Theorem ssel2 1503

Description: Membership relationships follow from a subclass relationship.

Assertion

Ref	Expression
ssel2	|- ((A (_ B /\ C e. A) -> C e. B)

Proof of Theorem ssel2

Step	Hyp	Ref	Expression
1		ssel 1502	. 2 |- (A (_ B -> (C e. A -> C e. B))
2	1	imp 277	1 |- ((A (_ B /\ C e. A) -> C e. B)

Syntax hints:   -> wi 2   /\ wa 196   e. wcel 1092   (_ wss 1487

This theorem is referenced by:  tz7.7 2224  onnmin 2270  onmindif 2312  onmindif2 2313  ordunisssuc 2334  limsssuc 2362  fundmen 3333  isfinite2 3437  suplem2pr 3956  uzwo 4605  nnwoOLD 4608  infxpidmlem5 4937  infxpidmlem7 4939  infxpidmlem8 4940  ocsh 5164  ocorth 5172  pjoml 5271  spansnsst 5476  spansncv 5542  sumdmdi 5785

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-in 1491  df-ss 1492

metamath.org