Theorem sbequ1 2097

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		pm3.4 570	. . 3 |- ( ( x = y /\ ph ) -> ( x = y -> ph ) )
2		19.8a 1955	. . 3 |- ( ( x = y /\ ph ) -> E. x ( x = y /\ ph ) )
3		df-sb 1806	. . 3 |- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
4	1, 2, 3	sylanbrc 677	. 2 |- ( ( x = y /\ ph ) -> [ y / x ] ph )
5	4	ex 441	1 |- ( x = y -> ( ph -> [ y / x ] ph ) )

Syntax hints:   -> wi 4   /\ wa 376   E.wex 1671   [wsb 1805

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1677  ax-4 1690  ax-5 1766  ax-6 1813  ax-7 1859  ax-12 1950

This theorem depends on definitions:  df-bi 190  df-an 378  df-ex 1672  df-sb 1806

This theorem is referenced by:  sbequ12  2098  dfsb2  2222  sbequi  2224  sbi1  2241  2eu6  2407  sb5ALT  36952  2pm13.193  36989  2pm13.193VD  37363  sb5ALTVD  37373