Metamath Proof Explorer


Theorem gsumxp2

Description: Write a group sum over a cartesian product as a double sum in two ways. This corresponds to the first equation in Lang p. 6. (Contributed by AV, 27-Dec-2023)

Ref Expression
Hypotheses gsumxp2.b B = Base G
gsumxp2.z 0 ˙ = 0 G
gsumxp2.g φ G CMnd
gsumxp2.a φ A V
gsumxp2.r φ C W
gsumxp2.f φ F : A × C B
gsumxp2.w φ finSupp 0 ˙ F
Assertion gsumxp2 φ G k C G j A j F k = G j A G k C j F k

Proof

Step Hyp Ref Expression
1 gsumxp2.b B = Base G
2 gsumxp2.z 0 ˙ = 0 G
3 gsumxp2.g φ G CMnd
4 gsumxp2.a φ A V
5 gsumxp2.r φ C W
6 gsumxp2.f φ F : A × C B
7 gsumxp2.w φ finSupp 0 ˙ F
8 6 fovrnda φ j A k C j F k B
9 7 fsuppimpd φ F supp 0 ˙ Fin
10 simpl φ j A k C φ
11 opelxpi j A k C j k A × C
12 11 ad2antlr φ j A k C ¬ j k supp 0 ˙ F j k A × C
13 simpr φ j A k C ¬ j k supp 0 ˙ F ¬ j k supp 0 ˙ F
14 12 13 eldifd φ j A k C ¬ j k supp 0 ˙ F j k A × C supp 0 ˙ F
15 ssidd φ F supp 0 ˙ F supp 0 ˙
16 4 5 xpexd φ A × C V
17 2 fvexi 0 ˙ V
18 17 a1i φ 0 ˙ V
19 6 15 16 18 suppssr φ j k A × C supp 0 ˙ F F j k = 0 ˙
20 10 14 19 syl2an2r φ j A k C ¬ j k supp 0 ˙ F F j k = 0 ˙
21 20 ex φ j A k C ¬ j k supp 0 ˙ F F j k = 0 ˙
22 df-br j supp 0 ˙ F k j k supp 0 ˙ F
23 22 notbii ¬ j supp 0 ˙ F k ¬ j k supp 0 ˙ F
24 df-ov j F k = F j k
25 24 eqeq1i j F k = 0 ˙ F j k = 0 ˙
26 21 23 25 3imtr4g φ j A k C ¬ j supp 0 ˙ F k j F k = 0 ˙
27 26 impr φ j A k C ¬ j supp 0 ˙ F k j F k = 0 ˙
28 1 2 3 4 5 8 9 27 gsumcom3 φ G j A G k C j F k = G k C G j A j F k
29 28 eqcomd φ G k C G j A j F k = G j A G k C j F k