Theorem reldisj 3419

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 3091	. . . 4 ⊢ (𝐴 ⊆ 𝐶 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶))
2		pm5.44 911	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))))
3		eldif 3085	. . . . . . 7 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))
4	3	imbi2i 225	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵)) ↔ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)))
5	2, 4	syl6bbr 197	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵))))
6	5	sps 1518	. . . 4 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵))))
7	1, 6	sylbi 120	. . 3 ⊢ (𝐴 ⊆ 𝐶 → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵))))
8	7	albidv 1797	. 2 ⊢ (𝐴 ⊆ 𝐶 → (∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵))))
9		disj1 3418	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵))
10		dfss2 3091	. 2 ⊢ (𝐴 ⊆ (𝐶 ∖ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵)))
11	8, 9, 10	3bitr4g 222	1 ⊢ (𝐴 ⊆ 𝐶 → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ↔ wb 104  ∀wal 1330   = wceq 1332   ∈ wcel 1481   ∖ cdif 3073   ∩ cin 3075   ⊆ wss 3076  ∅c0 3368

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122

This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-ral 2422  df-v 2691  df-dif 3078  df-in 3082  df-ss 3089  df-nul 3369

This theorem is referenced by:  disj2  3423  ssdifsn  3659  structcnvcnv  12014