Theorem reldisj 3476

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 3146	. . . 4 ⊢ (𝐴 ⊆ 𝐶 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶))
2		pm5.44 925	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))))
3		eldif 3140	. . . . . . 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 1537	. . . 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 1824	. 2 ⊢ (𝐴 ⊆ 𝐶 → (∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵))))
9		disj1 3475	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵))
10		dfss2 3146	. 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 1351   = wceq 1353   ∈ wcel 2148   ∖ cdif 3128   ∩ cin 3130   ⊆ wss 3131  ∅c0 3424

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 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-v 2741  df-dif 3133  df-in 3137  df-ss 3144  df-nul 3425

This theorem is referenced by:  disj2  3480  ssdifsn  3722  structcnvcnv  12480