Theorem nelss 3979

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 3907	. . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∈ 𝐵 → 𝐴 ∈ 𝐶))
2	1	com12 33	. 2 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ⊆ 𝐶 → 𝐴 ∈ 𝐶))
3	2	con3dimp 414	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∈ wcel 2147   ⊆ wss 3900

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-clel 2839  df-ss 3901

This theorem is used by:  nrelvOLD  5783  ordtr3  6427  smndex2dnrinv  19143  frlmssuvc2  22131  dflringlem  34096  1arithidom  34139  tfsconcatb0  44538  clsk1indlem1  45237  mapssbi  46431  fourierdlem10  47332  salgensscntex  47559