Theorem disjel 3551

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 3549	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵))
2		eleq2 2295	. . . 4 ⊢ (𝐴 = (𝐴 ∖ 𝐵) → (𝐶 ∈ 𝐴 ↔ 𝐶 ∈ (𝐴 ∖ 𝐵)))
3		eldifn 3332	. . . 4 ⊢ (𝐶 ∈ (𝐴 ∖ 𝐵) → ¬ 𝐶 ∈ 𝐵)
4	2, 3	biimtrdi 163	. . 3 ⊢ (𝐴 = (𝐴 ∖ 𝐵) → (𝐶 ∈ 𝐴 → ¬ 𝐶 ∈ 𝐵))
5	1, 4	sylbi 121	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∈ 𝐴 → ¬ 𝐶 ∈ 𝐵))
6	5	imp 124	1 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ∈ 𝐴) → ¬ 𝐶 ∈ 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   = wceq 1398   ∈ wcel 2202   ∖ cdif 3198   ∩ cin 3200  ∅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-nul 3497

This theorem is referenced by:  fvun1  5721  ctssdccl  7353  fsumsplit  12029  fprodsplitdc  12218  fprodsplit  12219