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 ''' A B A V

Proof

Step Hyp Ref Expression
1 afvprc ¬ A V F ''' A = V
2 nvelim F ''' A = V ¬ F ''' A B
3 1 2 syl ¬ A V ¬ F ''' A B
4 3 con4i F ''' A B A V