Metamath Proof Explorer


Theorem fressupp

Description: The restriction of a function to its support. (Contributed by Thierry Arnoux, 25-Jun-2024)

Ref Expression
Assertion fressupp Fun F F V Z W F supp Z F = F V × Z

Proof

Step Hyp Ref Expression
1 funrel Fun F Rel F
2 1 3ad2ant1 Fun F F V Z W Rel F
3 suppssdm F supp Z dom F
4 undif F supp Z dom F supp Z F dom F supp Z F = dom F
5 4 biimpi F supp Z dom F supp Z F dom F supp Z F = dom F
6 5 eqcomd F supp Z dom F dom F = supp Z F dom F supp Z F
7 3 6 mp1i Fun F F V Z W dom F = supp Z F dom F supp Z F
8 disjdif supp Z F dom F supp Z F =
9 8 a1i Fun F F V Z W supp Z F dom F supp Z F =
10 reldisjun Rel F dom F = supp Z F dom F supp Z F supp Z F dom F supp Z F = F = F supp Z F F dom F supp Z F
11 2 7 9 10 syl3anc Fun F F V Z W F = F supp Z F F dom F supp Z F
12 11 difeq1d Fun F F V Z W F F dom F supp Z F = F supp Z F F dom F supp Z F F dom F supp Z F
13 resss F dom F supp Z F F
14 sseqin2 F dom F supp Z F F F F dom F supp Z F = F dom F supp Z F
15 13 14 mpbi F F dom F supp Z F = F dom F supp Z F
16 suppiniseg Fun F F V Z W dom F supp Z F = F -1 Z
17 16 reseq2d Fun F F V Z W F dom F supp Z F = F F -1 Z
18 cnvrescnv F -1 Z -1 = F V × Z
19 funcnvres2 Fun F F -1 Z -1 = F F -1 Z
20 18 19 eqtr3id Fun F F V × Z = F F -1 Z
21 20 3ad2ant1 Fun F F V Z W F V × Z = F F -1 Z
22 17 21 eqtr4d Fun F F V Z W F dom F supp Z F = F V × Z
23 15 22 eqtrid Fun F F V Z W F F dom F supp Z F = F V × Z
24 indifbi F F dom F supp Z F = F V × Z F F dom F supp Z F = F V × Z
25 23 24 sylib Fun F F V Z W F F dom F supp Z F = F V × Z
26 8 reseq2i F supp Z F dom F supp Z F = F
27 resindi F supp Z F dom F supp Z F = F supp Z F F dom F supp Z F
28 res0 F =
29 26 27 28 3eqtr3i F supp Z F F dom F supp Z F =
30 undif5 F supp Z F F dom F supp Z F = F supp Z F F dom F supp Z F F dom F supp Z F = F supp Z F
31 29 30 mp1i Fun F F V Z W F supp Z F F dom F supp Z F F dom F supp Z F = F supp Z F
32 12 25 31 3eqtr3rd Fun F F V Z W F supp Z F = F V × Z