Metamath Proof Explorer


Theorem iuneqfzuzlem

Description: Lemma for iuneqfzuz : here, inclusion is proven; aiuneqfzuz uses this lemma twice, to prove equality. (Contributed by Glauco Siliprandi, 17-Aug-2020)

Ref Expression
Hypothesis iuneqfzuzlem.z Z=N
Assertion iuneqfzuzlem mZn=NmA=n=NmBnZAnZB

Proof

Step Hyp Ref Expression
1 iuneqfzuzlem.z Z=N
2 nfcv _mA
3 nfcsb1v _nm/nA
4 csbeq1a n=mA=m/nA
5 2 3 4 cbviun nZA=mZm/nA
6 5 eleq2i xnZAxmZm/nA
7 eliun xmZm/nAmZxm/nA
8 6 7 bitri xnZAmZxm/nA
9 8 biimpi xnZAmZxm/nA
10 9 adantl mZn=NmA=n=NmBxnZAmZxm/nA
11 nfra1 mmZn=NmA=n=NmB
12 nfv mxnZB
13 simp2 mZn=NmA=n=NmBmZxm/nAmZ
14 rspa mZn=NmA=n=NmBmZn=NmA=n=NmB
15 14 3adant3 mZn=NmA=n=NmBmZxm/nAn=NmA=n=NmB
16 simp3 mZn=NmA=n=NmBmZxm/nAxm/nA
17 id n=NmA=n=NmBn=NmA=n=NmB
18 fzssuz NmN
19 1 eqcomi N=Z
20 18 19 sseqtri NmZ
21 iunss1 NmZn=NmBnZB
22 20 21 mp1i n=NmA=n=NmBn=NmBnZB
23 17 22 eqsstrd n=NmA=n=NmBn=NmAnZB
24 23 3ad2ant2 mZn=NmA=n=NmBxm/nAn=NmAnZB
25 1 eleq2i mZmN
26 25 biimpi mZmN
27 eluzel2 mNN
28 26 27 syl mZN
29 eluzelz mNm
30 26 29 syl mZm
31 eluzle mNNm
32 26 31 syl mZNm
33 30 zred mZm
34 leid mmm
35 33 34 syl mZmm
36 28 30 30 32 35 elfzd mZmNm
37 nfcv _nx
38 37 3 nfel nxm/nA
39 4 eleq2d n=mxAxm/nA
40 38 39 rspce mNmxm/nAnNmxA
41 36 40 sylan mZxm/nAnNmxA
42 eliun xn=NmAnNmxA
43 41 42 sylibr mZxm/nAxn=NmA
44 43 3adant2 mZn=NmA=n=NmBxm/nAxn=NmA
45 24 44 sseldd mZn=NmA=n=NmBxm/nAxnZB
46 13 15 16 45 syl3anc mZn=NmA=n=NmBmZxm/nAxnZB
47 46 3exp mZn=NmA=n=NmBmZxm/nAxnZB
48 11 12 47 rexlimd mZn=NmA=n=NmBmZxm/nAxnZB
49 48 adantr mZn=NmA=n=NmBxnZAmZxm/nAxnZB
50 10 49 mpd mZn=NmA=n=NmBxnZAxnZB
51 50 ralrimiva mZn=NmA=n=NmBxnZAxnZB
52 dfss3 nZAnZBxnZAxnZB
53 51 52 sylibr mZn=NmA=n=NmBnZAnZB