Metamath Proof Explorer


Theorem nf5i

Description: Deduce that x is not free in ph from the definition. (Contributed by Mario Carneiro, 11-Aug-2016)

Ref Expression
Hypothesis nf5i.1
|- ( ph -> A. x ph )
Assertion nf5i
|- F/ x ph

Proof

Step Hyp Ref Expression
1 nf5i.1
 |-  ( ph -> A. x ph )
2 nf5-1
 |-  ( A. x ( ph -> A. x ph ) -> F/ x ph )
3 2 1 mpg
 |-  F/ x ph