Metamath Proof Explorer


Theorem el

Description: Every set is an element of some other set. See elALT for a shorter proof using more axioms. (Contributed by NM, 4-Jan-2002) (Proof shortened by Andrew Salmon, 25-Jul-2011)

Ref Expression
Assertion el
|- E. y x e. y

Proof

Step Hyp Ref Expression
1 zfpow
 |-  E. y A. z ( A. y ( y e. z -> y e. x ) -> z e. y )
2 ax9
 |-  ( z = x -> ( y e. z -> y e. x ) )
3 2 alrimiv
 |-  ( z = x -> A. y ( y e. z -> y e. x ) )
4 ax8
 |-  ( z = x -> ( z e. y -> x e. y ) )
5 3 4 embantd
 |-  ( z = x -> ( ( A. y ( y e. z -> y e. x ) -> z e. y ) -> x e. y ) )
6 5 spimvw
 |-  ( A. z ( A. y ( y e. z -> y e. x ) -> z e. y ) -> x e. y )
7 1 6 eximii
 |-  E. y x e. y