Theorem sbequ1 1542

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		df-sb 1536	. . 3 |- ([y / x]ph <-> ((x = y -> ph) /\ E.x(x = y /\ ph)))
2		pm3.4 358	. . 3 |- ((x = y /\ ph) -> (x = y -> ph))
3		19.8a 1376	. . 3 |- ((x = y /\ ph) -> E.x(x = y /\ ph))
4	1, 2, 3	sylanbrc 527	. 2 |- ((x = y /\ ph) -> [y / x]ph)
5	4	ex 402	1 |- (x = y -> (ph -> [y / x]ph))

Syntax hints:   -> wi 3   /\ wa 240   = wceq 1298  E.wex 1326  [wsbc 1534

This theorem is referenced by:  sbequ12 1545  dfsb2 1595  sbequi 1598  sbn 1601  sbi1 1602  hbsb4 1620  sb6rf 1635  sb6rfOLD 1636  mo 1787

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 1319

This theorem depends on definitions:  df-bi 164  df-an 242  df-ex 1327  df-sb 1536

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