Theorem sbal2 1749

Description: Move quantifier in and out of substitution.

Assertion

Ref	Expression
sbal2	|- (-. A.x x = y -> ([z / y]A.xph <-> A.x[z / y]ph))

Distinct variable groups:   y,z   x,z

Proof of Theorem sbal2

Step	Hyp	Ref	Expression
1		hbnae 1507	. . . 4 |- (-. A.x x = y -> A.y -. A.x x = y)
2		dveeq1 1745	. . . . . . 7 |- (-. A.x x = y -> (y = z -> A.x y = z))
3	2	alimi 1338	. . . . . 6 |- (A.x -. A.x x = y -> A.x(y = z -> A.x y = z))
4	3	hbnaes 1508	. . . . 5 |- (-. A.x x = y -> A.x(y = z -> A.x y = z))
5		19.21t 1473	. . . . 5 |- (A.x(y = z -> A.x y = z) -> (A.x(y = z -> ph) <-> (y = z -> A.xph)))
6	4, 5	syl 12	. . . 4 |- (-. A.x x = y -> (A.x(y = z -> ph) <-> (y = z -> A.xph)))
7	1, 6	albid 1459	. . 3 |- (-. A.x x = y -> (A.yA.x(y = z -> ph) <-> A.y(y = z -> A.xph)))
8		alcom 1379	. . 3 |- (A.yA.x(y = z -> ph) <-> A.xA.y(y = z -> ph))
9	7, 8	syl5rbbr 594	. 2 |- (-. A.x x = y -> (A.y(y = z -> A.xph) <-> A.xA.y(y = z -> ph)))
10		sb6 1644	. 2 |- ([z / y]A.xph <-> A.y(y = z -> A.xph))
11		sb6 1644	. . 3 |- ([z / y]ph <-> A.y(y = z -> ph))
12	11	albii 1346	. 2 |- (A.x[z / y]ph <-> A.xA.y(y = z -> ph))
13	9, 10, 12	3bitr4g 614	1 |- (-. A.x x = y -> ([z / y]A.xph <-> A.x[z / y]ph))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 163  A.wal 1296   = wceq 1298  [wsbc 1534

This theorem is referenced by:  axrepndlem2 6097

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-10 1308  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588

This theorem depends on definitions:  df-bi 164  df-an 242  df-ex 1327  df-sb 1536

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