Theorem sscon 4109

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 3943	. . . . 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 3927	. . 3 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))
5		eldif 3927	. . 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 3955	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   ∈ wcel 2109   ∖ cdif 3914   ⊆ wss 3917

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 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-v 3452  df-dif 3920  df-ss 3934

This theorem is referenced by:  sscond  4112  complss  4117  sscon34b  4270  sorpsscmpl  7713  sbthlem1  9057  sbthlem2  9058  cantnfp1lem1  9638  cantnfp1lem3  9640  isf34lem7  10339  isf34lem6  10340  setsres  17155  mplsubglem  21915  cctop  22900  clsval2  22944  ntrss  22949  hauscmplem  23300  ptbasin  23471  cfinfil  23787  csdfil  23788  uniioombllem5  25495  kur14lem6  35205  bj-2upln1upl  37019  dvasin  37705  readvrec2  42356  clsk3nimkb  44036  fourierdlem62  46173  caragendifcl  46519