Theorem sbequ2 1177

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		pm3.26 319	. . 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 1170	. 2 |- ([y / x]ph <-> ((x = y -> ph) /\ E.x(x = y /\ ph)))
4	2, 3	syl5ib 206	1 |- (x = y -> ([y / x]ph -> ph))

Syntax hints:   -> wi 3   /\ wa 223   = wceq 954  E.wex 978  [wsbc 1168

This theorem is referenced by:  stdpc7 1178  sbequ12 1179  dfsb2 1223  sbequi 1226  sbn 1229  sbi1 1230  hbsb4 1246  mo 1391  mopick 1431

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 147  df-an 225  df-sb 1170

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