Theorem sb6y 872

Description: Equivalence involving substitution with a variable not free.

Hypothesis

Ref	Expression
sb6y.1	|- (ph -> A.yph)

Assertion

Ref	Expression
sb6y	|- ([y / x]ph <-> A.x(x = y -> ph))

Proof of Theorem sb6y

Step	Hyp	Ref	Expression
1		sbequ2 864	. . . . 5 |- (x = y -> ([y / x]ph -> ph))
2	1	a4s 682	. . . 4 |- (A.x x = y -> ([y / x]ph -> ph))
3		ax-10 800	. . . . . . 7 |- (A.y y = x -> (A.yph -> A.xph))
4	3	eq4s 822	. . . . . 6 |- (A.x x = y -> (A.yph -> A.xph))
5		ax-1 3	. . . . . . 7 |- (ph -> (x = y -> ph))
6	5	19.20i 691	. . . . . 6 |- (A.xph -> A.x(x = y -> ph))
7	4, 6	syl6 23	. . . . 5 |- (A.x x = y -> (A.yph -> A.x(x = y -> ph)))
8		sb6y.1	. . . . 5 |- (ph -> A.yph)
9	7, 8	syl5 22	. . . 4 |- (A.x x = y -> (ph -> A.x(x = y -> ph)))
10	2, 9	syld 27	. . 3 |- (A.x x = y -> ([y / x]ph -> A.x(x = y -> ph)))
11		sb4 861	. . 3 |- (-. A.x x = y -> ([y / x]ph -> A.x(x = y -> ph)))
12	10, 11	pm2.61i 110	. 2 |- ([y / x]ph -> A.x(x = y -> ph))
13		sb2 859	. 2 |- (A.x(x = y -> ph) -> [y / x]ph)
14	12, 13	impbi 139	1 |- ([y / x]ph <-> A.x(x = y -> ph))

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

This theorem is referenced by:  sb5y 883

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

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

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