Theorem qseq1 8681

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 3288	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐴 𝑦 = [𝑥]𝐶 ↔ ∃𝑥 ∈ 𝐵 𝑦 = [𝑥]𝐶))
2	1	abbidv 2797	. 2 ⊢ (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥 ∈ 𝐴 𝑦 = [𝑥]𝐶} = {𝑦 ∣ ∃𝑥 ∈ 𝐵 𝑦 = [𝑥]𝐶})
3		df-qs 8628	. 2 ⊢ (𝐴 / 𝐶) = {𝑦 ∣ ∃𝑥 ∈ 𝐴 𝑦 = [𝑥]𝐶}
4		df-qs 8628	. 2 ⊢ (𝐵 / 𝐶) = {𝑦 ∣ ∃𝑥 ∈ 𝐵 𝑦 = [𝑥]𝐶}
5	2, 3, 4	3eqtr4g 2791	1 ⊢ (𝐴 = 𝐵 → (𝐴 / 𝐶) = (𝐵 / 𝐶))

Syntax hints:   → wi 4   = wceq 1541  {cab 2709  ∃wrex 3056  [cec 8620   / cqs 8621

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-rex 3057  df-qs 8628

This theorem is referenced by:  qseq1d  8684  qseq12  8686  pi1bas  24963  pstmval  33903  qseq1i  38323