Metamath Proof Explorer


Theorem nf5dv

Description: Apply the definition of not-free in a context. (Contributed by Mario Carneiro, 11-Aug-2016) df-nf changed. (Revised by Wolf Lammen, 18-Sep-2021) (Proof shortened by Wolf Lammen, 13-Jul-2022)

Ref Expression
Hypothesis nf5dv.1
|- ( ph -> ( ps -> A. x ps ) )
Assertion nf5dv
|- ( ph -> F/ x ps )

Proof

Step Hyp Ref Expression
1 nf5dv.1
 |-  ( ph -> ( ps -> A. x ps ) )
2 ax-5
 |-  ( ph -> A. x ph )
3 2 1 nf5dh
 |-  ( ph -> F/ x ps )