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)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 400   ∈ wcel 2142   ⊆ wss 3904

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144

This proof depends on definitions:  df-bi 210  df-an 401  df-ex 1809  df-clel 2837  df-ss 3921

This theorem is used by:  nrelvOLD  5786  ordtr3  6407  smndex2dnrinv  18983  frlmssuvc2  21956  dflringlem  33793  1arithidom  33836  tfsconcatb0  44099  clsk1indlem1  44799  mapssbi  45957  fourierdlem10  46859  salgensscntex  47086