Theorem sbequ2 1216

Description: An equality theorem for substitution.

Assertion

Ref	Expression
sbequ2	|- (x = y -> ([y / x]ph -> ph))

Proof of Theorem sbequ2

Step	Hyp	Ref	Expression
1		pm3.26 317	. . 3 |- (((x = y -> ph) /\ E.x(x = y /\ ph)) -> (x = y -> ph))
2	1	com12 11	. 2 |- (x = y -> (((x = y -> ph) /\ E.x(x = y /\ ph)) -> ph))
3		df-sb 1209	. 2 |- ([y / x]ph <-> ((x = y -> ph) /\ E.x(x = y /\ ph)))
4	2, 3	syl5ib 204	1 |- (x = y -> ([y / x]ph -> ph))

Syntax hints:   -> wi 3   /\ wa 221   = wceq 992  E.wex 1016  [wsbc 1207

This theorem is referenced by:  stdpc7 1217  sbequ12 1218  dfsb2 1262  sbequi 1265  sbn 1268  sbi1 1269  hbsb4 1286  mo 1432  mopick 1472

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7

This theorem depends on definitions:  df-bi 145  df-an 223  df-sb 1209

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