Theorem sbequ1 2009

Description: An equality theorem for substitution. (Contributed by NM, 16-May-1993.)

Assertion

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

Proof of Theorem sbequ1

Step	Hyp	Ref	Expression
1		pm3.4 559	. . 3 |- ( ( x = y /\ ph ) -> ( x = y -> ph ) )
2		19.8a 1875	. . 3 |- ( ( x = y /\ ph ) -> E. x ( x = y /\ ph ) )
3		df-sb 1758	. . 3 |- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
4	1, 2, 3	sylanbrc 662	. 2 |- ( ( x = y /\ ph ) -> [ y / x ] ph )
5	4	ex 432	1 |- ( x = y -> ( ph -> [ y / x ] ph ) )

Syntax hints:   -> wi 4   /\ wa 367   E.wex 1627   [wsb 1757

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1633  ax-4 1646  ax-5 1719  ax-6 1765  ax-7 1808  ax-12 1872

This theorem depends on definitions:  df-bi 185  df-an 369  df-ex 1628  df-sb 1758

This theorem is referenced by:  sbequ12  2010  dfsb2  2128  sbequi  2130  sbi1  2147  2eu6  2324  sb5ALT  33663  2pm13.193  33700  2pm13.193VD  34085  sb5ALTVD  34095