Theorem reldisj 3548

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		ssalel 3216	. . . 4 ⊢ (𝐴 ⊆ 𝐶 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶))
2		pm5.44 933	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))))
3		eldif 3210	. . . . . . 7 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))
4	3	imbi2i 226	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵)) ↔ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)))
5	2, 4	bitr4di 198	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵))))
6	5	sps 1586	. . . 4 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵))))
7	1, 6	sylbi 121	. . 3 ⊢ (𝐴 ⊆ 𝐶 → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵))))
8	7	albidv 1872	. 2 ⊢ (𝐴 ⊆ 𝐶 → (∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵))))
9		disj1 3547	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵))
10		ssalel 3216	. 2 ⊢ (𝐴 ⊆ (𝐶 ∖ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵)))
11	8, 9, 10	3bitr4g 223	1 ⊢ (𝐴 ⊆ 𝐶 → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105  ∀wal 1396   = wceq 1398   ∈ wcel 2202   ∖ cdif 3198   ∩ cin 3200   ⊆ wss 3201  ∅c0 3496

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 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-ral 2516  df-v 2805  df-dif 3203  df-in 3207  df-ss 3214  df-nul 3497

This theorem is referenced by:  disj2  3552  ssdifsn  3805  structcnvcnv  13159