Theorem sbequ1 1651

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		pm3.4 316	. . 3 |- ( ( x = y /\ ph ) -> ( x = y -> ph ) )
2		19.8a 1482	. . 3 |- ( ( x = y /\ ph ) -> E. x ( x = y /\ ph ) )
3		df-sb 1646	. . 3 |- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
4	1, 2, 3	sylanbrc 394	. 2 |- ( ( x = y /\ ph ) -> [ y / x ] ph )
5	4	ex 108	1 |- ( x = y -> ( ph -> [ y / x ] ph ) )

Syntax hints:   -> wi 4   /\ wa 97   E.wex 1381   [wsb 1645

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-4 1400

This theorem depends on definitions:  df-bi 110  df-sb 1646

This theorem is referenced by:  sbequ12  1654  sbequi  1720  sb6rf  1733  mo2n  1928  bj-bdfindes  10074  bj-findes  10106