Theorem hbs1f 874

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

Hypothesis

Ref	Expression
hbs1f.1	|- (ph -> A.xph)

Assertion

Ref	Expression
hbs1f	|- ([y / x]ph -> A.x[y / x]ph)

Proof of Theorem hbs1f

Step	Hyp	Ref	Expression
1		sb1 858	. . . 4 |- ([y / x]ph -> E.x(x = y /\ ph))
2		hbs1f.1	. . . . 5 |- (ph -> A.xph)
3	2	19.41 774	. . . 4 |- (E.x(x = y /\ ph) <-> (E.x x = y /\ ph))
4	1, 3	sylib 173	. . 3 |- ([y / x]ph -> (E.x x = y /\ ph))
5	4	pm3.27d 262	. 2 |- ([y / x]ph -> ph)
6		stdpc4 869	. . 3 |- (A.xph -> [y / x]ph)
7	6	a5i 687	. 2 |- (A.xph -> A.x[y / x]ph)
8	5, 2, 7	3syl 21	1 |- ([y / x]ph -> A.x[y / x]ph)

Syntax hints:   -> wi 2   /\ wa 196  A.wal 672  E.wex 678   = weq 797  [wsb 852

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-gen 677  ax-9 799

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853

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