Theorem sscon 4106

Description: Contraposition law for subsets. Exercise 15 of [TakeutiZaring] p. 22. (Contributed by NM, 22-Mar-1998.)

Assertion

Ref	Expression
sscon	⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))

Proof of Theorem sscon

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ssel 3940	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	con3d 152	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (¬ 𝑥 ∈ 𝐵 → ¬ 𝑥 ∈ 𝐴))
3	2	anim2d 612	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) → (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐴)))
4		eldif 3924	. . 3 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))
5		eldif 3924	. . 3 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐴) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐴))
6	3, 4, 5	3imtr4g 296	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ (𝐶 ∖ 𝐵) → 𝑥 ∈ (𝐶 ∖ 𝐴)))
7	6	ssrdv 3952	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   ∈ wcel 2109   ∖ cdif 3911   ⊆ wss 3914

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-v 3449  df-dif 3917  df-ss 3931

This theorem is referenced by:  sscond  4109  complss  4114  sscon34b  4267  sorpsscmpl  7710  sbthlem1  9051  sbthlem2  9052  cantnfp1lem1  9631  cantnfp1lem3  9633  isf34lem7  10332  isf34lem6  10333  setsres  17148  mplsubglem  21908  cctop  22893  clsval2  22937  ntrss  22942  hauscmplem  23293  ptbasin  23464  cfinfil  23780  csdfil  23781  uniioombllem5  25488  kur14lem6  35198  bj-2upln1upl  37012  dvasin  37698  readvrec2  42349  clsk3nimkb  44029  fourierdlem62  46166  caragendifcl  46512