Theorem disjel 3526

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 3524	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵))
2		eleq2 2273	. . . 4 ⊢ (𝐴 = (𝐴 ∖ 𝐵) → (𝐶 ∈ 𝐴 ↔ 𝐶 ∈ (𝐴 ∖ 𝐵)))
3		eldifn 3307	. . . 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 1375   ∈ wcel 2180   ∖ cdif 3174   ∩ cin 3176  ∅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-nul 3472

This theorem is referenced by:  fvun1  5673  ctssdccl  7246  fsumsplit  11884  fprodsplitdc  12073  fprodsplit  12074