Theorem hbs1f 1189

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 1176	. . . 4 |- ([y / x]ph -> E.x(x = y /\ ph))
2		hbs1f.1	. . . . 5 |- (ph -> A.xph)
3	2	19.41 1095	. . . 4 |- (E.x(x = y /\ ph) <-> (E.x x = y /\ ph))
4	1, 3	sylib 198	. . 3 |- ([y / x]ph -> (E.x x = y /\ ph))
5	4	pm3.27d 325	. 2 |- ([y / x]ph -> ph)
6		stdpc4 1185	. . 3 |- (A.xph -> [y / x]ph)
7	6	a5i 989	. 2 |- (A.xph -> A.x[y / x]ph)
8	5, 2, 7	3syl 20	1 |- ([y / x]ph -> A.x[y / x]ph)

Syntax hints:   -> wi 3   /\ wa 223  A.wal 954  E.wex 980  [wsbc 1170

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 963  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172

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