Theorem difdif 2166

Description: Double class difference. Exercise 11 of [TakeutiZaring] p. 22.

Assertion

Ref	Expression
difdif	|- (A \ (B \ A)) = A

Proof of Theorem difdif

Step	Hyp	Ref	Expression
1		eldif 2057	. . . . . 6 |- (x e. (B \ A) <-> (x e. B /\ -. x e. A))
2	1	negbii 187	. . . . 5 |- (-. x e. (B \ A) <-> -. (x e. B /\ -. x e. A))
3		iman 237	. . . . 5 |- ((x e. B -> x e. A) <-> -. (x e. B /\ -. x e. A))
4	2, 3	bitr4 176	. . . 4 |- (-. x e. (B \ A) <-> (x e. B -> x e. A))
5	4	anbi2i 480	. . 3 |- ((x e. A /\ -. x e. (B \ A)) <-> (x e. A /\ (x e. B -> x e. A)))
6		pm4.45im 332	. . 3 |- (x e. A <-> (x e. A /\ (x e. B -> x e. A)))
7	5, 6	bitr4 176	. 2 |- ((x e. A /\ -. x e. (B \ A)) <-> x e. A)
8	7	difeqri 2160	1 |- (A \ (B \ A)) = A

Syntax hints:  -. wn 2   -> wi 3   /\ wa 223   = wceq 956   e. wcel 958   \ cdif 2044

This theorem is referenced by:  dif0 2335

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-v 1812  df-dif 2049

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