Theorem sscon 3401

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 3268	. . . . 5 ⊢ (A ⊆ B → (x ∈ A → x ∈ B))
2	1	con3d 125	. . . 4 ⊢ (A ⊆ B → (¬ x ∈ B → ¬ x ∈ A))
3	2	anim2d 548	. . 3 ⊢ (A ⊆ B → ((x ∈ C ∧ ¬ x ∈ B) → (x ∈ C ∧ ¬ x ∈ A)))
4		eldif 3222	. . 3 ⊢ (x ∈ (C ∖ B) ↔ (x ∈ C ∧ ¬ x ∈ B))
5		eldif 3222	. . 3 ⊢ (x ∈ (C ∖ A) ↔ (x ∈ C ∧ ¬ x ∈ A))
6	3, 4, 5	3imtr4g 261	. 2 ⊢ (A ⊆ B → (x ∈ (C ∖ B) → x ∈ (C ∖ A)))
7	6	ssrdv 3279	1 ⊢ (A ⊆ B → (C ∖ B) ⊆ (C ∖ A))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 358   ∈ wcel 1710   ∖ cdif 3207   ⊆ wss 3258

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-ss 3260

This theorem is used by:  sscond  3404