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Theorem sb4a 1664
Description: A version of sb4 1695 that doesn't require a distinctor antecedent. (Contributed by NM, 2-Feb-2007.)
Assertion
Ref Expression
sb4a ([y / x]yφx(x = yφ))

Proof of Theorem sb4a
StepHypRef Expression
1 sb1 1631 . 2 ([y / x]yφx(x = y yφ))
2 equs5a 1657 . 2 (x(x = y yφ) → x(x = yφ))
31, 2syl 14 1 ([y / x]yφx(x = yφ))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wal 1226  wex 1362  [wsb 1627
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-gen 1318  ax-ie2 1364  ax-11 1378  ax-ial 1409
This theorem depends on definitions:  df-bi 110  df-sb 1628
This theorem is referenced by:  sb6f  1666  hbsb2a  1669
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