Theorem sb6x 1779

Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 12-Aug-2011.)

Hypothesis

Ref	Expression
sb6x.1	|- ( ph -> A. x ph )

Assertion

Ref	Expression
sb6x	|- ( [ y / x ] ph <-> A. x ( x = y -> ph ) )

Proof of Theorem sb6x

Step	Hyp	Ref	Expression
1		sb6x.1	. . 3 |- ( ph -> A. x ph )
2	1	sbh 1776	. 2 |- ( [ y / x ] ph <-> ph )
3		biidd 172	. . 3 |- ( x = y -> ( ph <-> ph ) )
4	1, 3	equsalh 1726	. 2 |- ( A. x ( x = y -> ph ) <-> ph )
5	2, 4	bitr4i 187	1 |- ( [ y / x ] ph <-> A. x ( x = y -> ph ) )

Syntax hints:   -> wi 4   <-> wb 105   A.wal 1351   [wsb 1762

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-ie1 1493  ax-ie2 1494  ax-4 1510  ax-i9 1530  ax-ial 1534

This theorem depends on definitions:  df-bi 117  df-sb 1763

This theorem is referenced by: (None)