Theorem nf5dh 2026

Description: Deduce that x is not free in ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016.) df-nf 1710 changed. (Revised by Wolf Lammen, 11-Oct-2021.)

Hypotheses

Ref	Expression
nf5dh.1	|- ( ph -> A. x ph )
nf5dh.2	|- ( ph -> ( ps -> A. x ps ) )

Assertion

Ref	Expression
nf5dh	|- ( ph -> F/ x ps )

Proof of Theorem nf5dh

Step	Hyp	Ref	Expression
1		nf5dh.1	. . 3 |- ( ph -> A. x ph )
2		nf5dh.2	. . 3 |- ( ph -> ( ps -> A. x ps ) )
3	1, 2	alrimih 1751	. 2 |- ( ph -> A. x ( ps -> A. x ps ) )
4		nf5-1 2023	. 2 |- ( A. x ( ps -> A. x ps ) -> F/ x ps )
5	3, 4	syl 17	1 |- ( ph -> F/ x ps )

Syntax hints:   -> wi 4   A.wal 1481   F/wnf 1708

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-10 2019

This theorem depends on definitions:  df-bi 197  df-ex 1705  df-nf 1710

This theorem is referenced by:  hbimd  2126  ax12indalem  34230  ax12inda2ALT  34231