Theorem qseq1 6700

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 2709	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐴 𝑦 = [𝑥]𝐶 ↔ ∃𝑥 ∈ 𝐵 𝑦 = [𝑥]𝐶))
2	1	abbidv 2327	. 2 ⊢ (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥 ∈ 𝐴 𝑦 = [𝑥]𝐶} = {𝑦 ∣ ∃𝑥 ∈ 𝐵 𝑦 = [𝑥]𝐶})
3		df-qs 6656	. 2 ⊢ (𝐴 / 𝐶) = {𝑦 ∣ ∃𝑥 ∈ 𝐴 𝑦 = [𝑥]𝐶}
4		df-qs 6656	. 2 ⊢ (𝐵 / 𝐶) = {𝑦 ∣ ∃𝑥 ∈ 𝐵 𝑦 = [𝑥]𝐶}
5	2, 3, 4	3eqtr4g 2267	1 ⊢ (𝐴 = 𝐵 → (𝐴 / 𝐶) = (𝐵 / 𝐶))

Syntax hints:   → wi 4   = wceq 1375  {cab 2195  ∃wrex 2489  [cec 6648   / cqs 6649

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 713  ax-5 1473  ax-7 1474  ax-gen 1475  ax-ie1 1519  ax-ie2 1520  ax-8 1530  ax-10 1531  ax-11 1532  ax-i12 1533  ax-bndl 1535  ax-4 1536  ax-17 1552  ax-i9 1556  ax-ial 1560  ax-i5r 1561  ax-ext 2191

This theorem depends on definitions:  df-bi 117  df-tru 1378  df-nf 1487  df-sb 1789  df-clab 2196  df-cleq 2202  df-clel 2205  df-nfc 2341  df-rex 2494  df-qs 6656

This theorem is referenced by: (None)