Metamath Proof Explorer


Theorem itgss

Description: Expand the set of an integral by adding zeroes outside the domain. (Contributed by Mario Carneiro, 11-Aug-2014) (Revised by Mario Carneiro, 23-Aug-2014)

Ref Expression
Hypotheses itgss.1 ⊢ φ → A ⊆ B
itgss.2 ⊢ φ ∧ x ∈ B ∖ A → C = 0
Assertion itgss ⊢ φ → ∫ A C dx = ∫ B C dx

Proof

Step Hyp Ref Expression
1 itgss.1 ⊢ φ → A ⊆ B
2 itgss.2 ⊢ φ ∧ x ∈ B ∖ A → C = 0
3 elfzelz ⊢ k ∈ 0 … 3 → k ∈ ℤ
4 iffalse ⊢ ¬ x ∈ A → if x ∈ A if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = 0
5 4 ad2antll ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B ∧ ¬ x ∈ A → if x ∈ A if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = 0
6 eldif ⊢ x ∈ B ∖ A ↔ x ∈ B ∧ ¬ x ∈ A
7 2 adantlr ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B ∖ A → C = 0
8 7 oveq1d ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B ∖ A → C i k = 0 i k
9 ax-icn ⊢ i ∈ ℂ
10 ine0 ⊢ i ≠ 0
11 expclz ⊢ i ∈ ℂ ∧ i ≠ 0 ∧ k ∈ ℤ → i k ∈ ℂ
12 9 10 11 mp3an12 ⊢ k ∈ ℤ → i k ∈ ℂ
13 expne0i ⊢ i ∈ ℂ ∧ i ≠ 0 ∧ k ∈ ℤ → i k ≠ 0
14 9 10 13 mp3an12 ⊢ k ∈ ℤ → i k ≠ 0
15 12 14 div0d ⊢ k ∈ ℤ → 0 i k = 0
16 15 ad2antlr ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B ∖ A → 0 i k = 0
17 8 16 eqtrd ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B ∖ A → C i k = 0
18 17 fveq2d ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B ∖ A → ℜ ⁡ C i k = ℜ ⁡ 0
19 re0 ⊢ ℜ ⁡ 0 = 0
20 18 19 eqtrdi ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B ∖ A → ℜ ⁡ C i k = 0
21 20 ifeq1d ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B ∖ A → if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 = if 0 ≤ ℜ ⁡ C i k 0 0
22 ifid ⊢ if 0 ≤ ℜ ⁡ C i k 0 0 = 0
23 21 22 eqtrdi ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B ∖ A → if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 = 0
24 6 23 sylan2br ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B ∧ ¬ x ∈ A → if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 = 0
25 5 24 eqtr4d ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B ∧ ¬ x ∈ A → if x ∈ A if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
26 25 expr ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B → ¬ x ∈ A → if x ∈ A if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
27 iftrue ⊢ x ∈ A → if x ∈ A if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
28 26 27 pm2.61d2 ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B → if x ∈ A if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
29 iftrue ⊢ x ∈ B → if x ∈ B if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
30 29 adantl ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B → if x ∈ B if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
31 28 30 eqtr4d ⊢ φ ∧ k ∈ ℤ ∧ x ∈ B → if x ∈ A if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = if x ∈ B if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0
32 1 adantr ⊢ φ ∧ k ∈ ℤ → A ⊆ B
33 32 sseld ⊢ φ ∧ k ∈ ℤ → x ∈ A → x ∈ B
34 33 con3dimp ⊢ φ ∧ k ∈ ℤ ∧ ¬ x ∈ B → ¬ x ∈ A
35 34 4 syl ⊢ φ ∧ k ∈ ℤ ∧ ¬ x ∈ B → if x ∈ A if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = 0
36 iffalse ⊢ ¬ x ∈ B → if x ∈ B if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = 0
37 36 adantl ⊢ φ ∧ k ∈ ℤ ∧ ¬ x ∈ B → if x ∈ B if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = 0
38 35 37 eqtr4d ⊢ φ ∧ k ∈ ℤ ∧ ¬ x ∈ B → if x ∈ A if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = if x ∈ B if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0
39 31 38 pm2.61dan ⊢ φ ∧ k ∈ ℤ → if x ∈ A if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0 = if x ∈ B if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0
40 ifan ⊢ if x ∈ A ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 = if x ∈ A if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0
41 ifan ⊢ if x ∈ B ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 = if x ∈ B if 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 0
42 39 40 41 3eqtr4g ⊢ φ ∧ k ∈ ℤ → if x ∈ A ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 = if x ∈ B ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
43 42 mpteq2dv ⊢ φ ∧ k ∈ ℤ → x ∈ ℝ ⟼ if x ∈ A ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 = x ∈ ℝ ⟼ if x ∈ B ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
44 43 fveq2d ⊢ φ ∧ k ∈ ℤ → ∫ 2 ⁡ x ∈ ℝ ⟼ if x ∈ A ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 = ∫ 2 ⁡ x ∈ ℝ ⟼ if x ∈ B ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
45 44 oveq2d ⊢ φ ∧ k ∈ ℤ → i k ⁢ ∫ 2 ⁡ x ∈ ℝ ⟼ if x ∈ A ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 = i k ⁢ ∫ 2 ⁡ x ∈ ℝ ⟼ if x ∈ B ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
46 3 45 sylan2 ⊢ φ ∧ k ∈ 0 … 3 → i k ⁢ ∫ 2 ⁡ x ∈ ℝ ⟼ if x ∈ A ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 = i k ⁢ ∫ 2 ⁡ x ∈ ℝ ⟼ if x ∈ B ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
47 46 sumeq2dv ⊢ φ → ∑ k = 0 3 i k ⁢ ∫ 2 ⁡ x ∈ ℝ ⟼ if x ∈ A ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0 = ∑ k = 0 3 i k ⁢ ∫ 2 ⁡ x ∈ ℝ ⟼ if x ∈ B ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
48 eqid ⊢ ℜ ⁡ C i k = ℜ ⁡ C i k
49 48 dfitg ⊢ ∫ A C dx = ∑ k = 0 3 i k ⁢ ∫ 2 ⁡ x ∈ ℝ ⟼ if x ∈ A ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
50 48 dfitg ⊢ ∫ B C dx = ∑ k = 0 3 i k ⁢ ∫ 2 ⁡ x ∈ ℝ ⟼ if x ∈ B ∧ 0 ≤ ℜ ⁡ C i k ℜ ⁡ C i k 0
51 47 49 50 3eqtr4g ⊢ φ → ∫ A C dx = ∫ B C dx