Metamath Proof Explorer


Theorem indsum

Description: Finite sum of a product with the indicator function / Cartesian product with the indicator function. Note: this theorem cannot be efficiently shortened using sumss2 , unless there are some additional auxiliary theorems like ( if ( x e. A , 1 , 0 ) x. B ) = if ( x e. A , B , 0 ) . (Contributed by Thierry Arnoux, 14-Aug-2017) (Proof shortened by AV, 11-Apr-2026)

Ref Expression
Hypotheses indsum.1 ⊢ φ → O ∈ Fin
indsum.2 ⊢ φ → A ⊆ O
indsum.3 ⊢ φ ∧ x ∈ O → B ∈ ℂ
Assertion indsum ⊢ φ → ∑ x ∈ O 𝟙 O ⁡ A ⁡ x ⁢ B = ∑ x ∈ A B

Proof

Step Hyp Ref Expression
1 indsum.1 ⊢ φ → O ∈ Fin
2 indsum.2 ⊢ φ → A ⊆ O
3 indsum.3 ⊢ φ ∧ x ∈ O → B ∈ ℂ
4 2 sselda ⊢ φ ∧ x ∈ A → x ∈ O
5 1 2 jca ⊢ φ → O ∈ Fin ∧ A ⊆ O
6 fvindre ⊢ O ∈ Fin ∧ A ⊆ O ∧ x ∈ O → 𝟙 O ⁡ A ⁡ x ∈ ℝ
7 5 6 sylan ⊢ φ ∧ x ∈ O → 𝟙 O ⁡ A ⁡ x ∈ ℝ
8 7 recnd ⊢ φ ∧ x ∈ O → 𝟙 O ⁡ A ⁡ x ∈ ℂ
9 8 3 mulcld ⊢ φ ∧ x ∈ O → 𝟙 O ⁡ A ⁡ x ⁢ B ∈ ℂ
10 4 9 syldan ⊢ φ ∧ x ∈ A → 𝟙 O ⁡ A ⁡ x ⁢ B ∈ ℂ
11 1 adantr ⊢ φ ∧ x ∈ O ∖ A → O ∈ Fin
12 2 adantr ⊢ φ ∧ x ∈ O ∖ A → A ⊆ O
13 simpr ⊢ φ ∧ x ∈ O ∖ A → x ∈ O ∖ A
14 ind0 ⊢ O ∈ Fin ∧ A ⊆ O ∧ x ∈ O ∖ A → 𝟙 O ⁡ A ⁡ x = 0
15 11 12 13 14 syl3anc ⊢ φ ∧ x ∈ O ∖ A → 𝟙 O ⁡ A ⁡ x = 0
16 15 oveq1d ⊢ φ ∧ x ∈ O ∖ A → 𝟙 O ⁡ A ⁡ x ⁢ B = 0 ⋅ B
17 difssd ⊢ φ → O ∖ A ⊆ O
18 17 sselda ⊢ φ ∧ x ∈ O ∖ A → x ∈ O
19 3 mul02d ⊢ φ ∧ x ∈ O → 0 ⋅ B = 0
20 18 19 syldan ⊢ φ ∧ x ∈ O ∖ A → 0 ⋅ B = 0
21 16 20 eqtrd ⊢ φ ∧ x ∈ O ∖ A → 𝟙 O ⁡ A ⁡ x ⁢ B = 0
22 2 10 21 1 fsumss ⊢ φ → ∑ x ∈ A 𝟙 O ⁡ A ⁡ x ⁢ B = ∑ x ∈ O 𝟙 O ⁡ A ⁡ x ⁢ B
23 1 adantr ⊢ φ ∧ x ∈ A → O ∈ Fin
24 2 adantr ⊢ φ ∧ x ∈ A → A ⊆ O
25 simpr ⊢ φ ∧ x ∈ A → x ∈ A
26 ind1 ⊢ O ∈ Fin ∧ A ⊆ O ∧ x ∈ A → 𝟙 O ⁡ A ⁡ x = 1
27 23 24 25 26 syl3anc ⊢ φ ∧ x ∈ A → 𝟙 O ⁡ A ⁡ x = 1
28 27 oveq1d ⊢ φ ∧ x ∈ A → 𝟙 O ⁡ A ⁡ x ⁢ B = 1 ⁢ B
29 3 mullidd ⊢ φ ∧ x ∈ O → 1 ⁢ B = B
30 4 29 syldan ⊢ φ ∧ x ∈ A → 1 ⁢ B = B
31 28 30 eqtrd ⊢ φ ∧ x ∈ A → 𝟙 O ⁡ A ⁡ x ⁢ B = B
32 31 sumeq2dv ⊢ φ → ∑ x ∈ A 𝟙 O ⁡ A ⁡ x ⁢ B = ∑ x ∈ A B
33 22 32 eqtr3d ⊢ φ → ∑ x ∈ O 𝟙 O ⁡ A ⁡ x ⁢ B = ∑ x ∈ A B