Theorem difeqri 1589

Description: Inference from membership to difference.

Hypothesis

Ref	Expression
difeqri.1	|- ((x e. A /\ -. x e. B) <-> x e. C)

Assertion

Ref	Expression
difeqri	|- (A \ B) = C

Distinct variable group(s):   x,A   x,B   x,C

Proof of Theorem difeqri

Step	Hyp	Ref	Expression
1		df-dif 1489	. 2 |- (A \ B) = {x | (x e. A /\ -. x e. B)}
2		difeqri.1	. . . 4 |- ((x e. A /\ -. x e. B) <-> x e. C)
3	2	bicomi 150	. . 3 |- (x e. C <-> (x e. A /\ -. x e. B))
4	3	biabri 1180	. 2 |- C = {x | (x e. A /\ -. x e. B)}
5	1, 4	eqtr4 1122	1 |- (A \ B) = C

Syntax hints:  -. wn 1   <-> wb 127   /\ wa 196  {cab 1090   = wceq 1091   e. wcel 1092   \ cdif 1484

This theorem is referenced by:  difdif 1595  ddif 1597  difab 1693

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-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-dif 1489

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