Theorem eqsbc1 2994

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2274. (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 2957	. 2 |- ( y = A -> ( [. y / x ]. x = B <-> [. A / x ]. x = B ) )
2		eqeq1 2177	. 2 |- ( y = A -> ( y = B <-> A = B ) )
3		sbsbc 2959	. . 3 |- ( [ y / x ] x = B <-> [. y / x ]. x = B )
4		eqsb1 2274	. . 3 |- ( [ y / x ] x = B <-> y = B )
5	3, 4	bitr3i 185	. 2 |- ( [. y / x ]. x = B <-> y = B )
6	1, 2, 5	vtoclbg 2791	1 |- ( A e. V -> ( [. A / x ]. x = B <-> A = B ) )

Syntax hints:   -> wi 4   <-> wb 104   = wceq 1348   [wsb 1755   e. wcel 2141   [.wsbc 2955

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-sbc 2956

This theorem is referenced by:  sbceqal  3010  eqsbc2  3015