Metamath Proof Explorer


Theorem fseqdom

Description: One half of fseqen . (Contributed by Mario Carneiro, 18-Nov-2014)

Ref Expression
Assertion fseqdom AVω×AnωAn

Proof

Step Hyp Ref Expression
1 omex ωV
2 ovex AnV
3 1 2 iunex nωAnV
4 xp1st xω×A1stxω
5 peano2 1stxωsuc1stxω
6 4 5 syl xω×Asuc1stxω
7 xp2nd xω×A2ndxA
8 fconst6g 2ndxAsuc1stx×2ndx:suc1stxA
9 7 8 syl xω×Asuc1stx×2ndx:suc1stxA
10 9 adantl AVxω×Asuc1stx×2ndx:suc1stxA
11 elmapg AVsuc1stxωsuc1stx×2ndxAsuc1stxsuc1stx×2ndx:suc1stxA
12 6 11 sylan2 AVxω×Asuc1stx×2ndxAsuc1stxsuc1stx×2ndx:suc1stxA
13 10 12 mpbird AVxω×Asuc1stx×2ndxAsuc1stx
14 oveq2 n=suc1stxAn=Asuc1stx
15 14 eliuni suc1stxωsuc1stx×2ndxAsuc1stxsuc1stx×2ndxnωAn
16 6 13 15 syl2an2 AVxω×Asuc1stx×2ndxnωAn
17 16 ex AVxω×Asuc1stx×2ndxnωAn
18 nsuceq0 suc1stx
19 fvex 2ndxV
20 19 snnz 2ndx
21 xp11 suc1stx2ndxsuc1stx×2ndx=suc1sty×2ndysuc1stx=suc1sty2ndx=2ndy
22 18 20 21 mp2an suc1stx×2ndx=suc1sty×2ndysuc1stx=suc1sty2ndx=2ndy
23 xp1st yω×A1styω
24 peano4 1stxω1styωsuc1stx=suc1sty1stx=1sty
25 4 23 24 syl2an xω×Ayω×Asuc1stx=suc1sty1stx=1sty
26 sneqbg 2ndxV2ndx=2ndy2ndx=2ndy
27 19 26 mp1i xω×Ayω×A2ndx=2ndy2ndx=2ndy
28 25 27 anbi12d xω×Ayω×Asuc1stx=suc1sty2ndx=2ndy1stx=1sty2ndx=2ndy
29 22 28 bitrid xω×Ayω×Asuc1stx×2ndx=suc1sty×2ndy1stx=1sty2ndx=2ndy
30 xpopth xω×Ayω×A1stx=1sty2ndx=2ndyx=y
31 29 30 bitrd xω×Ayω×Asuc1stx×2ndx=suc1sty×2ndyx=y
32 31 a1i AVxω×Ayω×Asuc1stx×2ndx=suc1sty×2ndyx=y
33 17 32 dom2d AVnωAnVω×AnωAn
34 3 33 mpi AVω×AnωAn