Theorem minel 4432

Description: A minimum element of a class has no elements in common with the class. (Contributed by NM, 22-Jun-1994.) (Proof shortened by JJ, 14-Jul-2021.)

Assertion

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

Proof of Theorem minel

Step	Hyp	Ref	Expression
1		inelcm 4431	. . . 4 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝐴 ∈ 𝐵) → (𝐶 ∩ 𝐵) ≠ ∅)
2	1	expcom 413	. . 3 ⊢ (𝐴 ∈ 𝐵 → (𝐴 ∈ 𝐶 → (𝐶 ∩ 𝐵) ≠ ∅))
3	2	necon2bd 2942	. 2 ⊢ (𝐴 ∈ 𝐵 → ((𝐶 ∩ 𝐵) = ∅ → ¬ 𝐴 ∈ 𝐶))
4	3	imp 406	1 ⊢ ((𝐴 ∈ 𝐵 ∧ (𝐶 ∩ 𝐵) = ∅) → ¬ 𝐴 ∈ 𝐶)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   = wceq 1540   ∈ wcel 2109   ≠ wne 2926   ∩ cin 3916  ∅c0 4299

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-ne 2927  df-v 3452  df-dif 3920  df-in 3924  df-nul 4300

This theorem is referenced by:  peano5  7872  fnsuppres  8173  domunfican  9279  unwdomg  9544  dfac5  10089  ccatval2  14550  mreexexlem2d  17613  hauspwpwf1  23881