Theorem preqr1 2477

Description: Reverse equality lemma for unordered pairs. If two unordered pairs have the same second element, the first elements are equal.

Hypotheses

Ref	Expression
preqr1.1	|- A e. V
preqr1.2	|- B e. V

Assertion

Ref	Expression
preqr1	|- ({A, C} = {B, C} -> A = B)

Proof of Theorem preqr1

Step	Hyp	Ref	Expression
1		preqr1.1	. . . . 5 |- A e. V
2	1	pri1 2446	. . . 4 |- A e. {A, C}
3		eleq2 1532	. . . 4 |- ({A, C} = {B, C} -> (A e. {A, C} <-> A e. {B, C}))
4	2, 3	mpbii 193	. . 3 |- ({A, C} = {B, C} -> A e. {B, C})
5	1	elpr 2420	. . 3 |- (A e. {B, C} <-> (A = B \/ A = C))
6	4, 5	sylib 198	. 2 |- ({A, C} = {B, C} -> (A = B \/ A = C))
7		preqr1.2	. . . . 5 |- B e. V
8	7	pri1 2446	. . . 4 |- B e. {B, C}
9		eleq2 1532	. . . 4 |- ({A, C} = {B, C} -> (B e. {A, C} <-> B e. {B, C}))
10	8, 9	mpbiri 194	. . 3 |- ({A, C} = {B, C} -> B e. {A, C})
11	7	elpr 2420	. . 3 |- (B e. {A, C} <-> (B = A \/ B = C))
12	10, 11	sylib 198	. 2 |- ({A, C} = {B, C} -> (B = A \/ B = C))
13		eqcom 1474	. 2 |- (A = B <-> B = A)
14		eqeq2 1481	. 2 |- (A = C -> (B = A <-> B = C))
15	6, 12, 13, 14	oplem1 771	1 |- ({A, C} = {B, C} -> A = B)

Syntax hints:   -> wi 3   \/ wo 222   = wceq 954   e. wcel 956  Vcvv 1807  {cpr 2406

This theorem is referenced by:  preqr2 2478  opthwiener 2802

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-10 964  ax-12 966  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-16 1208  ax-11o 1216  ax-ext 1457

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 979  df-sb 1170  df-clab 1462  df-cleq 1467  df-clel 1470  df-v 1808  df-un 2046  df-sn 2408  df-pr 2409

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