Theorem nelss 3158

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 3091	. . 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 624	1 |- ( ( A e. B /\ -. A e. C ) -> -. B C_ C )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 103   e. wcel 1480   C_ wss 3071

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-11 1484  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-in 3077  df-ss 3084

This theorem is referenced by: (None)