Theorem nelss 4000

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 3928	. . 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 2145   ⊆ wss 3902

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 1943  ax-6 2000  ax-7 2041  ax-8 2147

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-clel 2837  df-ss 3919

This theorem is used by:  nrelvOLD  5785  ordtr3  6408  smndex2dnrinv  19028  frlmssuvc2  22009  dflringlem  33891  1arithidom  33934  tfsconcatb0  44172  clsk1indlem1  44872  mapssbi  46030  fourierdlem10  46932  salgensscntex  47159