Theorem nssr 3288

Description: Negation of subclass relationship. One direction of Exercise 13 of [TakeutiZaring] p. 18. (Contributed by Jim Kingdon, 15-Jul-2018.)

Assertion

Ref	Expression
nssr	|- ( E. x ( x e. A /\ -. x e. B ) -> -. A C_ B )

Distinct variable groups:   x, A   x, B

Proof of Theorem nssr

Step	Hyp	Ref	Expression
1		exanaliim 1696	. 2 |- ( E. x ( x e. A /\ -. x e. B ) -> -. A. x ( x e. A -> x e. B ) )
2		ssalel 3216	. 2 |- ( A C_ B <-> A. x ( x e. A -> x e. B ) )
3	1, 2	sylnibr 684	1 |- ( E. x ( x e. A /\ -. x e. B ) -> -. A C_ B )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 104   A.wal 1396   E.wex 1541   e. wcel 2202   C_ wss 3201

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 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-11 1555  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-fal 1404  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-in 3207  df-ss 3214

This theorem is referenced by: (None)