Theorem reldisj 3523

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 3192	. . . 4 ⊢ (𝐴 ⊆ 𝐶 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶))
2		pm5.44 929	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))))
3		eldif 3186	. . . . . . 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 1563	. . . 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 1850	. 2 ⊢ (𝐴 ⊆ 𝐶 → (∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵))))
9		disj1 3522	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵))
10		ssalel 3192	. 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 1373   = wceq 1375   ∈ wcel 2180   ∖ cdif 3174   ∩ cin 3176   ⊆ wss 3177  ∅c0 3471

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 617  ax-in2 618  ax-io 713  ax-5 1473  ax-7 1474  ax-gen 1475  ax-ie1 1519  ax-ie2 1520  ax-8 1530  ax-10 1531  ax-11 1532  ax-i12 1533  ax-bndl 1535  ax-4 1536  ax-17 1552  ax-i9 1556  ax-ial 1560  ax-i5r 1561  ax-ext 2191

This theorem depends on definitions:  df-bi 117  df-tru 1378  df-nf 1487  df-sb 1789  df-clab 2196  df-cleq 2202  df-clel 2205  df-nfc 2341  df-ral 2493  df-v 2781  df-dif 3179  df-in 3183  df-ss 3190  df-nul 3472

This theorem is referenced by:  disj2  3527  ssdifsn  3775  structcnvcnv  13014