Theorem sscon 1599

Description: Contraposition law for subsets. Exercise 15 of [TakeutiZaring] p. 22.

Assertion

Ref	Expression
sscon	|- (A (_ B -> (C \ B) (_ (C \ A))

Proof of Theorem sscon

Step	Hyp	Ref	Expression
1		ssel 1502	. . . . 5 |- (A (_ B -> (x e. A -> x e. B))
2	1	con3d 87	. . . 4 |- (A (_ B -> (-. x e. B -> -. x e. A))
3	2	anim2d 433	. . 3 |- (A (_ B -> ((x e. C /\ -. x e. B) -> (x e. C /\ -. x e. A)))
4		eldif 1496	. . 3 |- (x e. (C \ B) <-> (x e. C /\ -. x e. B))
5		eldif 1496	. . 3 |- (x e. (C \ A) <-> (x e. C /\ -. x e. A))
6	3, 4, 5	3imtr4g 426	. 2 |- (A (_ B -> (x e. (C \ B) -> x e. (C \ A)))
7	6	ssrdv 1509	1 |- (A (_ B -> (C \ B) (_ (C \ A))

Syntax hints:  -. wn 1   -> wi 2   /\ wa 196   e. wcel 1092   \ cdif 1484   (_ wss 1487

This theorem is referenced by:  sbthlem1 3349  sbthlem2 3350

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-v 1349  df-dif 1489  df-in 1491  df-ss 1492

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