Metamath Proof Explorer


Theorem nf5di

Description: Since the converse holds by a1i , this inference shows that we can represent a not-free hypothesis with either F/ x ph (inference form) or ( ph -> F/ x ph ) (deduction form). (Contributed by NM, 17-Aug-2018) (Proof shortened by Wolf Lammen, 10-Jul-2019) Reduce axiom dependencies. (Revised by SN and Sky Winter, 23-Oct-2026)

Ref Expression
Hypothesis nf5di.1
|- ( ph -> F/ x ph )
Assertion nf5di
|- F/ x ph

Proof

Step Hyp Ref Expression
1 nf5di.1
 |-  ( ph -> F/ x ph )
2 hbe1
 |-  ( E. x ph -> A. x E. x ph )
3 1 nfrd
 |-  ( ph -> ( E. x ph -> A. x ph ) )
4 3 com12
 |-  ( E. x ph -> ( ph -> A. x ph ) )
5 2 4 eximdh
 |-  ( E. x ph -> ( E. x ph -> E. x A. x ph ) )
6 5 pm2.43i
 |-  ( E. x ph -> E. x A. x ph )
7 hbe1a
 |-  ( E. x A. x ph -> A. x ph )
8 6 7 syl
 |-  ( E. x ph -> A. x ph )
9 8 nfi
 |-  F/ x ph