Theorem nelss 4042

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 3970	. . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∈ 𝐵 → 𝐴 ∈ 𝐶))
2	1	com12 32	. 2 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ⊆ 𝐶 → 𝐴 ∈ 𝐶))
3	2	con3dimp 407	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 394   ∈ wcel 2098   ⊆ wss 3944

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100

This theorem depends on definitions:  df-bi 206  df-an 395  df-ex 1774  df-clel 2802  df-ss 3961

This theorem is referenced by:  nrelv  5802  ordtr3  6416  smndex2dnrinv  18875  frlmssuvc2  21746  1arithidom  33349  tfsconcatb0  42915  clsk1indlem1  43617  mapssbi  44725  fourierdlem10  45643  salgensscntex  45870