Theorem sscon 3338

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 3218	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	con3d 634	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (¬ 𝑥 ∈ 𝐵 → ¬ 𝑥 ∈ 𝐴))
3	2	anim2d 337	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) → (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐴)))
4		eldif 3206	. . 3 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))
5		eldif 3206	. . 3 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐴) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐴))
6	3, 4, 5	3imtr4g 205	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ (𝐶 ∖ 𝐵) → 𝑥 ∈ (𝐶 ∖ 𝐴)))
7	6	ssrdv 3230	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∖ 𝐵) ⊆ (𝐶 ∖ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ∈ wcel 2200   ∖ cdif 3194   ⊆ wss 3197

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 617  ax-in2 618  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-dif 3199  df-in 3203  df-ss 3210

This theorem is referenced by:  sscond  3341  sbthlem1  7132  sbthlem2  7133