Metamath Proof Explorer


Theorem itg1addlem2

Description: Lemma for itg1add . The function I represents the pieces into which we will break up the domain of the sum. Since it is infinite only when both i and j are zero, we arbitrarily define it to be zero there to simplify the sums that are manipulated in itg1addlem4 and itg1addlem5 . (Contributed by Mario Carneiro, 26-Jun-2014)

Ref Expression
Hypotheses i1fadd.1 ⊢ φ → F ∈ dom ⁡ ∫ 1
i1fadd.2 ⊢ φ → G ∈ dom ⁡ ∫ 1
itg1add.3 ⊢ I = i ∈ ℝ , j ∈ ℝ ⟼ if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j
Assertion itg1addlem2 ⊢ φ → I : ℝ 2 ⟶ ℝ

Proof

Step Hyp Ref Expression
1 i1fadd.1 ⊢ φ → F ∈ dom ⁡ ∫ 1
2 i1fadd.2 ⊢ φ → G ∈ dom ⁡ ∫ 1
3 itg1add.3 ⊢ I = i ∈ ℝ , j ∈ ℝ ⟼ if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j
4 iffalse ⊢ ¬ i = 0 ∧ j = 0 → if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j = vol ⁡ F -1 i ∩ G -1 j
5 4 adantl ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ ¬ i = 0 ∧ j = 0 → if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j = vol ⁡ F -1 i ∩ G -1 j
6 i1fima ⊢ F ∈ dom ⁡ ∫ 1 → F -1 i ∈ dom ⁡ vol
7 1 6 syl ⊢ φ → F -1 i ∈ dom ⁡ vol
8 i1fima ⊢ G ∈ dom ⁡ ∫ 1 → G -1 j ∈ dom ⁡ vol
9 2 8 syl ⊢ φ → G -1 j ∈ dom ⁡ vol
10 inmbl ⊢ F -1 i ∈ dom ⁡ vol ∧ G -1 j ∈ dom ⁡ vol → F -1 i ∩ G -1 j ∈ dom ⁡ vol
11 7 9 10 syl2anc ⊢ φ → F -1 i ∩ G -1 j ∈ dom ⁡ vol
12 11 ad2antrr ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ ¬ i = 0 ∧ j = 0 → F -1 i ∩ G -1 j ∈ dom ⁡ vol
13 mblvol ⊢ F -1 i ∩ G -1 j ∈ dom ⁡ vol → vol ⁡ F -1 i ∩ G -1 j = vol * ⁡ F -1 i ∩ G -1 j
14 12 13 syl ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ ¬ i = 0 ∧ j = 0 → vol ⁡ F -1 i ∩ G -1 j = vol * ⁡ F -1 i ∩ G -1 j
15 5 14 eqtrd ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ ¬ i = 0 ∧ j = 0 → if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j = vol * ⁡ F -1 i ∩ G -1 j
16 neorian ⊢ i ≠ 0 ∨ j ≠ 0 ↔ ¬ i = 0 ∧ j = 0
17 inss1 ⊢ F -1 i ∩ G -1 j ⊆ F -1 i
18 7 ad2antrr ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ i ≠ 0 → F -1 i ∈ dom ⁡ vol
19 mblss ⊢ F -1 i ∈ dom ⁡ vol → F -1 i ⊆ ℝ
20 18 19 syl ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ i ≠ 0 → F -1 i ⊆ ℝ
21 mblvol ⊢ F -1 i ∈ dom ⁡ vol → vol ⁡ F -1 i = vol * ⁡ F -1 i
22 18 21 syl ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ i ≠ 0 → vol ⁡ F -1 i = vol * ⁡ F -1 i
23 1 ad2antrr ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ i ≠ 0 → F ∈ dom ⁡ ∫ 1
24 simplrl ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ i ≠ 0 → i ∈ ℝ
25 simpr ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ i ≠ 0 → i ≠ 0
26 eldifsn ⊢ i ∈ ℝ ∖ 0 ↔ i ∈ ℝ ∧ i ≠ 0
27 24 25 26 sylanbrc ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ i ≠ 0 → i ∈ ℝ ∖ 0
28 i1fima2sn ⊢ F ∈ dom ⁡ ∫ 1 ∧ i ∈ ℝ ∖ 0 → vol ⁡ F -1 i ∈ ℝ
29 23 27 28 syl2anc ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ i ≠ 0 → vol ⁡ F -1 i ∈ ℝ
30 22 29 eqeltrrd ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ i ≠ 0 → vol * ⁡ F -1 i ∈ ℝ
31 ovolsscl ⊢ F -1 i ∩ G -1 j ⊆ F -1 i ∧ F -1 i ⊆ ℝ ∧ vol * ⁡ F -1 i ∈ ℝ → vol * ⁡ F -1 i ∩ G -1 j ∈ ℝ
32 17 20 30 31 mp3an2i ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ i ≠ 0 → vol * ⁡ F -1 i ∩ G -1 j ∈ ℝ
33 inss2 ⊢ F -1 i ∩ G -1 j ⊆ G -1 j
34 2 adantr ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ → G ∈ dom ⁡ ∫ 1
35 34 8 syl ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ → G -1 j ∈ dom ⁡ vol
36 35 adantr ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ j ≠ 0 → G -1 j ∈ dom ⁡ vol
37 mblss ⊢ G -1 j ∈ dom ⁡ vol → G -1 j ⊆ ℝ
38 36 37 syl ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ j ≠ 0 → G -1 j ⊆ ℝ
39 mblvol ⊢ G -1 j ∈ dom ⁡ vol → vol ⁡ G -1 j = vol * ⁡ G -1 j
40 36 39 syl ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ j ≠ 0 → vol ⁡ G -1 j = vol * ⁡ G -1 j
41 2 ad2antrr ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ j ≠ 0 → G ∈ dom ⁡ ∫ 1
42 simplrr ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ j ≠ 0 → j ∈ ℝ
43 simpr ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ j ≠ 0 → j ≠ 0
44 eldifsn ⊢ j ∈ ℝ ∖ 0 ↔ j ∈ ℝ ∧ j ≠ 0
45 42 43 44 sylanbrc ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ j ≠ 0 → j ∈ ℝ ∖ 0
46 i1fima2sn ⊢ G ∈ dom ⁡ ∫ 1 ∧ j ∈ ℝ ∖ 0 → vol ⁡ G -1 j ∈ ℝ
47 41 45 46 syl2anc ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ j ≠ 0 → vol ⁡ G -1 j ∈ ℝ
48 40 47 eqeltrrd ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ j ≠ 0 → vol * ⁡ G -1 j ∈ ℝ
49 ovolsscl ⊢ F -1 i ∩ G -1 j ⊆ G -1 j ∧ G -1 j ⊆ ℝ ∧ vol * ⁡ G -1 j ∈ ℝ → vol * ⁡ F -1 i ∩ G -1 j ∈ ℝ
50 33 38 48 49 mp3an2i ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ j ≠ 0 → vol * ⁡ F -1 i ∩ G -1 j ∈ ℝ
51 32 50 jaodan ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ i ≠ 0 ∨ j ≠ 0 → vol * ⁡ F -1 i ∩ G -1 j ∈ ℝ
52 16 51 sylan2br ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ ¬ i = 0 ∧ j = 0 → vol * ⁡ F -1 i ∩ G -1 j ∈ ℝ
53 15 52 eqeltrd ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ ∧ ¬ i = 0 ∧ j = 0 → if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j ∈ ℝ
54 53 ex ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ → ¬ i = 0 ∧ j = 0 → if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j ∈ ℝ
55 iftrue ⊢ i = 0 ∧ j = 0 → if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j = 0
56 0re ⊢ 0 ∈ ℝ
57 55 56 eqeltrdi ⊢ i = 0 ∧ j = 0 → if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j ∈ ℝ
58 54 57 pm2.61d2 ⊢ φ ∧ i ∈ ℝ ∧ j ∈ ℝ → if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j ∈ ℝ
59 58 ralrimivva ⊢ φ → ∀ i ∈ ℝ ∀ j ∈ ℝ if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j ∈ ℝ
60 3 fmpo ⊢ ∀ i ∈ ℝ ∀ j ∈ ℝ if i = 0 ∧ j = 0 0 vol ⁡ F -1 i ∩ G -1 j ∈ ℝ ↔ I : ℝ 2 ⟶ ℝ
61 59 60 sylib ⊢ φ → I : ℝ 2 ⟶ ℝ