Theorem nelss 4002

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 3930	. . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∈ 𝐵 → 𝐴 ∈ 𝐶))
2	1	com12 33	. 2 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ⊆ 𝐶 → 𝐴 ∈ 𝐶))
3	2	con3dimp 413	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 400   ∈ wcel 2141   ⊆ wss 3904

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1823  ax-4 1837  ax-5 1938  ax-6 1995  ax-7 2036  ax-8 2143

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1808  df-clel 2836  df-ss 3921

This theorem is referenced by:  nrelvOLD  5787  ordtr3  6407  smndex2dnrinv  18976  frlmssuvc2  21924  dflringlem  33750  1arithidom  33793  tfsconcatb0  44019  clsk1indlem1  44719  mapssbi  45877  fourierdlem10  46779  salgensscntex  47006