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 → (∃x ∈ A y = [x]C ↔ ∃x ∈ B y = [x]C))
2	1	abbidv 2468	. 2 ⊢ (A = B → {y ∣ ∃x ∈ A y = [x]C} = {y ∣ ∃x ∈ B y = [x]C})
3		df-qs 5952	. 2 ⊢ (A / C) = {y ∣ ∃x ∈ A y = [x]C}
4		df-qs 5952	. 2 ⊢ (B / C) = {y ∣ ∃x ∈ B y = [x]C}
5	2, 3, 4	3eqtr4g 2410	1 ⊢ (A = B → (A / C) = (B / C))

This proof depends on syntax axioms:   → wi 4   = wceq 1642  {cab 2339  ∃wrex 2616  [cec 5946   / cqs 5947

This proof depends on 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 proof 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 used by: (None)