Theorem disjel 3598

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 3596	. . 3 ⊢ ((A ∩ B) = ∅ ↔ A = (A ∖ B))
2		eleq2 2414	. . . 4 ⊢ (A = (A ∖ B) → (C ∈ A ↔ C ∈ (A ∖ B)))
3		eldifn 3390	. . . 4 ⊢ (C ∈ (A ∖ B) → ¬ C ∈ B)
4	2, 3	syl6bi 219	. . 3 ⊢ (A = (A ∖ B) → (C ∈ A → ¬ C ∈ B))
5	1, 4	sylbi 187	. 2 ⊢ ((A ∩ B) = ∅ → (C ∈ A → ¬ C ∈ B))
6	5	imp 418	1 ⊢ (((A ∩ B) = ∅ ∧ C ∈ A) → ¬ C ∈ B)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 358   = wceq 1642   ∈ wcel 1710   ∖ cdif 3207   ∩ cin 3209  ∅c0 3551

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ne 2519  df-ral 2620  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-nul 3552

This theorem is used by: (None)