Theorem reldisj 2317

Description: Two ways of saying that two classes are disjoint, using the complement of B relative to a universe C.

Assertion

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

Proof of Theorem reldisj

Step	Hyp	Ref	Expression
1		ssel 2066	. . . . 5 ⊢ (A ⊆ C → (x ∈ A → x ∈ C))
2	1	biantrurd 729	. . . 4 ⊢ (A ⊆ C → ((x ∈ A → ¬ x ∈ B) ↔ ((x ∈ A → x ∈ C) ⋀ (x ∈ A → ¬ x ∈ B))))
3		pm4.76 601	. . . . 5 ⊢ (((x ∈ A → x ∈ C) ⋀ (x ∈ A → ¬ x ∈ B)) ↔ (x ∈ A → (x ∈ C ⋀ ¬ x ∈ B)))
4		eldif 2060	. . . . . 6 ⊢ (x ∈ (C ∖ B) ↔ (x ∈ C ⋀ ¬ x ∈ B))
5	4	imbi2i 185	. . . . 5 ⊢ ((x ∈ A → x ∈ (C ∖ B)) ↔ (x ∈ A → (x ∈ C ⋀ ¬ x ∈ B)))
6	3, 5	bitr4 176	. . . 4 ⊢ (((x ∈ A → x ∈ C) ⋀ (x ∈ A → ¬ x ∈ B)) ↔ (x ∈ A → x ∈ (C ∖ B)))
7	2, 6	syl6bb 538	. . 3 ⊢ (A ⊆ C → ((x ∈ A → ¬ x ∈ B) ↔ (x ∈ A → x ∈ (C ∖ B))))
8	7	albidv 1280	. 2 ⊢ (A ⊆ C → (∀x(x ∈ A → ¬ x ∈ B) ↔ ∀x(x ∈ A → x ∈ (C ∖ B))))
9		disj1 2316	. 2 ⊢ ((A ∩ B) = ∅ ↔ ∀x(x ∈ A → ¬ x ∈ B))
10		dfss2 2061	. 2 ⊢ (A ⊆ (C ∖ B) ↔ ∀x(x ∈ A → x ∈ (C ∖ B)))
11	8, 9, 10	3bitr4g 557	1 ⊢ (A ⊆ C → ((A ∩ B) = ∅ ↔ A ⊆ (C ∖ B)))

Syntax hints:  ¬ wn 2   → wi 3   ↔ wb 146   ⋀ wa 223  ∀wal 956   = wceq 958   ∈ wcel 960   ∖ cdif 2047   ∩ cin 2049   ⊆ wss 2050  ∅c0 2283

This theorem is referenced by:  disj2 2320  elcls 7701  islp2 7744

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-ral 1652  df-v 1815  df-dif 2052  df-in 2054  df-ss 2056  df-nul 2284

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