Metamath Proof Explorer


Theorem ackbij1lem8

Description: Lemma for ackbij1 . (Contributed by Stefan O'Rear, 19-Nov-2014)

Ref Expression
Hypothesis ackbij.f ⊢ F = x ∈ 𝒫 ω ∩ Fin ⟼ card ⁡ ⋃ y ∈ x y × 𝒫 y
Assertion ackbij1lem8 ⊢ A ∈ ω → F ⁡ A = card ⁡ 𝒫 A

Proof

Step Hyp Ref Expression
1 ackbij.f ⊢ F = x ∈ 𝒫 ω ∩ Fin ⟼ card ⁡ ⋃ y ∈ x y × 𝒫 y
2 sneq ⊢ a = A → a = A
3 2 fveq2d ⊢ a = A → F ⁡ a = F ⁡ A
4 pweq ⊢ a = A → 𝒫 a = 𝒫 A
5 4 fveq2d ⊢ a = A → card ⁡ 𝒫 a = card ⁡ 𝒫 A
6 3 5 eqeq12d ⊢ a = A → F ⁡ a = card ⁡ 𝒫 a ↔ F ⁡ A = card ⁡ 𝒫 A
7 ackbij1lem4 ⊢ a ∈ ω → a ∈ 𝒫 ω ∩ Fin
8 1 ackbij1lem7 ⊢ a ∈ 𝒫 ω ∩ Fin → F ⁡ a = card ⁡ ⋃ y ∈ a y × 𝒫 y
9 7 8 syl ⊢ a ∈ ω → F ⁡ a = card ⁡ ⋃ y ∈ a y × 𝒫 y
10 vex ⊢ a ∈ V
11 sneq ⊢ y = a → y = a
12 pweq ⊢ y = a → 𝒫 y = 𝒫 a
13 11 12 xpeq12d ⊢ y = a → y × 𝒫 y = a × 𝒫 a
14 10 13 iunxsn ⊢ ⋃ y ∈ a y × 𝒫 y = a × 𝒫 a
15 14 fveq2i ⊢ card ⁡ ⋃ y ∈ a y × 𝒫 y = card ⁡ a × 𝒫 a
16 vpwex ⊢ 𝒫 a ∈ V
17 xpsnen2g ⊢ a ∈ V ∧ 𝒫 a ∈ V → a × 𝒫 a ≈ 𝒫 a
18 10 16 17 mp2an ⊢ a × 𝒫 a ≈ 𝒫 a
19 carden2b ⊢ a × 𝒫 a ≈ 𝒫 a → card ⁡ a × 𝒫 a = card ⁡ 𝒫 a
20 18 19 ax-mp ⊢ card ⁡ a × 𝒫 a = card ⁡ 𝒫 a
21 15 20 eqtri ⊢ card ⁡ ⋃ y ∈ a y × 𝒫 y = card ⁡ 𝒫 a
22 9 21 eqtrdi ⊢ a ∈ ω → F ⁡ a = card ⁡ 𝒫 a
23 6 22 vtoclga ⊢ A ∈ ω → F ⁡ A = card ⁡ 𝒫 A