Theorem nfd 1766

Description: Deduce that x is not free in ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016.)

Hypotheses

Ref	Expression
nfd.1	|- F/xph
nfd.2	|- (ph -> (ps -> A.xps))

Assertion

Ref	Expression
nfd	|- (ph -> F/xps)

Proof of Theorem nfd

Step	Hyp	Ref	Expression
1		nfd.1	. . 3 |- F/xph
2		nfd.2	. . 3 |- (ph -> (ps -> A.xps))
3	1, 2	alrimi 1765	. 2 |- (ph -> A.x(ps -> A.xps))
4		df-nf 1545	. 2 |- ( F/xps <-> A.x(ps -> A.xps))
5	3, 4	sylibr 203	1 |- (ph -> F/xps)

Syntax hints:   -> wi 4  A.wal 1540   F/wnf 1544

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-11 1746

This theorem depends on definitions:  df-bi 177  df-ex 1542  df-nf 1545

This theorem is referenced by:  nfdh  1767  nfnd  1791  nfndOLD  1792  nfald  1852  nfaldOLD  1853  nfeqf  1958  dvelimf  1997  a16nf  2051  nfsb2  2058  sbal2  2134  copsexg  4607