Theorem qseq2 5976

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

Assertion

Ref	Expression
qseq2	⊢ (A = B → (C / A) = (C / B))

Proof of Theorem qseq2

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

Step	Hyp	Ref	Expression
1		eceq2 5964	. . . . 5 ⊢ (A = B → [x]A = [x]B)
2	1	eqeq2d 2364	. . . 4 ⊢ (A = B → (y = [x]A ↔ y = [x]B))
3	2	rexbidv 2636	. . 3 ⊢ (A = B → (∃x ∈ C y = [x]A ↔ ∃x ∈ C y = [x]B))
4	3	abbidv 2468	. 2 ⊢ (A = B → {y ∣ ∃x ∈ C y = [x]A} = {y ∣ ∃x ∈ C y = [x]B})
5		df-qs 5952	. 2 ⊢ (C / A) = {y ∣ ∃x ∈ C y = [x]A}
6		df-qs 5952	. 2 ⊢ (C / B) = {y ∣ ∃x ∈ C y = [x]B}
7	4, 5, 6	3eqtr4g 2410	1 ⊢ (A = B → (C / A) = (C / B))

Syntax hints:   → wi 4   = wceq 1642  {cab 2339  ∃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-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-rex 2621  df-br 4641  df-ima 4728  df-ec 5948  df-qs 5952

This theorem is referenced by: (None)