Metamath Proof Explorer


Theorem r1om

Description: The set of hereditarily finite sets is countable. See ackbij2 for an explicit bijection that works without Infinity. See also r1omALT . (Contributed by Stefan O'Rear, 18-Nov-2014)

Ref Expression
Assertion r1om R1 ω ω

Proof

Step Hyp Ref Expression
1 omex ω V
2 limom Lim ω
3 r1lim ω V Lim ω R1 ω = a ω R1 a
4 1 2 3 mp2an R1 ω = a ω R1 a
5 r1fnon R1 Fn On
6 fnfun R1 Fn On Fun R1
7 funiunfv Fun R1 a ω R1 a = R1 ω
8 5 6 7 mp2b a ω R1 a = R1 ω
9 4 8 eqtri R1 ω = R1 ω
10 iuneq1 e = a f e f × 𝒫 f = f a f × 𝒫 f
11 sneq f = b f = b
12 pweq f = b 𝒫 f = 𝒫 b
13 11 12 xpeq12d f = b f × 𝒫 f = b × 𝒫 b
14 13 cbviunv f a f × 𝒫 f = b a b × 𝒫 b
15 10 14 eqtrdi e = a f e f × 𝒫 f = b a b × 𝒫 b
16 15 fveq2d e = a card f e f × 𝒫 f = card b a b × 𝒫 b
17 16 cbvmptv e 𝒫 ω Fin card f e f × 𝒫 f = a 𝒫 ω Fin card b a b × 𝒫 b
18 dmeq c = a dom c = dom a
19 18 pweqd c = a 𝒫 dom c = 𝒫 dom a
20 imaeq1 c = a c d = a d
21 20 fveq2d c = a e 𝒫 ω Fin card f e f × 𝒫 f c d = e 𝒫 ω Fin card f e f × 𝒫 f a d
22 19 21 mpteq12dv c = a d 𝒫 dom c e 𝒫 ω Fin card f e f × 𝒫 f c d = d 𝒫 dom a e 𝒫 ω Fin card f e f × 𝒫 f a d
23 imaeq2 d = b a d = a b
24 23 fveq2d d = b e 𝒫 ω Fin card f e f × 𝒫 f a d = e 𝒫 ω Fin card f e f × 𝒫 f a b
25 24 cbvmptv d 𝒫 dom a e 𝒫 ω Fin card f e f × 𝒫 f a d = b 𝒫 dom a e 𝒫 ω Fin card f e f × 𝒫 f a b
26 22 25 eqtrdi c = a d 𝒫 dom c e 𝒫 ω Fin card f e f × 𝒫 f c d = b 𝒫 dom a e 𝒫 ω Fin card f e f × 𝒫 f a b
27 26 cbvmptv c V d 𝒫 dom c e 𝒫 ω Fin card f e f × 𝒫 f c d = a V b 𝒫 dom a e 𝒫 ω Fin card f e f × 𝒫 f a b
28 eqid rec c V d 𝒫 dom c e 𝒫 ω Fin card f e f × 𝒫 f c d ω = rec c V d 𝒫 dom c e 𝒫 ω Fin card f e f × 𝒫 f c d ω
29 17 27 28 ackbij2 rec c V d 𝒫 dom c e 𝒫 ω Fin card f e f × 𝒫 f c d ω : R1 ω 1-1 onto ω
30 fvex R1 ω V
31 9 30 eqeltrri R1 ω V
32 31 f1oen rec c V d 𝒫 dom c e 𝒫 ω Fin card f e f × 𝒫 f c d ω : R1 ω 1-1 onto ω R1 ω ω
33 29 32 ax-mp R1 ω ω
34 9 33 eqbrtri R1 ω ω