Metamath Proof Explorer


Theorem ttrclselem1

Description: Lemma for ttrclse . Show that all finite ordinal function values of F are subsets of A . (Contributed by Scott Fenton, 31-Oct-2024)

Ref Expression
Hypothesis ttrclselem.1 F=recbVwbPredRAwPredRAX
Assertion ttrclselem1 NωFNA

Proof

Step Hyp Ref Expression
1 ttrclselem.1 F=recbVwbPredRAwPredRAX
2 nn0suc NωN=nωN=sucn
3 1 fveq1i FN=recbVwbPredRAwPredRAXN
4 fveq2 N=recbVwbPredRAwPredRAXN=recbVwbPredRAwPredRAX
5 3 4 eqtrid N=FN=recbVwbPredRAwPredRAX
6 rdg0g PredRAXVrecbVwbPredRAwPredRAX=PredRAX
7 predss PredRAXA
8 6 7 eqsstrdi PredRAXVrecbVwbPredRAwPredRAXA
9 rdg0n ¬PredRAXVrecbVwbPredRAwPredRAX=
10 0ss A
11 9 10 eqsstrdi ¬PredRAXVrecbVwbPredRAwPredRAXA
12 8 11 pm2.61i recbVwbPredRAwPredRAXA
13 5 12 eqsstrdi N=FNA
14 nnon nωnOn
15 nfcv _bPredRAX
16 nfcv _bn
17 nfmpt1 _bbVwbPredRAw
18 17 15 nfrdg _brecbVwbPredRAwPredRAX
19 1 18 nfcxfr _bF
20 19 16 nffv _bFn
21 nfcv _bPredRAt
22 20 21 nfiun _btFnPredRAt
23 predeq3 w=tPredRAw=PredRAt
24 23 cbviunv wbPredRAw=tbPredRAt
25 iuneq1 b=FntbPredRAt=tFnPredRAt
26 24 25 eqtrid b=FnwbPredRAw=tFnPredRAt
27 15 16 22 1 26 rdgsucmptf nOntFnPredRAtVFsucn=tFnPredRAt
28 iunss tFnPredRAtAtFnPredRAtA
29 predss PredRAtA
30 29 a1i tFnPredRAtA
31 28 30 mprgbir tFnPredRAtA
32 27 31 eqsstrdi nOntFnPredRAtVFsucnA
33 14 32 sylan nωtFnPredRAtVFsucnA
34 15 16 22 1 26 rdgsucmptnf ¬tFnPredRAtVFsucn=
35 34 10 eqsstrdi ¬tFnPredRAtVFsucnA
36 35 adantl nω¬tFnPredRAtVFsucnA
37 33 36 pm2.61dan nωFsucnA
38 fveq2 N=sucnFN=Fsucn
39 38 sseq1d N=sucnFNAFsucnA
40 37 39 syl5ibrcom nωN=sucnFNA
41 40 rexlimiv nωN=sucnFNA
42 13 41 jaoi N=nωN=sucnFNA
43 2 42 syl NωFNA