Theorem sb5 1268

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 1267	. 2 |- ([y / x]ph <-> A.x(x = y -> ph))
2		sb56 1266	. 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 954   = wceq 956  E.wex 980  [wsbc 1170

This theorem is referenced by:  2sb5 1335  dfsb7 1340  sbelx 1344

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 963  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-16 1210  ax-11o 1218

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172

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