Metamath Proof Explorer


Theorem mo0sn

Description: Two ways of expressing "at most one" element in a class. (Contributed by Zhi Wang, 19-Sep-2024)

Ref Expression
Assertion mo0sn *xxAA=yA=y

Proof

Step Hyp Ref Expression
1 nfv zxA
2 nfv xzA
3 eleq1w x=zxAzA
4 1 2 3 cbvmow *xxA*zzA
5 neq0 ¬A=zzA
6 5 anbi1i ¬A=*zzAzzA*zzA
7 df-eu ∃!zzAzzA*zzA
8 eu6 ∃!zzAyzzAz=y
9 6 7 8 3bitr2i ¬A=*zzAyzzAz=y
10 dfcleq A=yzzAzy
11 velsn zyz=y
12 11 bibi2i zAzyzAz=y
13 12 albii zzAzyzzAz=y
14 10 13 sylbbr zzAz=yA=y
15 14 eximi yzzAz=yyA=y
16 9 15 sylbi ¬A=*zzAyA=y
17 16 expcom *zzA¬A=yA=y
18 17 orrd *zzAA=yA=y
19 mo0 A=*zzA
20 mosn A=y*zzA
21 20 exlimiv yA=y*zzA
22 19 21 jaoi A=yA=y*zzA
23 18 22 impbii *zzAA=yA=y
24 4 23 bitri *xxAA=yA=y