Theorem sbal2 1005

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 group(s):   y,z   x,z

Proof of Theorem sbal2

Step	Hyp	Ref	Expression
1		eq6 826	. . . 4 |- (-. A.x x = y -> A.y -. A.x x = y)
2		ddeeq1 1001	. . . . . . 7 |- (-. A.x x = y -> (y = z -> A.x y = z))
3	2	19.20i 691	. . . . . 6 |- (A.x -. A.x x = y -> A.x(y = z -> A.x y = z))
4	3	eq6s 827	. . . . 5 |- (-. A.x x = y -> A.x(y = z -> A.x y = z))
5		19.21g 792	. . . . 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	biald 782	. . 3 |- (-. A.x x = y -> (A.yA.x(y = z -> ph) <-> A.y(y = z -> A.xph)))
8		alcom 715	. . 3 |- (A.yA.x(y = z -> ph) <-> A.xA.y(y = z -> ph))
9	7, 8	syl5rbbr 413	. 2 |- (-. A.x x = y -> (A.y(y = z -> A.xph) <-> A.xA.y(y = z -> ph)))
10		sb6 989	. 2 |- ([z / y]A.xph <-> A.y(y = z -> A.xph))
11		sb6 989	. . 3 |- ([z / y]ph <-> A.y(y = z -> ph))
12	11	bial 695	. 2 |- (A.x[z / y]ph <-> A.xA.y(y = z -> ph))
13	9, 10, 12	3bitr4g 428	1 |- (-. A.x x = y -> ([z / y]A.xph <-> A.x[z / y]ph))

Syntax hints:  -. wn 1   -> wi 2   <-> wb 127  A.wal 672   = weq 797  [wsb 852

This theorem is referenced by:  axrepndlem2 3739

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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