Theorem sbalyz 1999

Description: Move universal quantifier in and out of substitution. Identical to sbal 2000 except that it has an additional distinct variable constraint on y and z. (Contributed by Jim Kingdon, 29-Dec-2017.)

Assertion

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

Distinct variable group:   x, y, z

Allowed substitution hints:   ph( x, y, z)

Proof of Theorem sbalyz

Step	Hyp	Ref	Expression
1		nfa1 1541	. . . 4 |- F/ x A. x ph
2	1	nfsbxy 1942	. . 3 |- F/ x [ z / y ] A. x ph
3		ax-4 1510	. . . 4 |- ( A. x ph -> ph )
4	3	sbimi 1764	. . 3 |- ( [ z / y ] A. x ph -> [ z / y ] ph )
5	2, 4	alrimi 1522	. 2 |- ( [ z / y ] A. x ph -> A. x [ z / y ] ph )
6		sb6 1886	. . . . 5 |- ( [ z / y ] ph <-> A. y ( y = z -> ph ) )
7	6	albii 1470	. . . 4 |- ( A. x [ z / y ] ph <-> A. x A. y ( y = z -> ph ) )
8		alcom 1478	. . . 4 |- ( A. x A. y ( y = z -> ph ) <-> A. y A. x ( y = z -> ph ) )
9	7, 8	bitri 184	. . 3 |- ( A. x [ z / y ] ph <-> A. y A. x ( y = z -> ph ) )
10		nfv 1528	. . . . . 6 |- F/ x y = z
11	10	stdpc5 1584	. . . . 5 |- ( A. x ( y = z -> ph ) -> ( y = z -> A. x ph ) )
12	11	alimi 1455	. . . 4 |- ( A. y A. x ( y = z -> ph ) -> A. y ( y = z -> A. x ph ) )
13		sb2 1767	. . . 4 |- ( A. y ( y = z -> A. x ph ) -> [ z / y ] A. x ph )
14	12, 13	syl 14	. . 3 |- ( A. y A. x ( y = z -> ph ) -> [ z / y ] A. x ph )
15	9, 14	sylbi 121	. 2 |- ( A. x [ z / y ] ph -> [ z / y ] A. x ph )
16	5, 15	impbii 126	1 |- ( [ z / y ] A. x ph <-> A. x [ z / 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-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535

This theorem depends on definitions:  df-bi 117  df-nf 1461  df-sb 1763

This theorem is referenced by:  sbal  2000