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 ) ∈ 𝐵𝐴 ∈ 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 ) ∈ 𝐵𝐴 ∈ V )