Metamath Proof Explorer


Theorem divcl

Description: Closure law for division. (Contributed by NM, 21-Jul-2001) (Proof shortened by Mario Carneiro, 17-Feb-2014)

Ref Expression
Assertion divcl
|- ( ( A e. CC /\ B e. CC /\ B =/= 0 ) -> ( A / B ) e. CC )

Proof

Step Hyp Ref Expression
1 divval
 |-  ( ( A e. CC /\ B e. CC /\ B =/= 0 ) -> ( A / B ) = ( iota_ x e. CC ( B x. x ) = A ) )
2 receu
 |-  ( ( A e. CC /\ B e. CC /\ B =/= 0 ) -> E! x e. CC ( B x. x ) = A )
3 riotacl
 |-  ( E! x e. CC ( B x. x ) = A -> ( iota_ x e. CC ( B x. x ) = A ) e. CC )
4 2 3 syl
 |-  ( ( A e. CC /\ B e. CC /\ B =/= 0 ) -> ( iota_ x e. CC ( B x. x ) = A ) e. CC )
5 1 4 eqeltrd
 |-  ( ( A e. CC /\ B e. CC /\ B =/= 0 ) -> ( A / B ) e. CC )