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Metamath Proof Explorer

Theorem frpoins2fg

Description: Well-Founded Induction schema, using implicit substitution. (Contributed by Scott Fenton, 24-Aug-2022)

Ref Expression
Hypotheses frpoins2fg.1 
|- ( y e. A -> ( A. z e. Pred ( R , A , y ) ps -> ph ) )
frpoins2fg.2 
|- F/ y ps
frpoins2fg.3 
|- ( y = z -> ( ph <-> ps ) )
Assertion frpoins2fg 
|- ( ( R Fr A /\ R Po A /\ R Se A ) -> A. y e. A ph )

Proof

Step Hyp Ref Expression
1 frpoins2fg.1 
 |-  ( y e. A -> ( A. z e. Pred ( R , A , y ) ps -> ph ) )
2 frpoins2fg.2 
 |-  F/ y ps
3 frpoins2fg.3 
 |-  ( y = z -> ( ph <-> ps ) )
4 sbsbc 
 |-  ( [ z / y ] ph <-> [. z / y ]. ph )
5 2 3 sbiev 
 |-  ( [ z / y ] ph <-> ps )
6 4 5 bitr3i 
 |-  ( [. z / y ]. ph <-> ps )
7 6 ralbii 
 |-  ( A. z e. Pred ( R , A , y ) [. z / y ]. ph <-> A. z e. Pred ( R , A , y ) ps )
8 1 adantl 
 |-  ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ y e. A ) -> ( A. z e. Pred ( R , A , y ) ps -> ph ) )
9 7 8 biimtrid 
 |-  ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ y e. A ) -> ( A. z e. Pred ( R , A , y ) [. z / y ]. ph -> ph ) )
10 9 frpoinsg 
 |-  ( ( R Fr A /\ R Po A /\ R Se A ) -> A. y e. A ph )