Theorem qseq1 3225

Description: Equality theorem for quotient set.

Assertion

Ref	Expression
qseq1	⊢ (A = B → (A / C) = (B / C))

Proof of Theorem qseq1

Step	Hyp	Ref	Expression
1		rexeq 1325	. . 3 ⊢ (A = B → (∃x ∈ A y = [x]C ↔ ∃x ∈ B y = [x]C))
2	1	biabdv 1183	. 2 ⊢ (A = B → {y∣∃x ∈ A y = [x]C} = {y∣∃x ∈ B y = [x]C})
3		df-qs 3205	. 2 ⊢ (A / C) = {y∣∃x ∈ A y = [x]C}
4		df-qs 3205	. 2 ⊢ (B / C) = {y∣∃x ∈ B y = [x]C}
5	2, 3, 4	3eqtr4g 1147	1 ⊢ (A = B → (A / C) = (B / C))

Syntax hints:   → wi 2  {cab 1090   = wceq 1091  ∃wrex 1202  [cec 3198   / cqs 3199

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-rex 1206  df-qs 3205

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