Theorem sbequ8 1768

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		pm5.4 247	. . 3 |- ( ( x = y -> ( x = y -> ph ) ) <-> ( x = y -> ph ) )
2		simpl 107	. . . . . 6 |- ( ( x = y /\ ( x = y -> ph ) ) -> x = y )
3		pm3.35 339	. . . . . 6 |- ( ( x = y /\ ( x = y -> ph ) ) -> ph )
4	2, 3	jca 300	. . . . 5 |- ( ( x = y /\ ( x = y -> ph ) ) -> ( x = y /\ ph ) )
5		simpl 107	. . . . . 6 |- ( ( x = y /\ ph ) -> x = y )
6		pm3.4 326	. . . . . 6 |- ( ( x = y /\ ph ) -> ( x = y -> ph ) )
7	5, 6	jca 300	. . . . 5 |- ( ( x = y /\ ph ) -> ( x = y /\ ( x = y -> ph ) ) )
8	4, 7	impbii 124	. . . 4 |- ( ( x = y /\ ( x = y -> ph ) ) <-> ( x = y /\ ph ) )
9	8	exbii 1536	. . 3 |- ( E. x ( x = y /\ ( x = y -> ph ) ) <-> E. x ( x = y /\ ph ) )
10	1, 9	anbi12i 447	. 2 |- ( ( ( x = y -> ( x = y -> ph ) ) /\ E. x ( x = y /\ ( x = y -> ph ) ) ) <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
11		df-sb 1686	. 2 |- ( [ y / x ] ( x = y -> ph ) <-> ( ( x = y -> ( x = y -> ph ) ) /\ E. x ( x = y /\ ( x = y -> ph ) ) ) )
12		df-sb 1686	. 2 |- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
13	10, 11, 12	3bitr4ri 211	1 |- ( [ y / x ] ph <-> [ y / x ] ( x = y -> ph ) )

Syntax hints:   -> wi 4   /\ wa 102   <-> wb 103   E.wex 1421   [wsb 1685

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-4 1440  ax-ial 1467

This theorem depends on definitions:  df-bi 115  df-sb 1686

This theorem is referenced by:  sbidm  1772