Theorem disjdifg 4428

Description: A class does not intersect a relative complement of a superclass. (Contributed by NM, 24-Mar-1998.) Generalize from disjdif 4429. (Revised by BJ, 19-Jul-2026.)

Assertion

Ref	Expression
disjdifg	⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∩ (𝐶 ∖ 𝐵)) = ∅)

Proof of Theorem disjdifg

Step	Hyp	Ref	Expression
1		ssinss1 4145	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∩ 𝐶) ⊆ 𝐵)
2		inssdif0 4325	. 2 ⊢ ((𝐴 ∩ 𝐶) ⊆ 𝐵 ↔ (𝐴 ∩ (𝐶 ∖ 𝐵)) = ∅)
3	1, 2	sylib 221	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∩ (𝐶 ∖ 𝐵)) = ∅)

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ⊆ wss 3900   ∖ cdif 4048   ∩ cin 4092  ∅c0 4281

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-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-nul 4282

This theorem is used by:  disjdif  4429