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Metamath Proof Explorer

Theorem sb5ALT

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

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

Proof

Step Hyp Ref Expression
1 equsb1 
 |-  [ y / x ] x = y
2 sban 
 |-  ( [ y / x ] ( x = y /\ ph ) <-> ( [ y / x ] x = y /\ [ y / x ] ph ) )
3 2 simplbi2com 
 |-  ( [ y / x ] ph -> ( [ y / x ] x = y -> [ y / x ] ( x = y /\ ph ) ) )
4 1 3 mpi 
 |-  ( [ y / x ] ph -> [ y / x ] ( x = y /\ ph ) )
5 spsbe 
 |-  ( [ y / x ] ( x = y /\ ph ) -> E. x ( x = y /\ ph ) )
6 4 5 syl 
 |-  ( [ y / x ] ph -> E. x ( x = y /\ ph ) )
7 hbs1 
 |-  ( [ y / x ] ph -> A. x [ y / x ] ph )
8 simpr 
 |-  ( ( x = y /\ ph ) -> ph )
9 8 a1i 
 |-  ( E. x ( x = y /\ ph ) -> ( ( x = y /\ ph ) -> ph ) )
10 simpl 
 |-  ( ( x = y /\ ph ) -> x = y )
11 10 a1i 
 |-  ( E. x ( x = y /\ ph ) -> ( ( x = y /\ ph ) -> x = y ) )
12 sbequ1 
 |-  ( x = y -> ( ph -> [ y / x ] ph ) )
13 12 com12 
 |-  ( ph -> ( x = y -> [ y / x ] ph ) )
14 9 11 13 syl6c 
 |-  ( E. x ( x = y /\ ph ) -> ( ( x = y /\ ph ) -> [ y / x ] ph ) )
15 7 14 exlimexi 
 |-  ( E. x ( x = y /\ ph ) -> [ y / x ] ph )
16 6 15 impbii 
 |-  ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )