Theorem disjel 4410

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 4407	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵))
2		eleq2 2850	. . . 4 ⊢ (𝐴 = (𝐴 ∖ 𝐵) → (𝐶 ∈ 𝐴 ↔ 𝐶 ∈ (𝐴 ∖ 𝐵)))
3		eldifn 4079	. . . 4 ⊢ (𝐶 ∈ (𝐴 ∖ 𝐵) → ¬ 𝐶 ∈ 𝐵)
4	2, 3	biimtrdi 256	. . 3 ⊢ (𝐴 = (𝐴 ∖ 𝐵) → (𝐶 ∈ 𝐴 → ¬ 𝐶 ∈ 𝐵))
5	1, 4	sylbi 220	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐶 ∈ 𝐴 → ¬ 𝐶 ∈ 𝐵))
6	5	imp 412	1 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ∈ 𝐴) → ¬ 𝐶 ∈ 𝐵)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2145   ∖ cdif 3896   ∩ cin 3898  ∅c0 4279

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3078  df-v 3453  df-dif 3902  df-in 3906  df-nul 4280

This theorem is used by:  disjxun  5101  fvun1  6974  dedekindle  11467  fprodsplit  16126  prlngpln3  29420  unelldsys  34784  dvasin  38602