Metamath Proof Explorer


Theorem fullres2c

Description: Condition for a full functor to also be a full functor into the restriction. (Contributed by Mario Carneiro, 30-Jan-2017)

Ref Expression
Hypotheses ffthres2c.a
|- A = ( Base ` C )
ffthres2c.e
|- E = ( D |`s S )
ffthres2c.d
|- ( ph -> D e. Cat )
ffthres2c.r
|- ( ph -> S e. V )
ffthres2c.1
|- ( ph -> F : A --> S )
Assertion fullres2c
|- ( ph -> ( F ( C Full D ) G <-> F ( C Full E ) G ) )

Proof

Step Hyp Ref Expression
1 ffthres2c.a
 |-  A = ( Base ` C )
2 ffthres2c.e
 |-  E = ( D |`s S )
3 ffthres2c.d
 |-  ( ph -> D e. Cat )
4 ffthres2c.r
 |-  ( ph -> S e. V )
5 ffthres2c.1
 |-  ( ph -> F : A --> S )
6 1 2 3 4 5 funcres2c
 |-  ( ph -> ( F ( C Func D ) G <-> F ( C Func E ) G ) )
7 eqid
 |-  ( Hom ` D ) = ( Hom ` D )
8 2 7 resshom
 |-  ( S e. V -> ( Hom ` D ) = ( Hom ` E ) )
9 4 8 syl
 |-  ( ph -> ( Hom ` D ) = ( Hom ` E ) )
10 9 oveqd
 |-  ( ph -> ( ( F ` x ) ( Hom ` D ) ( F ` y ) ) = ( ( F ` x ) ( Hom ` E ) ( F ` y ) ) )
11 10 eqeq2d
 |-  ( ph -> ( ran ( x G y ) = ( ( F ` x ) ( Hom ` D ) ( F ` y ) ) <-> ran ( x G y ) = ( ( F ` x ) ( Hom ` E ) ( F ` y ) ) ) )
12 11 2ralbidv
 |-  ( ph -> ( A. x e. A A. y e. A ran ( x G y ) = ( ( F ` x ) ( Hom ` D ) ( F ` y ) ) <-> A. x e. A A. y e. A ran ( x G y ) = ( ( F ` x ) ( Hom ` E ) ( F ` y ) ) ) )
13 6 12 anbi12d
 |-  ( ph -> ( ( F ( C Func D ) G /\ A. x e. A A. y e. A ran ( x G y ) = ( ( F ` x ) ( Hom ` D ) ( F ` y ) ) ) <-> ( F ( C Func E ) G /\ A. x e. A A. y e. A ran ( x G y ) = ( ( F ` x ) ( Hom ` E ) ( F ` y ) ) ) ) )
14 1 7 isfull
 |-  ( F ( C Full D ) G <-> ( F ( C Func D ) G /\ A. x e. A A. y e. A ran ( x G y ) = ( ( F ` x ) ( Hom ` D ) ( F ` y ) ) ) )
15 eqid
 |-  ( Hom ` E ) = ( Hom ` E )
16 1 15 isfull
 |-  ( F ( C Full E ) G <-> ( F ( C Func E ) G /\ A. x e. A A. y e. A ran ( x G y ) = ( ( F ` x ) ( Hom ` E ) ( F ` y ) ) ) )
17 13 14 16 3bitr4g
 |-  ( ph -> ( F ( C Full D ) G <-> F ( C Full E ) G ) )