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Theorem disjel 3251
 Description: A set can't belong to both members of disjoint classes. (Contributed by NM, 28-Feb-2015.)
Assertion
Ref Expression
disjel (((AB) = ∅ 𝐶 A) → ¬ 𝐶 B)

Proof of Theorem disjel
StepHypRef Expression
1 disj3 3249 . . 3 ((AB) = ∅ ↔ A = (AB))
2 eleq2 2083 . . . 4 (A = (AB) → (𝐶 A𝐶 (AB)))
3 eldifn 3044 . . . 4 (𝐶 (AB) → ¬ 𝐶 B)
42, 3syl6bi 152 . . 3 (A = (AB) → (𝐶 A → ¬ 𝐶 B))
51, 4sylbi 114 . 2 ((AB) = ∅ → (𝐶 A → ¬ 𝐶 B))
65imp 115 1 (((AB) = ∅ 𝐶 A) → ¬ 𝐶 B)
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 97   = wceq 1228   ∈ wcel 1374   ∖ cdif 2891   ∩ cin 2893  ∅c0 3201 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 532  ax-in2 533  ax-io 617  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-10 1377  ax-11 1378  ax-i12 1379  ax-bnd 1380  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004 This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-nfc 2149  df-ral 2289  df-v 2537  df-dif 2897  df-in 2901  df-nul 3202 This theorem is referenced by:  fvun1  5164
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