Theorem sb6rf 1258

Description: Reversed substitution.

Hypothesis

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

Assertion

Ref	Expression
sb6rf	|- (ph <-> A.y(y = x -> [y / x]ph))

Proof of Theorem sb6rf

Step	Hyp	Ref	Expression
1		sb5rf.1	. . 3 |- (ph -> A.yph)
2		sbequ1 1176	. . . . 5 |- (x = y -> (ph -> [y / x]ph))
3	2	equcoms 1128	. . . 4 |- (y = x -> (ph -> [y / x]ph))
4	3	com12 11	. . 3 |- (ph -> (y = x -> [y / x]ph))
5	1, 4	19.21ai 996	. 2 |- (ph -> A.y(y = x -> [y / x]ph))
6		sb2 1175	. . . 4 |- (A.y(y = x -> [y / x]ph) -> [x / y][y / x]ph)
7		sbco 1250	. . . 4 |- ([x / y][y / x]ph <-> [x / y]ph)
8	6, 7	sylib 198	. . 3 |- (A.y(y = x -> [y / x]ph) -> [x / y]ph)
9	1	sbf 1184	. . 3 |- ([x / y]ph <-> ph)
10	8, 9	sylib 198	. 2 |- (A.y(y = x -> [y / x]ph) -> ph)
11	5, 10	impbi 157	1 |- (ph <-> A.y(y = x -> [y / x]ph))

Syntax hints:   -> wi 3   <-> wb 146  A.wal 952   = wceq 954  [wsbc 1168

This theorem is referenced by:  2sb6rf 1337  eu1 1390

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-10 964  ax-12 966  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-11o 1216

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 979  df-sb 1170

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