Metamath Proof Explorer


Theorem infm3lem

Description: Lemma for infm3 . (Contributed by NM, 14-Jun-2005)

Ref Expression
Assertion infm3lem
|- ( x e. RR -> E. y e. RR x = -u y )

Proof

Step Hyp Ref Expression
1 renegcl
 |-  ( x e. RR -> -u x e. RR )
2 recn
 |-  ( x e. RR -> x e. CC )
3 2 negnegd
 |-  ( x e. RR -> -u -u x = x )
4 3 eqcomd
 |-  ( x e. RR -> x = -u -u x )
5 negeq
 |-  ( y = -u x -> -u y = -u -u x )
6 5 rspceeqv
 |-  ( ( -u x e. RR /\ x = -u -u x ) -> E. y e. RR x = -u y )
7 1 4 6 syl2anc
 |-  ( x e. RR -> E. y e. RR x = -u y )