Theorem sbequ2 1175

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 1168	. 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 953  E.wex 977  [wsbc 1166

This theorem is referenced by:  stdpc7 1176  sbequ12 1177  dfsb2 1220  sbequi 1223  sbn 1226  sbi1 1227  hbsb4 1243  mo 1386  mopick 1426

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 1168

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