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 = ( 0g𝐺 )
gsumxp2.g ( 𝜑𝐺 ∈ CMnd )
gsumxp2.a ( 𝜑𝐴𝑉 )
gsumxp2.r ( 𝜑𝐶𝑊 )
gsumxp2.f ( 𝜑𝐹 : ( 𝐴 × 𝐶 ) ⟶ 𝐵 )
gsumxp2.w ( 𝜑𝐹 finSupp 0 )
Assertion gsumxp2 ( 𝜑 → ( 𝐺 Σg ( 𝑘𝐶 ↦ ( 𝐺 Σg ( 𝑗𝐴 ↦ ( 𝑗 𝐹 𝑘 ) ) ) ) ) = ( 𝐺 Σg ( 𝑗𝐴 ↦ ( 𝐺 Σg ( 𝑘𝐶 ↦ ( 𝑗 𝐹 𝑘 ) ) ) ) ) )

Proof

Step Hyp Ref Expression
1 gsumxp2.b 𝐵 = ( Base ‘ 𝐺 )
2 gsumxp2.z 0 = ( 0g𝐺 )
3 gsumxp2.g ( 𝜑𝐺 ∈ CMnd )
4 gsumxp2.a ( 𝜑𝐴𝑉 )
5 gsumxp2.r ( 𝜑𝐶𝑊 )
6 gsumxp2.f ( 𝜑𝐹 : ( 𝐴 × 𝐶 ) ⟶ 𝐵 )
7 gsumxp2.w ( 𝜑𝐹 finSupp 0 )
8 6 fovrnda ( ( 𝜑 ∧ ( 𝑗𝐴𝑘𝐶 ) ) → ( 𝑗 𝐹 𝑘 ) ∈ 𝐵 )
9 7 fsuppimpd ( 𝜑 → ( 𝐹 supp 0 ) ∈ Fin )
10 simpl ( ( 𝜑 ∧ ( 𝑗𝐴𝑘𝐶 ) ) → 𝜑 )
11 opelxpi ( ( 𝑗𝐴𝑘𝐶 ) → ⟨ 𝑗 , 𝑘 ⟩ ∈ ( 𝐴 × 𝐶 ) )
12 11 ad2antlr ( ( ( 𝜑 ∧ ( 𝑗𝐴𝑘𝐶 ) ) ∧ ¬ ⟨ 𝑗 , 𝑘 ⟩ ∈ ( 𝐹 supp 0 ) ) → ⟨ 𝑗 , 𝑘 ⟩ ∈ ( 𝐴 × 𝐶 ) )
13 simpr ( ( ( 𝜑 ∧ ( 𝑗𝐴𝑘𝐶 ) ) ∧ ¬ ⟨ 𝑗 , 𝑘 ⟩ ∈ ( 𝐹 supp 0 ) ) → ¬ ⟨ 𝑗 , 𝑘 ⟩ ∈ ( 𝐹 supp 0 ) )
14 12 13 eldifd ( ( ( 𝜑 ∧ ( 𝑗𝐴𝑘𝐶 ) ) ∧ ¬ ⟨ 𝑗 , 𝑘 ⟩ ∈ ( 𝐹 supp 0 ) ) → ⟨ 𝑗 , 𝑘 ⟩ ∈ ( ( 𝐴 × 𝐶 ) ∖ ( 𝐹 supp 0 ) ) )
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 ( 𝜑0 ∈ V )
19 6 15 16 18 suppssr ( ( 𝜑 ∧ ⟨ 𝑗 , 𝑘 ⟩ ∈ ( ( 𝐴 × 𝐶 ) ∖ ( 𝐹 supp 0 ) ) ) → ( 𝐹 ‘ ⟨ 𝑗 , 𝑘 ⟩ ) = 0 )
20 10 14 19 syl2an2r ( ( ( 𝜑 ∧ ( 𝑗𝐴𝑘𝐶 ) ) ∧ ¬ ⟨ 𝑗 , 𝑘 ⟩ ∈ ( 𝐹 supp 0 ) ) → ( 𝐹 ‘ ⟨ 𝑗 , 𝑘 ⟩ ) = 0 )
21 20 ex ( ( 𝜑 ∧ ( 𝑗𝐴𝑘𝐶 ) ) → ( ¬ ⟨ 𝑗 , 𝑘 ⟩ ∈ ( 𝐹 supp 0 ) → ( 𝐹 ‘ ⟨ 𝑗 , 𝑘 ⟩ ) = 0 ) )
22 df-br ( 𝑗 ( 𝐹 supp 0 ) 𝑘 ↔ ⟨ 𝑗 , 𝑘 ⟩ ∈ ( 𝐹 supp 0 ) )
23 22 notbii ( ¬ 𝑗 ( 𝐹 supp 0 ) 𝑘 ↔ ¬ ⟨ 𝑗 , 𝑘 ⟩ ∈ ( 𝐹 supp 0 ) )
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 ∧ ( 𝑗𝐴𝑘𝐶 ) ) → ( ¬ 𝑗 ( 𝐹 supp 0 ) 𝑘 → ( 𝑗 𝐹 𝑘 ) = 0 ) )
27 26 impr ( ( 𝜑 ∧ ( ( 𝑗𝐴𝑘𝐶 ) ∧ ¬ 𝑗 ( 𝐹 supp 0 ) 𝑘 ) ) → ( 𝑗 𝐹 𝑘 ) = 0 )
28 1 2 3 4 5 8 9 27 gsumcom3 ( 𝜑 → ( 𝐺 Σg ( 𝑗𝐴 ↦ ( 𝐺 Σg ( 𝑘𝐶 ↦ ( 𝑗 𝐹 𝑘 ) ) ) ) ) = ( 𝐺 Σg ( 𝑘𝐶 ↦ ( 𝐺 Σg ( 𝑗𝐴 ↦ ( 𝑗 𝐹 𝑘 ) ) ) ) ) )
29 28 eqcomd ( 𝜑 → ( 𝐺 Σg ( 𝑘𝐶 ↦ ( 𝐺 Σg ( 𝑗𝐴 ↦ ( 𝑗 𝐹 𝑘 ) ) ) ) ) = ( 𝐺 Σg ( 𝑗𝐴 ↦ ( 𝐺 Σg ( 𝑘𝐶 ↦ ( 𝑗 𝐹 𝑘 ) ) ) ) ) )