Theorem sbal2 2134

Description: Move quantifier in and out of substitution. (Contributed by NM, 2-Jan-2002.)

Assertion

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

Distinct variable groups:   y,z   x,z

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

Proof of Theorem sbal2

Step	Hyp	Ref	Expression
1		alcom 1737	. . 3 |- (A.yA.x(y = z -> ph) <-> A.xA.y(y = z -> ph))
2		nfnae 1956	. . . 4 |- F/y -. A.x x = y
3		nfnae 1956	. . . . . 6 |- F/x -. A.x x = y
4		dveeq1 2018	. . . . . 6 |- (-. A.x x = y -> (y = z -> A.x y = z))
5	3, 4	nfd 1766	. . . . 5 |- (-. A.x x = y -> F/x y = z)
6		19.21t 1795	. . . . 5 |- ( F/x y = z -> (A.x(y = z -> ph) <-> (y = z -> A.xph)))
7	5, 6	syl 15	. . . 4 |- (-. A.x x = y -> (A.x(y = z -> ph) <-> (y = z -> A.xph)))
8	2, 7	albid 1772	. . 3 |- (-. A.x x = y -> (A.yA.x(y = z -> ph) <-> A.y(y = z -> A.xph)))
9	1, 8	syl5rbbr 251	. 2 |- (-. A.x x = y -> (A.y(y = z -> A.xph) <-> A.xA.y(y = z -> ph)))
10		sb6 2099	. 2 |- ([z / y]A.xph <-> A.y(y = z -> A.xph))
11		sb6 2099	. . 3 |- ([z / y]ph <-> A.y(y = z -> ph))
12	11	albii 1566	. 2 |- (A.x[z / y]ph <-> A.xA.y(y = z -> ph))
13	9, 10, 12	3bitr4g 279	1 |- (-. A.x x = y -> ([z / y]A.xph <-> A.x[z / y]ph))

Syntax hints:  -. wn 3   -> wi 4   <-> wb 176  A.wal 1540   F/wnf 1544  [wsb 1648

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by: (None)