Theorem sbequ1 1918

Description: An equality theorem for substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

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

Proof of Theorem sbequ1

Step	Hyp	Ref	Expression
1		pm3.4 544	. . 3 |- ((x = y /\ ph) -> (x = y -> ph))
2		19.8a 1756	. . 3 |- ((x = y /\ ph) -> E.x(x = y /\ ph))
3		df-sb 1649	. . 3 |- ([y / x]ph <-> ((x = y -> ph) /\ E.x(x = y /\ ph)))
4	1, 2, 3	sylanbrc 645	. 2 |- ((x = y /\ ph) -> [y / x]ph)
5	4	ex 423	1 |- (x = y -> (ph -> [y / x]ph))

Syntax hints:   -> wi 4   /\ wa 358  E.wex 1541  [wsb 1648

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-11 1746

This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-sb 1649

This theorem is referenced by:  sbequ12  1919  dfsb2  2055  sbequi  2059  sbn  2062  sbi1  2063  sb6rf  2091  mo  2226