Theorem sb5ALT 38731

Description: Equivalence for substitution. Alternate proof of sb5 2430. This proof is sb5ALTVD 39149 automatically translated and minimized. (Contributed by Alan Sare, 21-Apr-2013.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
sb5ALT	|- ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )

Distinct variable group:   x, y

Allowed substitution hints:   ph( x, y)

Proof of Theorem sb5ALT

Step	Hyp	Ref	Expression
1		equsb1 2368	. . . 4 |- [ y / x ] x = y
2		sban 2399	. . . . 5 |- ( [ y / x ] ( x = y /\ ph ) <-> ( [ y / x ] x = y /\ [ y / x ] ph ) )
3	2	simplbi2com 657	. . . 4 |- ( [ y / x ] ph -> ( [ y / x ] x = y -> [ y / x ] ( x = y /\ ph ) ) )
4	1, 3	mpi 20	. . 3 |- ( [ y / x ] ph -> [ y / x ] ( x = y /\ ph ) )
5		spsbe 1884	. . 3 |- ( [ y / x ] ( x = y /\ ph ) -> E. x ( x = y /\ ph ) )
6	4, 5	syl 17	. 2 |- ( [ y / x ] ph -> E. x ( x = y /\ ph ) )
7		hbs1 2436	. . 3 |- ( [ y / x ] ph -> A. x [ y / x ] ph )
8		simpr 477	. . . . 5 |- ( ( x = y /\ ph ) -> ph )
9	8	a1i 11	. . . 4 |- ( E. x ( x = y /\ ph ) -> ( ( x = y /\ ph ) -> ph ) )
10		simpl 473	. . . . 5 |- ( ( x = y /\ ph ) -> x = y )
11	10	a1i 11	. . . 4 |- ( E. x ( x = y /\ ph ) -> ( ( x = y /\ ph ) -> x = y ) )
12		sbequ1 2110	. . . . 5 |- ( x = y -> ( ph -> [ y / x ] ph ) )
13	12	com12 32	. . . 4 |- ( ph -> ( x = y -> [ y / x ] ph ) )
14	9, 11, 13	syl6c 70	. . 3 |- ( E. x ( x = y /\ ph ) -> ( ( x = y /\ ph ) -> [ y / x ] ph ) )
15	7, 14	exlimexi 38730	. 2 |- ( E. x ( x = y /\ ph ) -> [ y / x ] ph )
16	6, 15	impbii 199	1 |- ( [ y / x ] ph <-> E. 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  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by: (None)