Theorem nfbid 1525

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 380	. 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 1522	. . 3 |- ( ph -> F/ x ( ps -> ch ) )
5	3, 2	nfimd 1522	. . 3 |- ( ph -> F/ x ( ch -> ps ) )
6	4, 5	nfand 1505	. 2 |- ( ph -> F/ x ( ( ps -> ch ) /\ ( ch -> ps ) ) )
7	1, 6	nfxfrd 1409	1 |- ( ph -> F/ x ( ps <-> ch ) )

Syntax hints:   -> wi 4   /\ wa 102   <-> wb 103   F/wnf 1394

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1381  ax-gen 1383  ax-4 1445  ax-ial 1472  ax-i5r 1473

This theorem depends on definitions:  df-bi 115  df-nf 1395

This theorem is referenced by:  nfbi  1526  nfeudv  1963  nfeqd  2243  nfiotadxy  4983  iota2df  5004  bdsepnft  11778  strcollnft  11879