Theorem sbequ8 1861

Description: Elimination of equality from antecedent after substitution. (Contributed by NM, 5-Aug-1993.) (Proof revised by Jim Kingdon, 20-Jan-2018.)

Assertion

Ref	Expression
sbequ8	|- ( [ y / x ] ph <-> [ y / x ] ( x = y -> ph ) )

Proof of Theorem sbequ8

Step	Hyp	Ref	Expression
1		pm5.4 249	. . 3 |- ( ( x = y -> ( x = y -> ph ) ) <-> ( x = y -> ph ) )
2		simpl 109	. . . . . 6 |- ( ( x = y /\ ( x = y -> ph ) ) -> x = y )
3		pm3.35 347	. . . . . 6 |- ( ( x = y /\ ( x = y -> ph ) ) -> ph )
4	2, 3	jca 306	. . . . 5 |- ( ( x = y /\ ( x = y -> ph ) ) -> ( x = y /\ ph ) )
5		simpl 109	. . . . . 6 |- ( ( x = y /\ ph ) -> x = y )
6		pm3.4 333	. . . . . 6 |- ( ( x = y /\ ph ) -> ( x = y -> ph ) )
7	5, 6	jca 306	. . . . 5 |- ( ( x = y /\ ph ) -> ( x = y /\ ( x = y -> ph ) ) )
8	4, 7	impbii 126	. . . 4 |- ( ( x = y /\ ( x = y -> ph ) ) <-> ( x = y /\ ph ) )
9	8	exbii 1619	. . 3 |- ( E. x ( x = y /\ ( x = y -> ph ) ) <-> E. x ( x = y /\ ph ) )
10	1, 9	anbi12i 460	. 2 |- ( ( ( x = y -> ( x = y -> ph ) ) /\ E. x ( x = y /\ ( x = y -> ph ) ) ) <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
11		df-sb 1777	. 2 |- ( [ y / x ] ( x = y -> ph ) <-> ( ( x = y -> ( x = y -> ph ) ) /\ E. x ( x = y /\ ( x = y -> ph ) ) ) )
12		df-sb 1777	. 2 |- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
13	10, 11, 12	3bitr4ri 213	1 |- ( [ y / x ] ph <-> [ y / x ] ( x = y -> ph ) )

Syntax hints:   -> wi 4   /\ wa 104   <-> wb 105   E.wex 1506   [wsb 1776

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-5 1461  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-4 1524  ax-ial 1548

This theorem depends on definitions:  df-bi 117  df-sb 1777

This theorem is referenced by:  sbidm  1865