Theorem ssdif0 3610

Description: Subclass expressed in terms of difference. Exercise 7 of [TakeutiZaring] p. 22. (Contributed by NM, 29-Apr-1994.)

Assertion

Ref	Expression
ssdif0	⊢ (A ⊆ B ↔ (A ∖ B) = ∅)

Proof of Theorem ssdif0

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		iman 413	. . . 4 ⊢ ((x ∈ A → x ∈ B) ↔ ¬ (x ∈ A ∧ ¬ x ∈ B))
2		eldif 3222	. . . 4 ⊢ (x ∈ (A ∖ B) ↔ (x ∈ A ∧ ¬ x ∈ B))
3	1, 2	xchbinxr 302	. . 3 ⊢ ((x ∈ A → x ∈ B) ↔ ¬ x ∈ (A ∖ B))
4	3	albii 1566	. 2 ⊢ (∀x(x ∈ A → x ∈ B) ↔ ∀x ¬ x ∈ (A ∖ B))
5		dfss2 3263	. 2 ⊢ (A ⊆ B ↔ ∀x(x ∈ A → x ∈ B))
6		eq0 3565	. 2 ⊢ ((A ∖ B) = ∅ ↔ ∀x ¬ x ∈ (A ∖ B))
7	4, 5, 6	3bitr4i 268	1 ⊢ (A ⊆ B ↔ (A ∖ B) = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540   = wceq 1642   ∈ wcel 1710   ∖ cdif 3207   ⊆ wss 3258  ∅c0 3551

This theorem was proved from 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 theorem 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-ne 2519  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-ss 3260  df-nul 3552

This theorem is referenced by:  vdif0  3611  pssdifn0  3612  difid  3619  difin0  3624  sfinltfin  4536