Metamath Proof Explorer


Theorem rereccli

Description: Closure law for reciprocal. (Contributed by NM, 30-Apr-2005)

Ref Expression
Hypotheses redivcl.1
|- A e. RR
rereccl.2
|- A =/= 0
Assertion rereccli
|- ( 1 / A ) e. RR

Proof

Step Hyp Ref Expression
1 redivcl.1
 |-  A e. RR
2 rereccl.2
 |-  A =/= 0
3 1 rerecclzi
 |-  ( A =/= 0 -> ( 1 / A ) e. RR )
4 2 3 ax-mp
 |-  ( 1 / A ) e. RR