Theorem opthreg 3455

Description: Theorem for alternate representation of ordered pairs, requiring Regularity. Exercise 34 of [Enderton] p. 207.

Hypotheses

Ref	Expression
preleq.1	|- A e. V
preleq.2	|- B e. V
preleq.3	|- C e. V
preleq.4	|- D e. V

Assertion

Ref	Expression
opthreg	|- ({A, {A, B}} = {C, {C, D}} -> (A = C /\ B = D))

Proof of Theorem opthreg

Step	Hyp	Ref	Expression
1		preleq.1	. . . . 5 |- A e. V
2	1	pri1 1841	. . . 4 |- A e. {A, B}
3		preleq.3	. . . . 5 |- C e. V
4	3	pri1 1841	. . . 4 |- C e. {C, D}
5	2, 4	pm3.2i 234	. . 3 |- (A e. {A, B} /\ C e. {C, D})
6		prex 1892	. . . 4 |- {A, B} e. V
7		prex 1892	. . . 4 |- {C, D} e. V
8	1, 6, 3, 7	preleq 3454	. . 3 |- (((A e. {A, B} /\ C e. {C, D}) /\ {A, {A, B}} = {C, {C, D}}) -> (A = C /\ {A, B} = {C, D}))
9	5, 8	mpan 518	. 2 |- ({A, {A, B}} = {C, {C, D}} -> (A = C /\ {A, B} = {C, D}))
10		preq1 1870	. . . . 5 |- (A = C -> {A, B} = {C, B})
11	10	cleq1d 1109	. . . 4 |- (A = C -> ({A, B} = {C, D} <-> {C, B} = {C, D}))
12		preleq.2	. . . . 5 |- B e. V
13		preleq.4	. . . . 5 |- D e. V
14	12, 13	prer2 1873	. . . 4 |- ({C, B} = {C, D} -> B = D)
15	11, 14	syl6bi 187	. . 3 |- (A = C -> ({A, B} = {C, D} -> B = D))
16	15	imdistani 340	. 2 |- ((A = C /\ {A, B} = {C, D}) -> (A = C /\ B = D))
17	9, 16	syl 12	1 |- ({A, {A, B}} = {C, {C, D}} -> (A = C /\ B = D))

Syntax hints:   -> wi 2   /\ wa 196   = wceq 1091   e. wcel 1092  Vcvv 1348  {cpr 1809

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-13 804  ax-14 805  ax-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077  ax-reg 1078

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-opab 2098  df-eprel 2122  df-fr 2169

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