Theorem sscon 4100

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 3934	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	con3d 153	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (¬ 𝑥 ∈ 𝐵 → ¬ 𝑥 ∈ 𝐴))
3	2	anim2d 624	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) → (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐴)))
4		eldif 3918	. . 3 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))
5		eldif 3918	. . 3 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐴) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐴))
6	3, 4, 5	3imtr4g 299	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ (𝐶 ∖ 𝐵) → 𝑥 ∈ (𝐶 ∖ 𝐴)))
7	6	ssrdv 3946	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∈ wcel 2146   ∖ cdif 3905   ⊆ wss 3908

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-v 3460  df-dif 3911  df-ss 3925

This theorem is used by:  sscond  4103  complss  4108  sscon34b  4260  sorpsscmpl  7744  sbthlem1  9085  sbthlem2  9086  cantnfp1lem1  9657  cantnfp1lem3  9659  isf34lem7  10381  isf34lem6  10382  setsres  17263  chnccat  18707  mplsubglem  22185  cctop  23200  clsval2  23244  ntrss  23249  hauscmplem  23600  ptbasin  23771  cfinfil  24087  csdfil  24088  uniioombllem5  25783  kur14lem6  35724  bj-2upln1upl  37701  dvasin  38396  readvrec2  43163  clsk3nimkb  44807  fourierdlem62  46923  caragendifcl  47269