Theorem minel 1743

Description: A minimum element of a class has no elements in common with the class.

Assertion

Ref	Expression
minel	⊢ ((A ∈ B ∧ (C ∩ B) = ∅) → ¬ A ∈ C)

Proof of Theorem minel

Step	Hyp	Ref	Expression
1		inelcm 1742	. . . . . 6 ⊢ ((A ∈ C ∧ A ∈ B) → ¬ (C ∩ B) = ∅)
2	1	con2i 89	. . . . 5 ⊢ ((C ∩ B) = ∅ → ¬ (A ∈ C ∧ A ∈ B))
3		imnan 207	. . . . 5 ⊢ ((A ∈ C → ¬ A ∈ B) ↔ ¬ (A ∈ C ∧ A ∈ B))
4	2, 3	sylibr 175	. . . 4 ⊢ ((C ∩ B) = ∅ → (A ∈ C → ¬ A ∈ B))
5	4	con2d 83	. . 3 ⊢ ((C ∩ B) = ∅ → (A ∈ B → ¬ A ∈ C))
6	5	com12 13	. 2 ⊢ (A ∈ B → ((C ∩ B) = ∅ → ¬ A ∈ C))
7	6	imp 277	1 ⊢ ((A ∈ B ∧ (C ∩ B) = ∅) → ¬ A ∈ C)

Syntax hints:  ¬ wn 1   → wi 2   ∧ wa 196   = wceq 1091   ∈ wcel 1092   ∩ cin 1486  ∅c0 1707

This theorem is referenced by:  peano5 2394

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-dif 1489  df-in 1491  df-nul 1708

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