Theorem sb6x 1184

Description: Equivalence involving substitution for a variable not free.

Hypothesis

Ref	Expression
sb6x.1	|- (ph -> A.xph)

Assertion

Ref	Expression
sb6x	|- ([y / x]ph <-> A.x(x = y -> ph))

Proof of Theorem sb6x

Step	Hyp	Ref	Expression
1		sb6x.1	. . . 4 |- (ph -> A.xph)
2	1	sbf 1182	. . 3 |- ([y / x]ph <-> ph)
3		ax-1 4	. . . 4 |- (ph -> (x = y -> ph))
4	1, 3	19.21ai 995	. . 3 |- (ph -> A.x(x = y -> ph))
5	2, 4	sylbi 199	. 2 |- ([y / x]ph -> A.x(x = y -> ph))
6		sb2 1173	. 2 |- (A.x(x = y -> ph) -> [y / x]ph)
7	5, 6	impbi 157	1 |- ([y / x]ph <-> A.x(x = y -> ph))

Syntax hints:   -> wi 3   <-> wb 146  A.wal 951   = wceq 953  [wsbc 1166

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 960  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 978  df-sb 1168

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