Theorem hbsb3 875

Description: If y is not free in ph, x is not free in [y / x]ph.

Hypothesis

Ref	Expression
hbsb3.1	|- (ph -> A.yph)

Assertion

Ref	Expression
hbsb3	|- ([y / x]ph -> A.x[y / x]ph)

Proof of Theorem hbsb3

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	. . . . . 6 |- (A.y y = x -> (A.yph -> A.xph))
4	3	eq4s 822	. . . . 5 |- (A.x x = y -> (A.yph -> A.xph))
5		hbsb3.1	. . . . 5 |- (ph -> A.yph)
6	4, 5	syl5 22	. . . 4 |- (A.x x = y -> (ph -> A.xph))
7	2, 6	syld 27	. . 3 |- (A.x x = y -> ([y / x]ph -> A.xph))
8		sbequ1 863	. . . 4 |- (x = y -> (ph -> [y / x]ph))
9	8	19.20ii 692	. . 3 |- (A.x x = y -> (A.xph -> A.x[y / x]ph))
10	7, 9	syld 27	. 2 |- (A.x x = y -> ([y / x]ph -> A.x[y / x]ph))
11		hbsb2 873	. 2 |- (-. A.x x = y -> ([y / x]ph -> A.x[y / x]ph))
12	10, 11	pm2.61i 110	1 |- ([y / x]ph -> A.x[y / x]ph)

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

This theorem is referenced by:  sbco2 913  sb8 918  mo 1020  axrepndlem1 3738  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

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

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