Metamath Proof Explorer


Theorem finnisoeu

Description: A finite totally ordered set has a unique order isomorphism to a finite ordinal. (Contributed by Stefan O'Rear, 16-Nov-2014) (Proof shortened by Mario Carneiro, 26-Jun-2015)

Ref Expression
Assertion finnisoeu R Or A A Fin ∃! f f Isom E , R card A A

Proof

Step Hyp Ref Expression
1 eqid OrdIso R A = OrdIso R A
2 1 oiexg A Fin OrdIso R A V
3 2 adantl R Or A A Fin OrdIso R A V
4 simpr R Or A A Fin A Fin
5 wofi R Or A A Fin R We A
6 1 oiiso A Fin R We A OrdIso R A Isom E , R dom OrdIso R A A
7 4 5 6 syl2anc R Or A A Fin OrdIso R A Isom E , R dom OrdIso R A A
8 1 oien A Fin R We A dom OrdIso R A A
9 4 5 8 syl2anc R Or A A Fin dom OrdIso R A A
10 ficardid A Fin card A A
11 10 adantl R Or A A Fin card A A
12 11 ensymd R Or A A Fin A card A
13 entr dom OrdIso R A A A card A dom OrdIso R A card A
14 9 12 13 syl2anc R Or A A Fin dom OrdIso R A card A
15 1 oion A Fin dom OrdIso R A On
16 15 adantl R Or A A Fin dom OrdIso R A On
17 ficardom A Fin card A ω
18 17 adantl R Or A A Fin card A ω
19 onomeneq dom OrdIso R A On card A ω dom OrdIso R A card A dom OrdIso R A = card A
20 16 18 19 syl2anc R Or A A Fin dom OrdIso R A card A dom OrdIso R A = card A
21 14 20 mpbid R Or A A Fin dom OrdIso R A = card A
22 isoeq4 dom OrdIso R A = card A OrdIso R A Isom E , R dom OrdIso R A A OrdIso R A Isom E , R card A A
23 21 22 syl R Or A A Fin OrdIso R A Isom E , R dom OrdIso R A A OrdIso R A Isom E , R card A A
24 7 23 mpbid R Or A A Fin OrdIso R A Isom E , R card A A
25 isoeq1 f = OrdIso R A f Isom E , R card A A OrdIso R A Isom E , R card A A
26 3 24 25 spcedv R Or A A Fin f f Isom E , R card A A
27 wemoiso2 R We A * f f Isom E , R card A A
28 5 27 syl R Or A A Fin * f f Isom E , R card A A
29 df-eu ∃! f f Isom E , R card A A f f Isom E , R card A A * f f Isom E , R card A A
30 26 28 29 sylanbrc R Or A A Fin ∃! f f Isom E , R card A A