Carneiro, 16-Nov-2014)

Ref Expression
Assertion pwwf ⊢ A ∈ ⋃ R1 On ↔ 𝒫 A ∈ ⋃ R1 On

Proof

Step Hyp Ref Expression
1 r1rankidb ⊢ A ∈ ⋃ R1 On → A ⊆ R1 ⁡ rank ⁡ A
2 1 sspwd ⊢ A ∈ ⋃ R1 On → 𝒫 A ⊆ 𝒫 R1 ⁡ rank ⁡ A
3 rankdmr1 ⊢ rank ⁡ A ∈ dom ⁡ R1
4 r1sucg ⊢ rank ⁡ A ∈ dom ⁡ R1 → R1 ⁡ suc ⁡ rank ⁡ A = 𝒫 R1 ⁡ rank ⁡ A
5 3 4 ax-mp ⊢ R1 ⁡ suc ⁡ rank ⁡ A = 𝒫 R1 ⁡ rank ⁡ A
6 2 5 sseqtrrdi ⊢ A ∈ ⋃ R1 On → 𝒫 A ⊆ R1 ⁡ suc ⁡ rank ⁡ A
7 fvex ⊢ R1 ⁡ suc ⁡ rank ⁡ A ∈ V
8 7 elpw2 ⊢ 𝒫 A ∈ 𝒫 R1 ⁡ suc ⁡ rank ⁡ A ↔ 𝒫 A ⊆ R1 ⁡ suc ⁡ rank ⁡ A
9 6 8 sylibr ⊢ A ∈ ⋃ R1 On → 𝒫 A ∈ 𝒫 R1 ⁡ suc ⁡ rank ⁡ A
10 r1funlim ⊢ Fun ⁡ R1 ∧ Lim ⁡ dom ⁡ R1
11 10 simpri ⊢ Lim ⁡ dom ⁡ R1
12 limsuc ⊢ Lim ⁡ dom ⁡ R1 → rank ⁡ A ∈ dom ⁡ R1 ↔ suc ⁡ rank ⁡ A ∈ dom ⁡ R1
13 11 12 ax-mp ⊢ rank ⁡ A ∈ dom ⁡ R1 ↔ suc ⁡ rank ⁡ A ∈ dom ⁡ R1
14 3 13 mpbi ⊢ suc ⁡ rank ⁡ A ∈ dom ⁡ R1
15 r1sucg ⊢ suc ⁡ rank ⁡ A ∈ dom ⁡ R1 → R1 ⁡ suc ⁡ suc ⁡ rank ⁡ A = 𝒫 R1 ⁡ suc ⁡ rank ⁡ A
16 14 15 ax-mp ⊢ R1 ⁡ suc ⁡ suc ⁡ rank ⁡ A = 𝒫 R1 ⁡ suc ⁡ rank ⁡ A
17 9 16 eleqtrrdi ⊢ A ∈ ⋃ R1 On → 𝒫 A ∈ R1 ⁡ suc ⁡ suc ⁡ rank ⁡ A
18 r1elwf ⊢ 𝒫 A ∈ R1 ⁡ suc ⁡ suc ⁡ rank ⁡ A → 𝒫 A ∈ ⋃ R1 On
19 17 18 syl ⊢ A ∈ ⋃ R1 On → 𝒫 A ∈ ⋃ R1 On
20 r1elssi ⊢ 𝒫 A ∈ ⋃ R1 On → 𝒫 A ⊆ ⋃ R1 On
21 pwexr ⊢ 𝒫 A ∈ ⋃ R1 On → A ∈ V
22 pwidg ⊢ A ∈ V → A ∈ 𝒫 A
23 21 22 syl ⊢ 𝒫 A ∈ ⋃ R1 On → A ∈ 𝒫 A
24 20 23 sseldd ⊢ 𝒫 A ∈ ⋃ R1 On → A ∈ ⋃ R1 On
25 19 24 impbii ⊢ A ∈ ⋃ R1 On ↔ 𝒫 A ∈ ⋃ R1 On