Theorem disjdifg 3598

Description: A class and its relative complement are disjoint. (Contributed by NM, 24-Mar-1998.) Generalize from disjdif 3599. (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 3460	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∩ 𝐶) ⊆ 𝐵)
2		inssdif0im 3592	. 2 ⊢ ((𝐴 ∩ 𝐶) ⊆ 𝐵 → (𝐴 ∩ (𝐶 ∖ 𝐵)) = ∅)
3	1, 2	syl 14	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∩ (𝐶 ∖ 𝐵)) = ∅)

Syntax hints:   → wi 4   = wceq 1402   ∖ cdif 3217   ∩ cin 3219   ⊆ wss 3220  ∅c0 3520

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-dif 3222  df-in 3226  df-ss 3233  df-nul 3521

This theorem is referenced by:  disjdif  3599