Metamath Proof Explorer


Theorem mo4f

Description: At-most-one quantifier expressed using implicit substitution. Note that the disjoint variable condition on y , ph can be replaced by the nonfreeness hypothesis |- F/ y ph with essentially the same proof. (Contributed by NM, 10-Apr-2004) Remove dependency on ax-13 . (Revised by Wolf Lammen, 19-Jan-2023)

Ref Expression
Hypotheses mo4f.1
|- F/ x ps
mo4f.2
|- ( x = y -> ( ph <-> ps ) )
Assertion mo4f
|- ( E* x ph <-> A. x A. y ( ( ph /\ ps ) -> x = y ) )

Proof

Step Hyp Ref Expression
1 mo4f.1
 |-  F/ x ps
2 mo4f.2
 |-  ( x = y -> ( ph <-> ps ) )
3 nfv
 |-  F/ y ph
4 3 mo3
 |-  ( E* x ph <-> A. x A. y ( ( ph /\ [ y / x ] ph ) -> x = y ) )
5 1 2 sbiev
 |-  ( [ y / x ] ph <-> ps )
6 5 anbi2i
 |-  ( ( ph /\ [ y / x ] ph ) <-> ( ph /\ ps ) )
7 6 imbi1i
 |-  ( ( ( ph /\ [ y / x ] ph ) -> x = y ) <-> ( ( ph /\ ps ) -> x = y ) )
8 7 2albii
 |-  ( A. x A. y ( ( ph /\ [ y / x ] ph ) -> x = y ) <-> A. x A. y ( ( ph /\ ps ) -> x = y ) )
9 4 8 bitri
 |-  ( E* x ph <-> A. x A. y ( ( ph /\ ps ) -> x = y ) )