Theorem eqsbc3r 2507

Description: eqsbc3 2494 with set variable on right side of equals sign. This proof was automatically generated from the virtual deduction proof eqsbc3rVD 16664 using a translation program. (Contributed by Alan Sare, 24-Oct-2011.)

Assertion

Ref	Expression
eqsbc3r	|- (A e. B -> ([A / x]C = x <-> C = A))

Distinct variable groups:   x,C   x,A

Proof of Theorem eqsbc3r

Step	Hyp	Ref	Expression
1		eqcom 1886	. . . . . 6 |- (C = x <-> x = C)
2	1	sbcbii 2506	. . . . 5 |- (A e. B -> ([A / x]C = x <-> [A / x]x = C))
3	2	biimpd 170	. . . 4 |- (A e. B -> ([A / x]C = x -> [A / x]x = C))
4		eqsbc3 2494	. . . 4 |- (A e. B -> ([A / x]x = C <-> A = C))
5	3, 4	sylibd 219	. . 3 |- (A e. B -> ([A / x]C = x -> A = C))
6		eqcom 1886	. . 3 |- (A = C <-> C = A)
7	5, 6	syl6ib 229	. 2 |- (A e. B -> ([A / x]C = x -> C = A))
8		idd 75	. . . . 5 |- (A e. B -> (C = A -> C = A))
9	8, 6	syl6ibr 230	. . . 4 |- (A e. B -> (C = A -> A = C))
10	9, 4	sylibrd 221	. . 3 |- (A e. B -> (C = A -> [A / x]x = C))
11	10, 2	sylibrd 221	. 2 |- (A e. B -> (C = A -> [A / x]C = x))
12	7, 11	impbid 574	1 |- (A e. B -> ([A / x]C = x <-> C = A))

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

This theorem is referenced by:  sbcoreleleq 5830  sbcoreleleqVD 16683

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-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-v 2294  df-sbc 2454

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