Metamath Proof Explorer


Theorem fin23lem41

Description: Lemma for fin23 . A set which satisfies the descending sequence condition must be III-finite. (Contributed by Stefan O'Rear, 2-Nov-2014)

Ref Expression
Hypothesis fin23lem40.f F = g | a 𝒫 g ω x ω a suc x a x ran a ran a
Assertion fin23lem41 A F A FinIII

Proof

Step Hyp Ref Expression
1 fin23lem40.f F = g | a 𝒫 g ω x ω a suc x a x ran a ran a
2 brdomi ω 𝒫 A b b : ω 1-1 𝒫 A
3 1 fin23lem33 A F c d d : ω 1-1 V ran d A c d : ω 1-1 V ran c d ran d
4 3 adantl b : ω 1-1 𝒫 A A F c d d : ω 1-1 V ran d A c d : ω 1-1 V ran c d ran d
5 ssv 𝒫 A V
6 f1ss b : ω 1-1 𝒫 A 𝒫 A V b : ω 1-1 V
7 5 6 mpan2 b : ω 1-1 𝒫 A b : ω 1-1 V
8 7 ad2antrr b : ω 1-1 𝒫 A A F d d : ω 1-1 V ran d A c d : ω 1-1 V ran c d ran d b : ω 1-1 V
9 f1f b : ω 1-1 𝒫 A b : ω 𝒫 A
10 frn b : ω 𝒫 A ran b 𝒫 A
11 uniss ran b 𝒫 A ran b 𝒫 A
12 9 10 11 3syl b : ω 1-1 𝒫 A ran b 𝒫 A
13 unipw 𝒫 A = A
14 12 13 sseqtrdi b : ω 1-1 𝒫 A ran b A
15 14 ad2antrr b : ω 1-1 𝒫 A A F d d : ω 1-1 V ran d A c d : ω 1-1 V ran c d ran d ran b A
16 f1eq1 d = e d : ω 1-1 V e : ω 1-1 V
17 rneq d = e ran d = ran e
18 17 unieqd d = e ran d = ran e
19 18 sseq1d d = e ran d A ran e A
20 16 19 anbi12d d = e d : ω 1-1 V ran d A e : ω 1-1 V ran e A
21 fveq2 d = e c d = c e
22 f1eq1 c d = c e c d : ω 1-1 V c e : ω 1-1 V
23 21 22 syl d = e c d : ω 1-1 V c e : ω 1-1 V
24 21 rneqd d = e ran c d = ran c e
25 24 unieqd d = e ran c d = ran c e
26 25 18 psseq12d d = e ran c d ran d ran c e ran e
27 23 26 anbi12d d = e c d : ω 1-1 V ran c d ran d c e : ω 1-1 V ran c e ran e
28 20 27 imbi12d d = e d : ω 1-1 V ran d A c d : ω 1-1 V ran c d ran d e : ω 1-1 V ran e A c e : ω 1-1 V ran c e ran e
29 28 cbvalvw d d : ω 1-1 V ran d A c d : ω 1-1 V ran c d ran d e e : ω 1-1 V ran e A c e : ω 1-1 V ran c e ran e
30 29 biimpi d d : ω 1-1 V ran d A c d : ω 1-1 V ran c d ran d e e : ω 1-1 V ran e A c e : ω 1-1 V ran c e ran e
31 30 adantl b : ω 1-1 𝒫 A A F d d : ω 1-1 V ran d A c d : ω 1-1 V ran c d ran d e e : ω 1-1 V ran e A c e : ω 1-1 V ran c e ran e
32 eqid rec c b ω = rec c b ω
33 1 8 15 31 32 fin23lem39 b : ω 1-1 𝒫 A A F d d : ω 1-1 V ran d A c d : ω 1-1 V ran c d ran d ¬ A F
34 4 33 exlimddv b : ω 1-1 𝒫 A A F ¬ A F
35 34 pm2.01da b : ω 1-1 𝒫 A ¬ A F
36 35 exlimiv b b : ω 1-1 𝒫 A ¬ A F
37 2 36 syl ω 𝒫 A ¬ A F
38 37 con2i A F ¬ ω 𝒫 A
39 pwexg A F 𝒫 A V
40 isfin4-2 𝒫 A V 𝒫 A FinIV ¬ ω 𝒫 A
41 39 40 syl A F 𝒫 A FinIV ¬ ω 𝒫 A
42 38 41 mpbird A F 𝒫 A FinIV
43 isfin3 A FinIII 𝒫 A FinIV
44 42 43 sylibr A F A FinIII