Theorem sb6a 1905

Description: Equivalence for substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
sb6a	|- ( [ y / x ] ph <-> A. x ( x = y -> [ x / y ] ph ) )

Distinct variable group:   x, y

Allowed substitution hints:   ph( x, y)

Proof of Theorem sb6a

Step	Hyp	Ref	Expression
1		sb6 1807	. 2 |- ( [ y / x ] ph <-> A. x ( x = y -> ph ) )
2		sbequ12 1694	. . . . 5 |- ( y = x -> ( ph <-> [ x / y ] ph ) )
3	2	equcoms 1634	. . . 4 |- ( x = y -> ( ph <-> [ x / y ] ph ) )
4	3	pm5.74i 178	. . 3 |- ( ( x = y -> ph ) <-> ( x = y -> [ x / y ] ph ) )
5	4	albii 1399	. 2 |- ( A. x ( x = y -> ph ) <-> A. x ( x = y -> [ x / y ] ph ) )
6	1, 5	bitri 182	1 |- ( [ y / x ] ph <-> A. x ( x = y -> [ x / y ] ph ) )

Syntax hints:   -> wi 4   <-> wb 103   A.wal 1282   [wsb 1685

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 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467

This theorem depends on definitions:  df-bi 115  df-sb 1686

This theorem is referenced by: (None)