Theorem qseq1 8780

Description: Equality theorem for quotient set. (Contributed by NM, 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 3310	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐴 𝑦 = [𝑥]𝐶 ↔ ∃𝑥 ∈ 𝐵 𝑦 = [𝑥]𝐶))
2	1	abbidv 2794	. 2 ⊢ (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥 ∈ 𝐴 𝑦 = [𝑥]𝐶} = {𝑦 ∣ ∃𝑥 ∈ 𝐵 𝑦 = [𝑥]𝐶})
3		df-qs 8731	. 2 ⊢ (𝐴 / 𝐶) = {𝑦 ∣ ∃𝑥 ∈ 𝐴 𝑦 = [𝑥]𝐶}
4		df-qs 8731	. 2 ⊢ (𝐵 / 𝐶) = {𝑦 ∣ ∃𝑥 ∈ 𝐵 𝑦 = [𝑥]𝐶}
5	2, 3, 4	3eqtr4g 2790	1 ⊢ (𝐴 = 𝐵 → (𝐴 / 𝐶) = (𝐵 / 𝐶))

Syntax hints:   → wi 4   = wceq 1533  {cab 2702  ∃wrex 3059  [cec 8723   / cqs 8724

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-9 2108  ax-ext 2696

This theorem depends on definitions:  df-bi 206  df-an 395  df-ex 1774  df-sb 2060  df-clab 2703  df-cleq 2717  df-rex 3060  df-qs 8731

This theorem is referenced by:  qseq1d  8783  qseq12  8785  pi1bas  25009  pstmval  33624  qseq1i  37889