Theorem nelss 4048

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 3976	. . 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 2104   ⊆ wss 3949

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 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-ext 2701

This theorem depends on definitions:  df-bi 206  df-an 395  df-tru 1542  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-v 3474  df-in 3956  df-ss 3966

This theorem is referenced by:  nrelv  5801  ordtr3  6410  smndex2dnrinv  18834  frlmssuvc2  21571  tfsconcatb0  42398  clsk1indlem1  43100  mapssbi  44212  fourierdlem10  45133  salgensscntex  45360