Theorem qseq1d 8766

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 8763	. 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 1570   / cqs 8702

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 1943  ax-6 2000  ax-7 2041  ax-9 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-rex 3093  df-qs 8709

This theorem is used by:  fracbas  33657  n0elim  39425