Theorem sb5rf 1258

Description: Reversed substitution.

Hypothesis

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

Assertion

Ref	Expression
sb5rf	|- (ph <-> E.y(y = x /\ [y / x]ph))

Proof of Theorem sb5rf

Step	Hyp	Ref	Expression
1		sb5rf.1	. . . 4 |- (ph -> A.yph)
2	1	sbid2 1252	. . 3 |- ([x / y][y / x]ph <-> ph)
3		sb1 1175	. . 3 |- ([x / y][y / x]ph -> E.y(y = x /\ [y / x]ph))
4	2, 3	sylbir 201	. 2 |- (ph -> E.y(y = x /\ [y / x]ph))
5		sbequ12r 1181	. . . 4 |- (y = x -> ([y / x]ph <-> ph))
6	5	biimpa 416	. . 3 |- ((y = x /\ [y / x]ph) -> ph)
7	1, 6	19.23ai 1063	. 2 |- (E.y(y = x /\ [y / x]ph) -> ph)
8	4, 7	impbi 157	1 |- (ph <-> E.y(y = x /\ [y / x]ph))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  A.wal 953   = wceq 955  E.wex 979  [wsbc 1169

This theorem is referenced by:  2sb5rf 1337

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 961  ax-gen 962  ax-8 963  ax-10 965  ax-12 967  ax-4 972  ax-5o 974  ax-6o 977  ax-9o 1122  ax-10o 1139  ax-11o 1217

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 980  df-sb 1171

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