Theorem disjel 3579

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 3577	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵))
2		eleq2 2302	. . . 4 ⊢ (𝐴 = (𝐴 ∖ 𝐵) → (𝐶 ∈ 𝐴 ↔ 𝐶 ∈ (𝐴 ∖ 𝐵)))
3		eldifn 3352	. . . 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 1402   ∈ wcel 2209   ∖ cdif 3217   ∩ cin 3219  ∅c0 3520

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 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-v 2823  df-dif 3222  df-in 3226  df-nul 3521

This theorem is referenced by:  fvun1  5766  ctssdccl  7445  fsumsplit  12157  fprodsplitdc  12346  fprodsplit  12347