Theorem sbequ1 1822

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		pm3.4 531	. . 3 |- ((x = y /\ ph) -> (x = y -> ph))
2		19.8a 1665	. . 3 |- ((x = y /\ ph) -> E.x(x = y /\ ph))
3		df-sb 1816	. . 3 |- ([y / x]ph <-> ((x = y -> ph) /\ E.x(x = y /\ ph)))
4	1, 2, 3	sylanbrc 664	. 2 |- ((x = y /\ ph) -> [y / x]ph)
5	4	ex 398	1 |- (x = y -> (ph -> [y / x]ph))

Syntax hints:   -> wi 3   /\ wa 337   = wceq 1586  E.wex 1615  [wsbc 1814

This theorem is referenced by:  sbequ12 1825  dfsb2 1871  sbequi 1874  sbn 1877  sbi1 1878  hbsb4 1895  sb6rf 1907  mo 2053  sb5ALT 17304  sb5ALTVD 17559

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 1608

This theorem depends on definitions:  df-bi 220  df-an 339  df-ex 1616  df-sb 1816

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