Theorem sbceqal 2502

Description: Set theory version of sbeqal1 16355. (Contributed by Andrew Salmon, 28-Jun-2011.)

Assertion

Ref	Expression
sbceqal	|- (A e. C -> (A.x(x = A -> x = B) -> A = B))

Distinct variable groups:   x,B   x,A

Proof of Theorem sbceqal

Step	Hyp	Ref	Expression
1		a4sbc 2457	. 2 |- (A e. C -> (A.x(x = A -> x = B) -> [A / x](x = A -> x = B)))
2		sbcimg 2496	. . 3 |- (A e. C -> ([A / x](x = A -> x = B) <-> ([A / x]x = A -> [A / x]x = B)))
3		eqid 1884	. . . . 5 |- A = A
4		eqeq1 1890	. . . . . 6 |- (x = A -> (x = A <-> A = A))
5	4	sbcieg 2484	. . . . 5 |- (A e. C -> ([A / x]x = A <-> A = A))
6	3, 5	mpbiri 211	. . . 4 |- (A e. C -> [A / x]x = A)
7		pm5.5 804	. . . 4 |- ([A / x]x = A -> (([A / x]x = A -> [A / x]x = B) <-> [A / x]x = B))
8	6, 7	syl 12	. . 3 |- (A e. C -> (([A / x]x = A -> [A / x]x = B) <-> [A / x]x = B))
9		eqsbc3 2494	. . 3 |- (A e. C -> ([A / x]x = B <-> A = B))
10	2, 8, 9	3bitrd 603	. 2 |- (A e. C -> ([A / x](x = A -> x = B) <-> A = B))
11	1, 10	sylibd 219	1 |- (A e. C -> (A.x(x = A -> x = B) -> A = B))

Syntax hints:   -> wi 3   <-> wb 163  A.wal 1296   = wceq 1298   e. wcel 1300  [wsbc 1534

This theorem is referenced by:  sbeqalb 2503

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-9 1307  ax-10 1308  ax-11 1309  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-or 241  df-an 242  df-3an 860  df-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-v 2294  df-sbc 2454

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