Theorem qseq1 5975

Description: Equality theorem for quotient set. (Contributed by set.mm contributors, 23-Jul-1995.)

Assertion

Ref	Expression
qseq1	|- (A = B -> (A/.C) = (B/.C))

Proof of Theorem qseq1

Dummy variables x y are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		rexeq 2809	. . 3 |- (A = B -> (E.x e. A y = [x]C <-> E.x e. B y = [x]C))
2	1	abbidv 2468	. 2 |- (A = B -> {y | E.x e. A y = [x]C} = {y | E.x e. B y = [x]C})
3		df-qs 5952	. 2 |- (A/.C) = {y | E.x e. A y = [x]C}
4		df-qs 5952	. 2 |- (B/.C) = {y | E.x e. B y = [x]C}
5	2, 3, 4	3eqtr4g 2410	1 |- (A = B -> (A/.C) = (B/.C))

Syntax hints:   -> wi 4   = wceq 1642  {cab 2339  E.wrex 2616  [cec 5946  /.cqs 5947

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-rex 2621  df-qs 5952

This theorem is referenced by: (None)