Theorem qseq2d 8685

Description: Equality theorem for quotient set, deduction form. (Contributed by Peter Mazsa, 27-May-2021.)

Hypothesis

Ref	Expression
qseq1d.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
qseq2d	⊢ (𝜑 → (𝐶 / 𝐴) = (𝐶 / 𝐵))

Proof of Theorem qseq2d

Step	Hyp	Ref	Expression
1		qseq1d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		qseq2 8682	. 2 ⊢ (𝐴 = 𝐵 → (𝐶 / 𝐴) = (𝐶 / 𝐵))
3	1, 2	syl 17	1 ⊢ (𝜑 → (𝐶 / 𝐴) = (𝐶 / 𝐵))

Syntax hints:   → wi 4   = wceq 1541   / 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-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rex 3057  df-rab 3396  df-v 3438  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4284  df-if 4476  df-sn 4577  df-pr 4579  df-op 4583  df-br 5092  df-opab 5154  df-cnv 5624  df-dm 5626  df-rn 5627  df-res 5628  df-ima 5629  df-ec 8624  df-qs 8628

This theorem is referenced by:  qustriv  33324  opprqusbas  33448  qsdrngi  33455  pstmval  33903  prjspnval2  42650  0prjspn  42660