Theorem nfd2 2015

Description: Deduce that x is not free in ps in a context. (Contributed by Wolf Lammen, 16-Sep-2021.)

Hypothesis

Ref	Expression
nfd2.1	|- ( ph -> ( E. x ps -> A. x ps ) )

Assertion

Ref	Expression
nfd2	|- ( ph -> F/ x ps )

Proof of Theorem nfd2

Step	Hyp	Ref	Expression
1		nfd2.1	. 2 |- ( ph -> ( E. x ps -> A. x ps ) )
2		nf2 1661	. 2 |- ( F/ x ps <-> ( E. x ps -> A. x ps ) )
3	1, 2	sylibr 133	1 |- ( ph -> F/ x ps )

Syntax hints:   -> wi 4   A.wal 1346   F/wnf 1453   E.wex 1485

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-gen 1442  ax-ie2 1487  ax-4 1503  ax-ial 1527

This theorem depends on definitions:  df-bi 116  df-nf 1454

This theorem is referenced by:  nf5-1  2017