Theorem disjdif 3382

Description: A class and its relative complement are disjoint. Theorem 38 of [Suppes] p. 29. (Contributed by NM, 24-Mar-1998.)

Assertion

Ref	Expression
disjdif	⊢ (𝐴 ∩ (𝐵 ∖ 𝐴)) = ∅

Proof of Theorem disjdif

Step	Hyp	Ref	Expression
1		inss1 3243	. 2 ⊢ (𝐴 ∩ 𝐵) ⊆ 𝐴
2		inssdif0im 3377	. 2 ⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐴 → (𝐴 ∩ (𝐵 ∖ 𝐴)) = ∅)
3	1, 2	ax-mp 7	1 ⊢ (𝐴 ∩ (𝐵 ∖ 𝐴)) = ∅

Syntax hints:   = wceq 1299   ∖ cdif 3018   ∩ cin 3020   ⊆ wss 3021  ∅c0 3310

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 584  ax-in2 585  ax-io 671  ax-5 1391  ax-7 1392  ax-gen 1393  ax-ie1 1437  ax-ie2 1438  ax-8 1450  ax-10 1451  ax-11 1452  ax-i12 1453  ax-bndl 1454  ax-4 1455  ax-17 1474  ax-i9 1478  ax-ial 1482  ax-i5r 1483  ax-ext 2082

This theorem depends on definitions:  df-bi 116  df-tru 1302  df-nf 1405  df-sb 1704  df-clab 2087  df-cleq 2093  df-clel 2096  df-nfc 2229  df-v 2643  df-dif 3023  df-in 3027  df-ss 3034  df-nul 3311

This theorem is referenced by:  ssdifin0  3391  difdifdirss  3394  fvsnun1  5549  fvsnun2  5550  phplem2  6676  unfiin  6743  xpfi  6747  sbthlem7  6779  sbthlemi8  6780  exmidfodomrlemim  6966  fihashssdif  10405  zfz1isolem1  10424  fsumlessfi  11068  setsfun  11776  setsfun0  11777  setsslid  11791