Theorem sbequ8 1885

Description: Elimination of equality from antecedent after substitution. (Contributed by NM, 5-Aug-1993.) Reduce dependencies on axioms. (Revised by Wolf Lammen, 28-Jul-2018.)

Assertion

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

Proof of Theorem sbequ8

Step	Hyp	Ref	Expression
1		pm5.4 377	. . . 4 |- ( ( x = y -> ( x = y -> ph ) ) <-> ( x = y -> ph ) )
2	1	bicomi 214	. . 3 |- ( ( x = y -> ph ) <-> ( x = y -> ( x = y -> ph ) ) )
3		abai 836	. . . 4 |- ( ( x = y /\ ph ) <-> ( x = y /\ ( x = y -> ph ) ) )
4	3	exbii 1774	. . 3 |- ( E. x ( x = y /\ ph ) <-> E. x ( x = y /\ ( x = y -> ph ) ) )
5	2, 4	anbi12i 733	. 2 |- ( ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) <-> ( ( x = y -> ( x = y -> ph ) ) /\ E. x ( x = y /\ ( x = y -> ph ) ) ) )
6		df-sb 1881	. 2 |- ( [ y / x ] ph <-> ( ( x = y -> ph ) /\ E. x ( x = y /\ ph ) ) )
7		df-sb 1881	. 2 |- ( [ y / x ] ( x = y -> ph ) <-> ( ( x = y -> ( x = y -> ph ) ) /\ E. x ( x = y /\ ( x = y -> ph ) ) ) )
8	5, 6, 7	3bitr4i 292	1 |- ( [ y / x ] ph <-> [ y / x ] ( x = y -> ph ) )

Syntax hints:   -> wi 4   <-> wb 196   /\ 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

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

This theorem is referenced by: (None)