Theorem nfbid 1599

Description: If in a context x is not free in ps and ch, then it is not free in ( ps <-> ch ). (Contributed by Mario Carneiro, 24-Sep-2016.) (Proof shortened by Wolf Lammen, 29-Dec-2017.)

Hypotheses

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

Assertion

Ref	Expression
nfbid	|- ( ph -> F/ x ( ps <-> ch ) )

Proof of Theorem nfbid

Step	Hyp	Ref	Expression
1		dfbi2 388	. 2 |- ( ( ps <-> ch ) <-> ( ( ps -> ch ) /\ ( ch -> ps ) ) )
2		nfbid.1	. . . 4 |- ( ph -> F/ x ps )
3		nfbid.2	. . . 4 |- ( ph -> F/ x ch )
4	2, 3	nfimd 1596	. . 3 |- ( ph -> F/ x ( ps -> ch ) )
5	3, 2	nfimd 1596	. . 3 |- ( ph -> F/ x ( ch -> ps ) )
6	4, 5	nfand 1579	. 2 |- ( ph -> F/ x ( ( ps -> ch ) /\ ( ch -> ps ) ) )
7	1, 6	nfxfrd 1486	1 |- ( ph -> F/ x ( ps <-> ch ) )

Syntax hints:   -> wi 4   /\ wa 104   <-> wb 105   F/wnf 1471

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1458  ax-gen 1460  ax-4 1521  ax-ial 1545  ax-i5r 1546

This theorem depends on definitions:  df-bi 117  df-nf 1472

This theorem is referenced by:  nfbi  1600  nfeudv  2053  nfeqd  2347  nfiotadw  5199  iota2df  5221  bdsepnft  15117