Theorem disj2 3480

Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 17-May-1998.)

Assertion

Ref	Expression
disj2	⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))

Proof of Theorem disj2

Step	Hyp	Ref	Expression
1		ssv 3179	. 2 ⊢ 𝐴 ⊆ V
2		reldisj 3476	. 2 ⊢ (𝐴 ⊆ V → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵)))
3	1, 2	ax-mp 5	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))

Syntax hints:   ↔ wb 105   = wceq 1353  Vcvv 2739   ∖ cdif 3128   ∩ cin 3130   ⊆ wss 3131  ∅c0 3424

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 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-v 2741  df-dif 3133  df-in 3137  df-ss 3144  df-nul 3425

This theorem is referenced by:  ssindif0im  3484  intirr  5017  setsresg  12502  setscom  12504