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