Theorem eqsbc1 3025

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2297. (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 2987	. 2 |- ( y = A -> ( [. y / x ]. x = B <-> [. A / x ]. x = B ) )
2		eqeq1 2200	. 2 |- ( y = A -> ( y = B <-> A = B ) )
3		sbsbc 2989	. . 3 |- ( [ y / x ] x = B <-> [. y / x ]. x = B )
4		eqsb1 2297	. . 3 |- ( [ y / x ] x = B <-> y = B )
5	3, 4	bitr3i 186	. 2 |- ( [. y / x ]. x = B <-> y = B )
6	1, 2, 5	vtoclbg 2821	1 |- ( A e. V -> ( [. A / x ]. x = B <-> A = B ) )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1364   [wsb 1773   e. wcel 2164   [.wsbc 2985

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-v 2762  df-sbc 2986

This theorem is referenced by:  sbceqal  3041  eqsbc2  3046