Theorem nfbid 1588

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 1585	. . 3 |- ( ph -> F/ x ( ps -> ch ) )
5	3, 2	nfimd 1585	. . 3 |- ( ph -> F/ x ( ch -> ps ) )
6	4, 5	nfand 1568	. 2 |- ( ph -> F/ x ( ( ps -> ch ) /\ ( ch -> ps ) ) )
7	1, 6	nfxfrd 1475	1 |- ( ph -> F/ x ( ps <-> ch ) )

Syntax hints:   -> wi 4   /\ wa 104   <-> wb 105   F/wnf 1460

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 1447  ax-gen 1449  ax-4 1510  ax-ial 1534  ax-i5r 1535

This theorem depends on definitions:  df-bi 117  df-nf 1461

This theorem is referenced by:  nfbi  1589  nfeudv  2041  nfeqd  2334  nfiotadw  5182  iota2df  5203  bdsepnft  14642