Theorem sbequ1 2084

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		pm3.4 564	. . 3 |- ( ( x = y /\ ph ) -> ( x = y -> ph ) )
2		19.8a 1937	. . 3 |- ( ( x = y /\ ph ) -> E. x ( x = y /\ ph ) )
3		df-sb 1800	. . 3 |- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
4	1, 2, 3	sylanbrc 671	. 2 |- ( ( x = y /\ ph ) -> [ y / x ] ph )
5	4	ex 436	1 |- ( x = y -> ( ph -> [ y / x ] ph ) )

Syntax hints:   -> wi 4   /\ wa 371   E.wex 1665   [wsb 1799

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1671  ax-4 1684  ax-5 1760  ax-6 1807  ax-7 1853  ax-12 1935

This theorem depends on definitions:  df-bi 189  df-an 373  df-ex 1666  df-sb 1800

This theorem is referenced by:  sbequ12  2085  dfsb2  2204  sbequi  2206  sbi1  2223  2eu6  2389  sb5ALT  36893  2pm13.193  36930  2pm13.193VD  37310  sb5ALTVD  37320