Theorem sbequ1 1977

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		pm3.4 561	. . 3 |- ( ( x = y /\ ph ) -> ( x = y -> ph ) )
2		19.8a 1843	. . 3 |- ( ( x = y /\ ph ) -> E. x ( x = y /\ ph ) )
3		df-sb 1727	. . 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 434	1 |- ( x = y -> ( ph -> [ y / x ] ph ) )

Syntax hints:   -> wi 4   /\ wa 369   E.wex 1599   [wsb 1726

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1605  ax-4 1618  ax-5 1691  ax-6 1734  ax-7 1776  ax-12 1840

This theorem depends on definitions:  df-bi 185  df-an 371  df-ex 1600  df-sb 1727

This theorem is referenced by:  sbequ12  1978  dfsb2  2100  sbequi  2102  sbi1  2119  2eu6  2369  sb5ALT  33163  2pm13.193  33193  2pm13.193VD  33571  sb5ALTVD  33581