Theorem disj2 3300

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 3020	. 2 ⊢ 𝐴 ⊆ V
2		reldisj 3296	. 2 ⊢ (𝐴 ⊆ V → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵)))
3	1, 2	ax-mp 7	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))

Syntax hints:   ↔ wb 103   = wceq 1285  Vcvv 2602   ∖ cdif 2971   ∩ cin 2973   ⊆ wss 2974  ∅c0 3252

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 577  ax-in2 578  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064

This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-nfc 2209  df-ral 2354  df-v 2604  df-dif 2976  df-in 2980  df-ss 2987  df-nul 3253

This theorem is referenced by:  ssindif0im  3304  intirr  4735