Metamath Proof Explorer


Theorem nfimd

Description: If in a context x is not free in ps and ch , then it is not free in ( ps -> ch ) . Deduction form of nfim . (Contributed by Mario Carneiro, 24-Sep-2016) (Proof shortened by Wolf Lammen, 30-Dec-2017) df-nf changed. (Revised by Wolf Lammen, 18-Sep-2021) Eliminate curried form of nfimt . (Revised by Wolf Lammen, 10-Jul-2022)

Ref Expression
Hypotheses nfimd.1
|- ( ph -> F/ x ps )
nfimd.2
|- ( ph -> F/ x ch )
Assertion nfimd
|- ( ph -> F/ x ( ps -> ch ) )

Proof

Step Hyp Ref Expression
1 nfimd.1
 |-  ( ph -> F/ x ps )
2 nfimd.2
 |-  ( ph -> F/ x ch )
3 19.35
 |-  ( E. x ( ps -> ch ) <-> ( A. x ps -> E. x ch ) )
4 3 biimpi
 |-  ( E. x ( ps -> ch ) -> ( A. x ps -> E. x ch ) )
5 1 nfrd
 |-  ( ph -> ( E. x ps -> A. x ps ) )
6 2 nfrd
 |-  ( ph -> ( E. x ch -> A. x ch ) )
7 5 6 imim12d
 |-  ( ph -> ( ( A. x ps -> E. x ch ) -> ( E. x ps -> A. x ch ) ) )
8 19.38
 |-  ( ( E. x ps -> A. x ch ) -> A. x ( ps -> ch ) )
9 4 7 8 syl56
 |-  ( ph -> ( E. x ( ps -> ch ) -> A. x ( ps -> ch ) ) )
10 9 nfd
 |-  ( ph -> F/ x ( ps -> ch ) )