Theorem eqsbc1 3000

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2279. (Contributed by Andrew Salmon, 29-Jun-2011.)

Assertion

Ref	Expression
eqsbc1	|- ( A e. V -> ( [. A / x ]. x = B <-> A = B ) )

Distinct variable group:   x, B

Allowed substitution hints:   A( x)   V( x)

Proof of Theorem eqsbc1

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfsbcq 2962	. 2 |- ( y = A -> ( [. y / x ]. x = B <-> [. A / x ]. x = B ) )
2		eqeq1 2182	. 2 |- ( y = A -> ( y = B <-> A = B ) )
3		sbsbc 2964	. . 3 |- ( [ y / x ] x = B <-> [. y / x ]. x = B )
4		eqsb1 2279	. . 3 |- ( [ y / x ] x = B <-> y = B )
5	3, 4	bitr3i 186	. 2 |- ( [. y / x ]. x = B <-> y = B )
6	1, 2, 5	vtoclbg 2796	1 |- ( A e. V -> ( [. A / x ]. x = B <-> A = B ) )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1353   [wsb 1760   e. wcel 2146   [.wsbc 2960

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1445  ax-7 1446  ax-gen 1447  ax-ie1 1491  ax-ie2 1492  ax-8 1502  ax-10 1503  ax-11 1504  ax-i12 1505  ax-bndl 1507  ax-4 1508  ax-17 1524  ax-i9 1528  ax-ial 1532  ax-i5r 1533  ax-ext 2157

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1459  df-sb 1761  df-clab 2162  df-cleq 2168  df-clel 2171  df-nfc 2306  df-v 2737  df-sbc 2961

This theorem is referenced by:  sbceqal  3016  eqsbc2  3021