Theorem eqsbc1 3038

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2309. (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 3000	. 2 |- ( y = A -> ( [. y / x ]. x = B <-> [. A / x ]. x = B ) )
2		eqeq1 2212	. 2 |- ( y = A -> ( y = B <-> A = B ) )
3		sbsbc 3002	. . 3 |- ( [ y / x ] x = B <-> [. y / x ]. x = B )
4		eqsb1 2309	. . 3 |- ( [ y / x ] x = B <-> y = B )
5	3, 4	bitr3i 186	. 2 |- ( [. y / x ]. x = B <-> y = B )
6	1, 2, 5	vtoclbg 2834	1 |- ( A e. V -> ( [. A / x ]. x = B <-> A = B ) )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1373   [wsb 1785   e. wcel 2176   [.wsbc 2998

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 711  ax-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-10 1528  ax-11 1529  ax-i12 1530  ax-bndl 1532  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-ext 2187

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1484  df-sb 1786  df-clab 2192  df-cleq 2198  df-clel 2201  df-nfc 2337  df-v 2774  df-sbc 2999

This theorem is referenced by:  sbceqal  3054  eqsbc2  3059