Metamath Proof Explorer


Theorem hashgval

Description: The value of the # function in terms of the mapping G from _om to NN0 . The proof avoids the use of ax-ac . (Contributed by Paul Chapman, 22-Jun-2011) (Revised by Mario Carneiro, 26-Dec-2014)

Ref Expression
Hypothesis hashgval.1 G = rec x V x + 1 0 ω
Assertion hashgval A Fin G card A = A

Proof

Step Hyp Ref Expression
1 hashgval.1 G = rec x V x + 1 0 ω
2 resundir rec x V x + 1 0 ω card V Fin × +∞ Fin = rec x V x + 1 0 ω card Fin V Fin × +∞ Fin
3 eqid rec x V x + 1 0 ω = rec x V x + 1 0 ω
4 eqid rec x V x + 1 0 ω card = rec x V x + 1 0 ω card
5 3 4 hashkf rec x V x + 1 0 ω card : Fin 0
6 ffn rec x V x + 1 0 ω card : Fin 0 rec x V x + 1 0 ω card Fn Fin
7 fnresdm rec x V x + 1 0 ω card Fn Fin rec x V x + 1 0 ω card Fin = rec x V x + 1 0 ω card
8 5 6 7 mp2b rec x V x + 1 0 ω card Fin = rec x V x + 1 0 ω card
9 incom V Fin Fin = Fin V Fin
10 disjdif Fin V Fin =
11 9 10 eqtri V Fin Fin =
12 pnfex +∞ V
13 12 fconst V Fin × +∞ : V Fin +∞
14 ffn V Fin × +∞ : V Fin +∞ V Fin × +∞ Fn V Fin
15 fnresdisj V Fin × +∞ Fn V Fin V Fin Fin = V Fin × +∞ Fin =
16 13 14 15 mp2b V Fin Fin = V Fin × +∞ Fin =
17 11 16 mpbi V Fin × +∞ Fin =
18 8 17 uneq12i rec x V x + 1 0 ω card Fin V Fin × +∞ Fin = rec x V x + 1 0 ω card
19 un0 rec x V x + 1 0 ω card = rec x V x + 1 0 ω card
20 18 19 eqtri rec x V x + 1 0 ω card Fin V Fin × +∞ Fin = rec x V x + 1 0 ω card
21 2 20 eqtri rec x V x + 1 0 ω card V Fin × +∞ Fin = rec x V x + 1 0 ω card
22 df-hash . = rec x V x + 1 0 ω card V Fin × +∞
23 22 reseq1i . Fin = rec x V x + 1 0 ω card V Fin × +∞ Fin
24 1 coeq1i G card = rec x V x + 1 0 ω card
25 21 23 24 3eqtr4i . Fin = G card
26 25 fveq1i . Fin A = G card A
27 cardf2 card : x | y On y x On
28 ffun card : x | y On y x On Fun card
29 27 28 ax-mp Fun card
30 finnum A Fin A dom card
31 fvco Fun card A dom card G card A = G card A
32 29 30 31 sylancr A Fin G card A = G card A
33 26 32 syl5eq A Fin . Fin A = G card A
34 fvres A Fin . Fin A = A
35 33 34 eqtr3d A Fin G card A = A