Theorem sbequ1 1768

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		pm3.4 333	. . 3 |- ( ( x = y /\ ph ) -> ( x = y -> ph ) )
2		19.8a 1590	. . 3 |- ( ( x = y /\ ph ) -> E. x ( x = y /\ ph ) )
3		df-sb 1763	. . 3 |- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
4	1, 2, 3	sylanbrc 417	. 2 |- ( ( x = y /\ ph ) -> [ y / x ] ph )
5	4	ex 115	1 |- ( x = y -> ( ph -> [ y / x ] ph ) )

Syntax hints:   -> wi 4   /\ wa 104   E.wex 1492   [wsb 1762

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-4 1510

This theorem depends on definitions:  df-bi 117  df-sb 1763

This theorem is referenced by:  sbequ12  1771  sbequi  1839  sb6rf  1853  mo2n  2054  bj-bdfindes  14786  bj-findes  14818