Theorem nelss 4012

Description: Demonstrate by witnesses that two classes lack a subclass relation. (Contributed by Stefan O'Rear, 5-Feb-2015.)

Assertion

Ref	Expression
nelss	⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶)

Proof of Theorem nelss

Step	Hyp	Ref	Expression
1		ssel 3940	. . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∈ 𝐵 → 𝐴 ∈ 𝐶))
2	1	com12 32	. 2 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ⊆ 𝐶 → 𝐴 ∈ 𝐶))
3	2	con3dimp 409	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 396   ∈ wcel 2106   ⊆ wss 3913

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1544  df-ex 1782  df-sb 2068  df-clab 2709  df-cleq 2723  df-clel 2809  df-v 3448  df-in 3920  df-ss 3930

This theorem is referenced by:  nrelv  5761  ordtr3  6367  smndex2dnrinv  18739  frlmssuvc2  21238  tfsconcatb0  41737  clsk1indlem1  42439  mapssbi  43555  fourierdlem10  44478  salgensscntex  44705