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 ∈ A → x ∈ B))
2	1	con3d 87	. . . 4 ⊢ (A ⊆ B → (¬ x ∈ B → ¬ x ∈ A))
3	2	anim2d 433	. . 3 ⊢ (A ⊆ B → ((x ∈ C ∧ ¬ x ∈ B) → (x ∈ C ∧ ¬ x ∈ A)))
4		eldif 1496	. . 3 ⊢ (x ∈ (C ∖ B) ↔ (x ∈ C ∧ ¬ x ∈ B))
5		eldif 1496	. . 3 ⊢ (x ∈ (C ∖ A) ↔ (x ∈ C ∧ ¬ x ∈ A))
6	3, 4, 5	3imtr4g 426	. 2 ⊢ (A ⊆ B → (x ∈ (C ∖ B) → x ∈ (C ∖ A)))
7	6	ssrdv 1509	1 ⊢ (A ⊆ B → (C ∖ B) ⊆ (C ∖ A))

Syntax hints:  ¬ wn 1   → wi 2   ∧ wa 196   ∈ 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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