Theorem sbequ1 1215

Description: An equality theorem for substitution.

Assertion

Ref	Expression
sbequ1	|- (x = y -> (ph -> [y / x]ph))

Proof of Theorem sbequ1

Step	Hyp	Ref	Expression
1		pm3.4 329	. . . 4 |- ((x = y /\ ph) -> (x = y -> ph))
2		19.8a 1065	. . . 4 |- ((x = y /\ ph) -> E.x(x = y /\ ph))
3	1, 2	jca 286	. . 3 |- ((x = y /\ ph) -> ((x = y -> ph) /\ E.x(x = y /\ ph)))
4		df-sb 1209	. . 3 |- ([y / x]ph <-> ((x = y -> ph) /\ E.x(x = y /\ ph)))
5	3, 4	sylibr 198	. 2 |- ((x = y /\ ph) -> [y / x]ph)
6	5	ex 371	1 |- (x = y -> (ph -> [y / x]ph))

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

This theorem is referenced by:  sbequ12 1218  dfsb2 1262  sbequi 1265  sbn 1268  sbi1 1269  hbsb4 1286  sb6rf 1298  mo 1432

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 1009

This theorem depends on definitions:  df-bi 145  df-an 223  df-ex 1017  df-sb 1209

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