Theorem sb6xOLD 1554

Description: Equivalence involving substitution for a variable not free.

Hypothesis

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

Assertion

Ref	Expression
sb6xOLD	|- ([y / x]ph <-> A.x(x = y -> ph))

Proof of Theorem sb6xOLD

Step	Hyp	Ref	Expression
1		sb6x.1	. . . 4 |- (ph -> A.xph)
2	1	sbf 1551	. . 3 |- ([y / x]ph <-> ph)
3		ax-1 4	. . . 4 |- (ph -> (x = y -> ph))
4	1, 3	19.21ai 1345	. . 3 |- (ph -> A.x(x = y -> ph))
5	2, 4	sylbi 216	. 2 |- ([y / x]ph -> A.x(x = y -> ph))
6		sb2 1541	. 2 |- (A.x(x = y -> ph) -> [y / x]ph)
7	5, 6	impbii 174	1 |- ([y / x]ph <-> A.x(x = y -> ph))

Syntax hints:   -> wi 3   <-> wb 163  A.wal 1296   = wceq 1298  [wsbc 1534

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 1305  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481

This theorem depends on definitions:  df-bi 164  df-an 242  df-ex 1327  df-sb 1536

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