Theorem eqsbc1 3004

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2281. (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 2966	. 2 |- ( y = A -> ( [. y / x ]. x = B <-> [. A / x ]. x = B ) )
2		eqeq1 2184	. 2 |- ( y = A -> ( y = B <-> A = B ) )
3		sbsbc 2968	. . 3 |- ( [ y / x ] x = B <-> [. y / x ]. x = B )
4		eqsb1 2281	. . 3 |- ( [ y / x ] x = B <-> y = B )
5	3, 4	bitr3i 186	. 2 |- ( [. y / x ]. x = B <-> y = B )
6	1, 2, 5	vtoclbg 2800	1 |- ( A e. V -> ( [. A / x ]. x = B <-> A = B ) )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1353   [wsb 1762   e. wcel 2148   [.wsbc 2964

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 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2741  df-sbc 2965

This theorem is referenced by:  sbceqal  3020  eqsbc2  3025