Metamath Proof Explorer


Theorem rexdif1en

Description: If a set is equinumerous to a nonzero ordinal, then there exists an element in that set such that removing it leaves the set equinumerous to the predecessor of that ordinal. (Contributed by BTernaryTau, 26-Aug-2024) Generalize to all ordinals and avoid ax-un . (Revised by BTernaryTau, 5-Jan-2025)

Ref Expression
Assertion rexdif1en M On A suc M x A A x M

Proof

Step Hyp Ref Expression
1 encv A suc M A V suc M V
2 1 simpld A suc M A V
3 breng A V suc M V A suc M f f : A 1-1 onto suc M
4 1 3 syl A suc M A suc M f f : A 1-1 onto suc M
5 4 ibi A suc M f f : A 1-1 onto suc M
6 sucidg M On M suc M
7 f1ocnvdm f : A 1-1 onto suc M M suc M f -1 M A
8 7 ancoms M suc M f : A 1-1 onto suc M f -1 M A
9 6 8 sylan M On f : A 1-1 onto suc M f -1 M A
10 9 adantll A V M On f : A 1-1 onto suc M f -1 M A
11 vex f V
12 dif1enlem f V A V M On f : A 1-1 onto suc M A f -1 M M
13 11 12 mp3anl1 A V M On f : A 1-1 onto suc M A f -1 M M
14 sneq x = f -1 M x = f -1 M
15 14 difeq2d x = f -1 M A x = A f -1 M
16 15 breq1d x = f -1 M A x M A f -1 M M
17 16 rspcev f -1 M A A f -1 M M x A A x M
18 10 13 17 syl2anc A V M On f : A 1-1 onto suc M x A A x M
19 18 ex A V M On f : A 1-1 onto suc M x A A x M
20 19 exlimdv A V M On f f : A 1-1 onto suc M x A A x M
21 5 20 syl5 A V M On A suc M x A A x M
22 2 21 sylan A suc M M On A suc M x A A x M
23 22 ancoms M On A suc M A suc M x A A x M
24 23 syldbl2 M On A suc M x A A x M