Theorem disjdifg 4431

Description: A class does not intersect a relative complement of a superclass. (Contributed by NM, 24-Mar-1998.) Generalize from disjdif 4432. (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 4197	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∩ 𝐶) ⊆ 𝐵)
2		inssdif0 4328	. 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 1569   ∖ cdif 3901   ∩ cin 3903   ⊆ wss 3904  ∅c0 4285

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-fal 1582  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3416  df-v 3456  df-dif 3907  df-in 3911  df-ss 3921  df-nul 4286

This theorem is used by:  disjdif  4432