Metamath Proof Explorer


Theorem rerecclzi

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

Ref Expression
Hypothesis redivcl.1
|- A e. RR
Assertion rerecclzi
|- ( A =/= 0 -> ( 1 / A ) e. RR )

Proof

Step Hyp Ref Expression
1 redivcl.1
 |-  A e. RR
2 rereccl
 |-  ( ( A e. RR /\ A =/= 0 ) -> ( 1 / A ) e. RR )
3 1 2 mpan
 |-  ( A =/= 0 -> ( 1 / A ) e. RR )