Theorem sscon 4090

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 3925	. . . . 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 3909	. . 3 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))
5		eldif 3909	. . 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 3937	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∈ wcel 2145   ∖ cdif 3896   ⊆ wss 3899

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 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3453  df-dif 3902  df-ss 3916

This theorem is used by:  sscond  4093  complss  4098  sscon34b  4250  sorpsscmpl  7739  sbthlem1  9090  sbthlem2  9091  cantnfp1lem1  9663  cantnfp1lem3  9665  isf34lem7  10438  isf34lem6  10439  setsres  17336  chnccat  18780  mplsubglem  22286  cctop  23304  clsval2  23348  ntrss  23353  hauscmplem  23704  ptbasin  23876  cfinfil  24192  csdfil  24193  uniioombllem5  25888  kur14lem6  35945  bj-2upln1upl  37907  dvasin  38590  readvrec2  43380  clsk3nimkb  44999  fourierdlem62  47122  caragendifcl  47468