Theorem eqsbc2 3104

Description: Substitution for the right-hand side in an equality. This proof was automatically generated from the virtual deduction proof eqsbc2VD in set.mm using a translation program. (Contributed by Alan Sare, 24-Oct-2011.)

Assertion

Ref	Expression
eqsbc2	|- (A e. B -> ( [.A / x ].C = x <-> C = A))

Distinct variable groups:   x,C   x,A

Allowed substitution hint:   B(x)

Proof of Theorem eqsbc2

Step	Hyp	Ref	Expression
1		eqcom 2355	. . . . . 6 |- (C = x <-> x = C)
2	1	sbcbii 3102	. . . . 5 |- ( [.A / x ].C = x <-> [.A / x ].x = C)
3	2	biimpi 186	. . . 4 |- ( [.A / x ].C = x -> [.A / x ].x = C)
4		eqsbc1 3086	. . . 4 |- (A e. B -> ( [.A / x ].x = C <-> A = C))
5	3, 4	syl5ib 210	. . 3 |- (A e. B -> ( [.A / x ].C = x -> A = C))
6		eqcom 2355	. . 3 |- (A = C <-> C = A)
7	5, 6	syl6ib 217	. 2 |- (A e. B -> ( [.A / x ].C = x -> C = A))
8		idd 21	. . . . 5 |- (A e. B -> (C = A -> C = A))
9	8, 6	syl6ibr 218	. . . 4 |- (A e. B -> (C = A -> A = C))
10	9, 4	sylibrd 225	. . 3 |- (A e. B -> (C = A -> [.A / x ].x = C))
11	10, 2	syl6ibr 218	. 2 |- (A e. B -> (C = A -> [.A / x ].C = x))
12	7, 11	impbid 183	1 |- (A e. B -> ( [.A / x ].C = x <-> C = A))

Syntax hints:   -> wi 4   <-> wb 176   = wceq 1642   e. wcel 1710   [.wsbc 3047

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-sbc 3048

This theorem is referenced by: (None)