Theorem nelss 3288

Description: Demonstrate by witnesses that two classes lack a subclass relation. (Contributed by Stefan O'Rear, 5-Feb-2015.)

Assertion

Ref	Expression
nelss	|- ( ( A e. B /\ -. A e. C ) -> -. B C_ C )

Proof of Theorem nelss

Step	Hyp	Ref	Expression
1		ssel 3221	. . 3 |- ( B C_ C -> ( A e. B -> A e. C ) )
2	1	com12 30	. 2 |- ( A e. B -> ( B C_ C -> A e. C ) )
3	2	con3dimp 640	1 |- ( ( A e. B /\ -. A e. C ) -> -. B C_ C )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 104   e. wcel 2202   C_ wss 3200

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 619  ax-in2 620  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-11 1554  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-in 3206  df-ss 3213

This theorem is referenced by: (None)