Theorem minel 4415

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 4414	. . . 4 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝐴 ∈ 𝐵) → (𝐶 ∩ 𝐵) ≠ ∅)
2	1	expcom 413	. . 3 ⊢ (𝐴 ∈ 𝐵 → (𝐴 ∈ 𝐶 → (𝐶 ∩ 𝐵) ≠ ∅))
3	2	necon2bd 2945	. 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 1541   ∈ wcel 2113   ≠ wne 2929   ∩ cin 3897  ∅c0 4282

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-ne 2930  df-v 3439  df-dif 3901  df-in 3905  df-nul 4283

This theorem is referenced by:  peano5  7831  fnsuppres  8129  domunfican  9215  unwdomg  9479  dfac5  10029  ccatval2  14489  mreexexlem2d  17555  hauspwpwf1  23905