Theorem difdisj 1758

Description: A class and its relative complement are disjoint. Theorem 38 of [Suppes] p. 29.

Assertion

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

Proof of Theorem difdisj

Step	Hyp	Ref	Expression
1		inss1 1657	. 2 ⊢ (A ∩ B) ⊆ A
2		inssdif0 1754	. 2 ⊢ ((A ∩ B) ⊆ A ↔ (A ∩ (B ∖ A)) = ∅)
3	1, 2	mpbi 164	1 ⊢ (A ∩ (B ∖ A)) = ∅

Syntax hints:   = wceq 1091   ∖ cdif 1484   ∩ cin 1486   ⊆ wss 1487  ∅c0 1707

This theorem is referenced by:  undifv 1760  difdifdir 1765  undom 3342  sbthlem7 3355  sbthlem8 3356  mapdom2lem 3388  mapdom2 3389  limensuci 3401  phplem3 3405  pssnn 3428  unfi 3441  fodomb 3615  ruclem6 4890  ruclem7 4891  infxpidmlem11 4943

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-dif 1489  df-in 1491  df-ss 1492  df-nul 1708

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