Metamath Proof Explorer


Theorem afvvv

Description: If a function's value at an argument is a set, the argument is also a set. (Contributed by Alexander van der Vekens, 25-May-2017)

Ref Expression
Assertion afvvv F'''ABAV

Proof

Step Hyp Ref Expression
1 afvprc ¬AVF'''A=V
2 nvelim F'''A=V¬F'''AB
3 1 2 syl ¬AV¬F'''AB
4 3 con4i F'''ABAV