Theorem disjdifg 4424

Description: A class does not intersect a relative complement of a superclass. (Contributed by NM, 24-Mar-1998.) Generalize from disjdif 4425. (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 4142	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∩ 𝐶) ⊆ 𝐵)
2		inssdif0 4321	. 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   ⊆ wss 3895   ∖ cdif 4045   ∩ cin 4089  ∅c0 4278

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-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-nul 4279

This theorem is used by:  disjdif  4425