Theorem sb5 1266

Description: Equivalence for substitution. Similar to Theorem 6.1 of [Quine] p. 40.

Assertion

Ref	Expression
sb5	|- ([y / x]ph <-> E.x(x = y /\ ph))

Distinct variable group:   x,y

Proof of Theorem sb5

Step	Hyp	Ref	Expression
1		sb6 1265	. 2 |- ([y / x]ph <-> A.x(x = y -> ph))
2		sb56 1264	. 2 |- (E.x(x = y /\ ph) <-> A.x(x = y -> ph))
3	1, 2	bitr4 176	1 |- ([y / x]ph <-> E.x(x = y /\ ph))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  A.wal 952   = wceq 954  E.wex 978  [wsbc 1168

This theorem is referenced by:  2sb5 1333  dfsb7 1338  sbelx 1342

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 961  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-16 1208  ax-11o 1216

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

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