Metamath Proof Explorer

Theorem rereccld

Description: Closure law for reciprocal. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses redivcld.1 
|- ( ph -> A e. RR )
rereccld.2 
|- ( ph -> A =/= 0 )
Assertion rereccld 
|- ( ph -> ( 1 / A ) e. RR )

Proof

Step Hyp Ref Expression
1 redivcld.1 
 |-  ( ph -> A e. RR )
2 rereccld.2 
 |-  ( ph -> A =/= 0 )
3 rereccl 
 |-  ( ( A e. RR /\ A =/= 0 ) -> ( 1 / A ) e. RR )
4 1 2 3 syl2anc 
 |-  ( ph -> ( 1 / A ) e. RR )