Theorem qseq1d 8783

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
qseq1d	⊢ (𝜑 → (𝐴 / 𝐶) = (𝐵 / 𝐶))

Proof of Theorem qseq1d

Step	Hyp	Ref	Expression
1		qseq1d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		qseq1 8780	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 / 𝐶) = (𝐵 / 𝐶))
3	1, 2	syl 18	1 ⊢ (𝜑 → (𝐴 / 𝐶) = (𝐵 / 𝐶))

This proof depends on syntax axioms:   → wi 4   = wceq 1569   / cqs 8719

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-rex 3087  df-qs 8726

This theorem is used by:  fracbas  33913  n0elim  39760