Theorem sbhb 2000

Description: Two ways of expressing " x is (effectively) not free in ph." (Contributed by NM, 29-May-2009.)

Assertion

Ref	Expression
sbhb	|- ( ( ph -> A. x ph ) <-> A. y ( ph -> [ y / x ] ph ) )

Distinct variable group:   ph, y

Allowed substitution hint:   ph( x)

Proof of Theorem sbhb

Step	Hyp	Ref	Expression
1		ax-17 1579	. . . 4 |- ( ph -> A. y ph )
2	1	sb8h 1907	. . 3 |- ( A. x ph <-> A. y [ y / x ] ph )
3	2	imbi2i 226	. 2 |- ( ( ph -> A. x ph ) <-> ( ph -> A. y [ y / x ] ph ) )
4		19.21v 1926	. 2 |- ( A. y ( ph -> [ y / x ] ph ) <-> ( ph -> A. y [ y / x ] ph ) )
5	3, 4	bitr4i 187	1 |- ( ( ph -> A. x ph ) <-> A. y ( ph -> [ y / x ] ph ) )

Syntax hints:   -> wi 4   <-> wb 105   A.wal 1400   [wsb 1815

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 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-11 1559  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588

This theorem depends on definitions:  df-bi 117  df-nf 1514  df-sb 1816

This theorem is referenced by:  sbnf2  2041