Theorem qseq2d 8812

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 8809	. 2 ⊢ (𝐴 = 𝐵 → (𝐶 / 𝐴) = (𝐶 / 𝐵))
3	1, 2	syl 18	1 ⊢ (𝜑 → (𝐶 / 𝐴) = (𝐶 / 𝐵))

This proof depends on syntax axioms:   → wi 4   = wceq 1570   / cqs 8747

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5172  df-cnv 5664  df-dm 5666  df-rn 5667  df-res 5668  df-ima 5669  df-ec 8750  df-qs 8754

This theorem is used by:  qustriv  19428  opprqusbas  34096  qsdrngi  34103  pstmval  34610  prjspnval2  43850  0prjspn  43868