Theorem disjel 3448

Description: A set can't belong to both members of disjoint classes. (Contributed by NM, 28-Feb-2015.)

Assertion

Ref	Expression
disjel	⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ∈ 𝐴) → ¬ 𝐶 ∈ 𝐵)

Proof of Theorem disjel

Step	Hyp	Ref	Expression
1		disj3 3446	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵))
2		eleq2 2221	. . . 4 ⊢ (𝐴 = (𝐴 ∖ 𝐵) → (𝐶 ∈ 𝐴 ↔ 𝐶 ∈ (𝐴 ∖ 𝐵)))
3		eldifn 3230	. . . 4 ⊢ (𝐶 ∈ (𝐴 ∖ 𝐵) → ¬ 𝐶 ∈ 𝐵)
4	2, 3	syl6bi 162	. . 3 ⊢ (𝐴 = (𝐴 ∖ 𝐵) → (𝐶 ∈ 𝐴 → ¬ 𝐶 ∈ 𝐵))
5	1, 4	sylbi 120	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∈ 𝐴 → ¬ 𝐶 ∈ 𝐵))
6	5	imp 123	1 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ∈ 𝐴) → ¬ 𝐶 ∈ 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   = wceq 1335   ∈ wcel 2128   ∖ cdif 3099   ∩ cin 3101  ∅c0 3394

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 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-10 1485  ax-11 1486  ax-i12 1487  ax-bndl 1489  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139

This theorem depends on definitions:  df-bi 116  df-tru 1338  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-clel 2153  df-nfc 2288  df-ral 2440  df-v 2714  df-dif 3104  df-in 3108  df-nul 3395

This theorem is referenced by:  fvun1  5535  ctssdccl  7056  fsumsplit  11308  fprodsplitdc  11497  fprodsplit  11498