Theorem sb6rfOLD 1636

Description: Reversed substitution.

Hypothesis

Ref	Expression
sb5rf.1	|- (ph -> A.yph)

Assertion

Ref	Expression
sb6rfOLD	|- (ph <-> A.y(y = x -> [y / x]ph))

Proof of Theorem sb6rfOLD

Step	Hyp	Ref	Expression
1		sb5rf.1	. . 3 |- (ph -> A.yph)
2		sbequ1 1542	. . . . 5 |- (x = y -> (ph -> [y / x]ph))
3	2	equcoms 1489	. . . 4 |- (y = x -> (ph -> [y / x]ph))
4	3	com12 14	. . 3 |- (ph -> (y = x -> [y / x]ph))
5	1, 4	19.21ai 1345	. 2 |- (ph -> A.y(y = x -> [y / x]ph))
6		sb2 1541	. . . 4 |- (A.y(y = x -> [y / x]ph) -> [x / y][y / x]ph)
7		sbco 1625	. . . 4 |- ([x / y][y / x]ph <-> [x / y]ph)
8	6, 7	sylib 215	. . 3 |- (A.y(y = x -> [y / x]ph) -> [x / y]ph)
9	1	sbf 1551	. . 3 |- ([x / y]ph <-> ph)
10	8, 9	sylib 215	. 2 |- (A.y(y = x -> [y / x]ph) -> ph)
11	5, 10	impbii 174	1 |- (ph <-> A.y(y = x -> [y / x]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-7 1304  ax-gen 1305  ax-8 1306  ax-10 1308  ax-12 1310  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-11o 1588

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

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