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 B B 0 A B

Proof

Step Hyp Ref Expression
1 divval A B B 0 A B = ι x | B x = A
2 receu A B B 0 ∃! x B x = A
3 riotacl ∃! x B x = A ι x | B x = A
4 2 3 syl A B B 0 ι x | B x = A
5 1 4 eqeltrd A B B 0 A B