Metamath Proof Explorer


Theorem fuchomOLD

Description: Obsolete proof of fuchom as of 14-Oct-2024. (Contributed by Mario Carneiro, 6-Jan-2017) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypotheses fucbas.q Q = C FuncCat D
fuchom.n N = C Nat D
Assertion fuchomOLD N = Hom Q

Proof

Step Hyp Ref Expression
1 fucbas.q Q = C FuncCat D
2 fuchom.n N = C Nat D
3 eqid C Func D = C Func D
4 eqid Base C = Base C
5 eqid comp D = comp D
6 simpl C Cat D Cat C Cat
7 simpr C Cat D Cat D Cat
8 eqid comp Q = comp Q
9 1 3 2 4 5 6 7 8 fuccofval C Cat D Cat comp Q = v C Func D × C Func D , h C Func D 1 st v / f 2 nd v / g b g N h , a f N g x Base C b x 1 st f x 1 st g x comp D 1 st h x a x
10 1 3 2 4 5 6 7 9 fucval C Cat D Cat Q = Base ndx C Func D Hom ndx N comp ndx comp Q
11 catstr Base ndx C Func D Hom ndx N comp ndx comp Q Struct 1 15
12 homid Hom = Slot Hom ndx
13 snsstp2 Hom ndx N Base ndx C Func D Hom ndx N comp ndx comp Q
14 2 ovexi N V
15 14 a1i C Cat D Cat N V
16 eqid Hom Q = Hom Q
17 10 11 12 13 15 16 strfv3 C Cat D Cat Hom Q = N
18 17 eqcomd C Cat D Cat N = Hom Q
19 df-hom Hom = Slot 14
20 19 str0 = Hom
21 2 natffn N Fn C Func D × C Func D
22 funcrcl f C Func D C Cat D Cat
23 22 con3i ¬ C Cat D Cat ¬ f C Func D
24 23 eq0rdv ¬ C Cat D Cat C Func D =
25 24 xpeq2d ¬ C Cat D Cat C Func D × C Func D = C Func D ×
26 xp0 C Func D × =
27 25 26 eqtrdi ¬ C Cat D Cat C Func D × C Func D =
28 27 fneq2d ¬ C Cat D Cat N Fn C Func D × C Func D N Fn
29 21 28 mpbii ¬ C Cat D Cat N Fn
30 fn0 N Fn N =
31 29 30 sylib ¬ C Cat D Cat N =
32 fnfuc FuncCat Fn Cat × Cat
33 32 fndmi dom FuncCat = Cat × Cat
34 33 ndmov ¬ C Cat D Cat C FuncCat D =
35 1 34 eqtrid ¬ C Cat D Cat Q =
36 35 fveq2d ¬ C Cat D Cat Hom Q = Hom
37 20 31 36 3eqtr4a ¬ C Cat D Cat N = Hom Q
38 18 37 pm2.61i N = Hom Q