Metamath Proof Explorer


Theorem opthhausdorff0

Description: Justification theorem for the ordered pair definition of Felix Hausdorff in "Grundzüge der Mengenlehre" ("Basics of Set Theory"), 1914, p. 32: <. A , B >._H = { { A , O } , { B , T } } . Hausdorff used 1 and 2 instead of O and T , but actually, any two different fixed sets will do (e.g., O = (/) and T = { (/) } , see 0nep0 ). Furthermore, Hausdorff demanded that O and T are both different from A as well as B , which is actually not necessary if all involved classes exist as sets (i.e. are not proper classes), in contrast to opthhausdorff . See df-op for other ordered pair definitions. (Contributed by AV, 12-Jun-2022)

Ref Expression
Hypotheses opthhausdorff0.a AV
opthhausdorff0.b BV
opthhausdorff0.c CV
opthhausdorff0.d DV
opthhausdorff0.1 OV
opthhausdorff0.2 TV
opthhausdorff0.3 OT
Assertion opthhausdorff0 AOBT=CODTA=CB=D

Proof

Step Hyp Ref Expression
1 opthhausdorff0.a AV
2 opthhausdorff0.b BV
3 opthhausdorff0.c CV
4 opthhausdorff0.d DV
5 opthhausdorff0.1 OV
6 opthhausdorff0.2 TV
7 opthhausdorff0.3 OT
8 prex AOV
9 prex BTV
10 prex COV
11 prex DTV
12 8 9 10 11 preq12b AOBT=CODTAO=COBT=DTAO=DTBT=CO
13 1 3 preqr1 AO=COA=C
14 2 4 preqr1 BT=DTB=D
15 13 14 anim12i AO=COBT=DTA=CB=D
16 1 5 4 6 preq12b AO=DTA=DO=TA=TO=D
17 eqneqall O=TOTBT=COA=CB=D
18 7 17 mpi O=TBT=COA=CB=D
19 18 adantl A=DO=TBT=COA=CB=D
20 2 6 3 5 preq12b BT=COB=CT=OB=OT=C
21 eqneqall O=TOTA=TO=DA=CB=D
22 7 21 mpi O=TA=TO=DA=CB=D
23 22 eqcoms T=OA=TO=DA=CB=D
24 23 adantl B=CT=OA=TO=DA=CB=D
25 simpl A=TO=DA=T
26 simpr B=OT=CT=C
27 25 26 sylan9eqr B=OT=CA=TO=DA=C
28 simpl B=OT=CB=O
29 simpr A=TO=DO=D
30 28 29 sylan9eq B=OT=CA=TO=DB=D
31 27 30 jca B=OT=CA=TO=DA=CB=D
32 31 ex B=OT=CA=TO=DA=CB=D
33 24 32 jaoi B=CT=OB=OT=CA=TO=DA=CB=D
34 20 33 sylbi BT=COA=TO=DA=CB=D
35 34 com12 A=TO=DBT=COA=CB=D
36 19 35 jaoi A=DO=TA=TO=DBT=COA=CB=D
37 16 36 sylbi AO=DTBT=COA=CB=D
38 37 imp AO=DTBT=COA=CB=D
39 15 38 jaoi AO=COBT=DTAO=DTBT=COA=CB=D
40 12 39 sylbi AOBT=CODTA=CB=D
41 preq1 A=CAO=CO
42 41 adantr A=CB=DAO=CO
43 preq1 B=DBT=DT
44 43 adantl A=CB=DBT=DT
45 42 44 preq12d A=CB=DAOBT=CODT
46 40 45 impbii AOBT=CODTA=CB=D