Theorem reldisj 3595

Description: Two ways of saying that two classes are disjoint, using the complement of B relative to a universe C. (Contributed by NM, 15-Feb-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

Ref	Expression
reldisj	⊢ (A ⊆ C → ((A ∩ B) = ∅ ↔ A ⊆ (C ∖ B)))

Proof of Theorem reldisj

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfss2 3263	. . . 4 ⊢ (A ⊆ C ↔ ∀x(x ∈ A → x ∈ C))
2		pm5.44 877	. . . . . 6 ⊢ ((x ∈ A → x ∈ C) → ((x ∈ A → ¬ x ∈ B) ↔ (x ∈ A → (x ∈ C ∧ ¬ x ∈ B))))
3		eldif 3222	. . . . . . 7 ⊢ (x ∈ (C ∖ B) ↔ (x ∈ C ∧ ¬ x ∈ B))
4	3	imbi2i 303	. . . . . 6 ⊢ ((x ∈ A → x ∈ (C ∖ B)) ↔ (x ∈ A → (x ∈ C ∧ ¬ x ∈ B)))
5	2, 4	syl6bbr 254	. . . . 5 ⊢ ((x ∈ A → x ∈ C) → ((x ∈ A → ¬ x ∈ B) ↔ (x ∈ A → x ∈ (C ∖ B))))
6	5	sps 1754	. . . 4 ⊢ (∀x(x ∈ A → x ∈ C) → ((x ∈ A → ¬ x ∈ B) ↔ (x ∈ A → x ∈ (C ∖ B))))
7	1, 6	sylbi 187	. . 3 ⊢ (A ⊆ C → ((x ∈ A → ¬ x ∈ B) ↔ (x ∈ A → x ∈ (C ∖ B))))
8	7	albidv 1625	. 2 ⊢ (A ⊆ C → (∀x(x ∈ A → ¬ x ∈ B) ↔ ∀x(x ∈ A → x ∈ (C ∖ B))))
9		disj1 3594	. 2 ⊢ ((A ∩ B) = ∅ ↔ ∀x(x ∈ A → ¬ x ∈ B))
10		dfss2 3263	. 2 ⊢ (A ⊆ (C ∖ B) ↔ ∀x(x ∈ A → x ∈ (C ∖ B)))
11	8, 9, 10	3bitr4g 279	1 ⊢ (A ⊆ C → ((A ∩ B) = ∅ ↔ A ⊆ (C ∖ B)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540   = wceq 1642   ∈ wcel 1710   ∖ cdif 3207   ∩ cin 3209   ⊆ wss 3258  ∅c0 3551

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-ne 2519  df-ral 2620  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-ss 3260  df-nul 3552

This theorem is used by:  disj2  3599