Theorem sbal2 1338

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 1130	. . . 4 |- (-. A.x x = y -> A.y -. A.x x = y)
2		dveeq1 1335	. . . . . . 7 |- (-. A.x x = y -> (y = z -> A.x y = z))
3	2	19.20i 968	. . . . . 6 |- (A.x -. A.x x = y -> A.x(y = z -> A.x y = z))
4	3	hbnaes 1131	. . . . 5 |- (-. A.x x = y -> A.x(y = z -> A.x y = z))
5		19.21t 1091	. . . . 5 |- (A.x(y = z -> A.x y = z) -> (A.x(y = z -> ph) <-> (y = z -> A.xph)))
6	4, 5	syl 10	. . . 4 |- (-. A.x x = y -> (A.x(y = z -> ph) <-> (y = z -> A.xph)))
7	1, 6	albid 1080	. . 3 |- (-. A.x x = y -> (A.yA.x(y = z -> ph) <-> A.y(y = z -> A.xph)))
8		alcom 1008	. . 3 |- (A.yA.x(y = z -> ph) <-> A.xA.y(y = z -> ph))
9	7, 8	syl5rbbr 533	. 2 |- (-. A.x x = y -> (A.y(y = z -> A.xph) <-> A.xA.y(y = z -> ph)))
10		sb6 1251	. 2 |- ([z / y]A.xph <-> A.y(y = z -> A.xph))
11		sb6 1251	. . 3 |- ([z / y]ph <-> A.y(y = z -> ph))
12	11	albii 975	. 2 |- (A.x[z / y]ph <-> A.xA.y(y = z -> ph))
13	9, 10, 12	3bitr4g 553	1 |- (-. A.x x = y -> ([z / y]A.xph <-> A.x[z / y]ph))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146  A.wal 950   = wceq 1099  [wsbc 1153

This theorem is referenced by:  axrepndlem2 4868

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 951  ax-5 952  ax-6 953  ax-7 954  ax-gen 955  ax-8 1101  ax-9 1102  ax-10 1103  ax-12 1104  ax-17 1190  ax-16 1194  ax-11o 1202

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 957  df-sb 1155

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