Theorem sb4a 1199

Description: A version of sb4 1223 that doesn't require a distinctor antecedent.

Assertion

Ref	Expression
sb4a	|- ([y / x]A.yph -> A.x(x = y -> ph))

Proof of Theorem sb4a

Step	Hyp	Ref	Expression
1		sb1 1176	. 2 |- ([y / x]A.yph -> E.x(x = y /\ A.yph))
2		equs5a 1197	. 2 |- (E.x(x = y /\ A.yph) -> A.x(x = y -> ph))
3	1, 2	syl 10	1 |- ([y / x]A.yph -> A.x(x = y -> ph))

Syntax hints:   -> wi 3   /\ wa 223  A.wal 954   = wceq 956  E.wex 980  [wsbc 1170

This theorem is referenced by:  sb6f 1201  hbsb2a 1204

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 963  ax-11 967  ax-4 973  ax-5o 975  ax-6o 978

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

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