Theorem sbequ1 2110

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		pm3.4 584	. . 3 |- ( ( x = y /\ ph ) -> ( x = y -> ph ) )
2		19.8a 2052	. . 3 |- ( ( x = y /\ ph ) -> E. x ( x = y /\ ph ) )
3		df-sb 1881	. . 3 |- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
4	1, 2, 3	sylanbrc 698	. 2 |- ( ( x = y /\ ph ) -> [ y / x ] ph )
5	4	ex 450	1 |- ( x = y -> ( ph -> [ y / x ] ph ) )

Syntax hints:   -> wi 4   /\ wa 384   E.wex 1704   [wsb 1880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-12 2047

This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1705  df-sb 1881

This theorem is referenced by:  sbequ12  2111  dfsb2  2373  sbequi  2375  sbi1  2392  2eu6  2558  sb5ALT  38731  2pm13.193  38768  2pm13.193VD  39139  sb5ALTVD  39149