Metamath Proof Explorer


Theorem stoweidlem13

Description: Lemma for stoweid . This lemma is used to prove the statement abs( f(t) - g(t) ) < 2 epsilon, in the last step of the proof in BrosowskiDeutsh p. 92. (Contributed by Glauco Siliprandi, 20-Apr-2017)

Ref Expression
Hypotheses stoweidlem13.1 ⊢ φ → E ∈ ℝ +
stoweidlem13.2 ⊢ φ → X ∈ ℝ
stoweidlem13.3 ⊢ φ → Y ∈ ℝ
stoweidlem13.4 ⊢ φ → j ∈ ℝ
stoweidlem13.5 ⊢ φ → j − 4 3 ⁢ E < X
stoweidlem13.6 ⊢ φ → X ≤ j − 1 3 ⁢ E
stoweidlem13.7 ⊢ φ → j − 4 3 ⁢ E < Y
stoweidlem13.8 ⊢ φ → Y < j + 1 3 ⁢ E
Assertion stoweidlem13 ⊢ φ → Y − X < 2 ⁢ E

Proof

Step Hyp Ref Expression
1 stoweidlem13.1 ⊢ φ → E ∈ ℝ +
2 stoweidlem13.2 ⊢ φ → X ∈ ℝ
3 stoweidlem13.3 ⊢ φ → Y ∈ ℝ
4 stoweidlem13.4 ⊢ φ → j ∈ ℝ
5 stoweidlem13.5 ⊢ φ → j − 4 3 ⁢ E < X
6 stoweidlem13.6 ⊢ φ → X ≤ j − 1 3 ⁢ E
7 stoweidlem13.7 ⊢ φ → j − 4 3 ⁢ E < Y
8 stoweidlem13.8 ⊢ φ → Y < j + 1 3 ⁢ E
9 3 2 resubcld ⊢ φ → Y − X ∈ ℝ
10 2re ⊢ 2 ∈ ℝ
11 1 rpred ⊢ φ → E ∈ ℝ
12 remulcl ⊢ 2 ∈ ℝ ∧ E ∈ ℝ → 2 ⁢ E ∈ ℝ
13 10 11 12 sylancr ⊢ φ → 2 ⁢ E ∈ ℝ
14 3 recnd ⊢ φ → Y ∈ ℂ
15 2 recnd ⊢ φ → X ∈ ℂ
16 14 15 negsubdi2d ⊢ φ → − Y − X = X − Y
17 2 3 resubcld ⊢ φ → X − Y ∈ ℝ
18 1red ⊢ φ → 1 ∈ ℝ
19 18 11 remulcld ⊢ φ → 1 ⁢ E ∈ ℝ
20 3re ⊢ 3 ∈ ℝ
21 3ne0 ⊢ 3 ≠ 0
22 20 21 rereccli ⊢ 1 3 ∈ ℝ
23 22 a1i ⊢ φ → 1 3 ∈ ℝ
24 4 23 resubcld ⊢ φ → j − 1 3 ∈ ℝ
25 24 11 remulcld ⊢ φ → j − 1 3 ⁢ E ∈ ℝ
26 25 3 resubcld ⊢ φ → j − 1 3 ⁢ E − Y ∈ ℝ
27 4re ⊢ 4 ∈ ℝ
28 27 20 21 3pm3.2i ⊢ 4 ∈ ℝ ∧ 3 ∈ ℝ ∧ 3 ≠ 0
29 redivcl ⊢ 4 ∈ ℝ ∧ 3 ∈ ℝ ∧ 3 ≠ 0 → 4 3 ∈ ℝ
30 28 29 mp1i ⊢ φ → 4 3 ∈ ℝ
31 4 30 resubcld ⊢ φ → j − 4 3 ∈ ℝ
32 31 11 remulcld ⊢ φ → j − 4 3 ⁢ E ∈ ℝ
33 25 32 resubcld ⊢ φ → j − 1 3 ⁢ E − j − 4 3 ⁢ E ∈ ℝ
34 2 25 3 6 lesub1dd ⊢ φ → X − Y ≤ j − 1 3 ⁢ E − Y
35 32 3 25 7 ltsub2dd ⊢ φ → j − 1 3 ⁢ E − Y < j − 1 3 ⁢ E − j − 4 3 ⁢ E
36 17 26 33 34 35 lelttrd ⊢ φ → X − Y < j − 1 3 ⁢ E − j − 4 3 ⁢ E
37 4 recnd ⊢ φ → j ∈ ℂ
38 23 recnd ⊢ φ → 1 3 ∈ ℂ
39 37 38 subcld ⊢ φ → j − 1 3 ∈ ℂ
40 30 recnd ⊢ φ → 4 3 ∈ ℂ
41 37 40 subcld ⊢ φ → j − 4 3 ∈ ℂ
42 11 recnd ⊢ φ → E ∈ ℂ
43 39 41 42 subdird ⊢ φ → j - 1 3 - j − 4 3 ⁢ E = j − 1 3 ⁢ E − j − 4 3 ⁢ E
44 37 38 37 40 sub4d ⊢ φ → j - 1 3 - j − 4 3 = j - j - 1 3 − 4 3
45 37 37 subcld ⊢ φ → j − j ∈ ℂ
46 45 38 40 subsub2d ⊢ φ → j - j - 1 3 − 4 3 = j − j + 4 3 - 1 3
47 44 46 eqtrd ⊢ φ → j - 1 3 - j − 4 3 = j − j + 4 3 - 1 3
48 47 oveq1d ⊢ φ → j - 1 3 - j − 4 3 ⁢ E = j − j + 4 3 - 1 3 ⁢ E
49 43 48 eqtr3d ⊢ φ → j − 1 3 ⁢ E − j − 4 3 ⁢ E = j − j + 4 3 - 1 3 ⁢ E
50 36 49 breqtrd ⊢ φ → X − Y < j − j + 4 3 - 1 3 ⁢ E
51 37 subidd ⊢ φ → j − j = 0
52 51 oveq1d ⊢ φ → j − j + 4 3 - 1 3 = 0 + 4 3 - 1 3
53 4cn ⊢ 4 ∈ ℂ
54 3cn ⊢ 3 ∈ ℂ
55 53 54 21 divcli ⊢ 4 3 ∈ ℂ
56 ax-1cn ⊢ 1 ∈ ℂ
57 56 54 21 divcli ⊢ 1 3 ∈ ℂ
58 1div1e1 ⊢ 1 1 = 1
59 58 oveq2i ⊢ 1 3 + 1 1 = 1 3 + 1
60 ax-1ne0 ⊢ 1 ≠ 0
61 56 54 56 56 21 60 divadddivi ⊢ 1 3 + 1 1 = 1 ⋅ 1 + 1 ⋅ 3 3 ⋅ 1
62 59 61 eqtr3i ⊢ 1 3 + 1 = 1 ⋅ 1 + 1 ⋅ 3 3 ⋅ 1
63 54 56 addcomi ⊢ 3 + 1 = 1 + 3
64 df-4 ⊢ 4 = 3 + 1
65 1t1e1 ⊢ 1 ⋅ 1 = 1
66 54 mullidi ⊢ 1 ⋅ 3 = 3
67 65 66 oveq12i ⊢ 1 ⋅ 1 + 1 ⋅ 3 = 1 + 3
68 63 64 67 3eqtr4ri ⊢ 1 ⋅ 1 + 1 ⋅ 3 = 4
69 68 oveq1i ⊢ 1 ⋅ 1 + 1 ⋅ 3 3 ⋅ 1 = 4 3 ⋅ 1
70 3t1e3 ⊢ 3 ⋅ 1 = 3
71 70 oveq2i ⊢ 4 3 ⋅ 1 = 4 3
72 62 69 71 3eqtri ⊢ 1 3 + 1 = 4 3
73 55 57 56 72 subaddrii ⊢ 4 3 − 1 3 = 1
74 73 oveq2i ⊢ 0 + 4 3 - 1 3 = 0 + 1
75 1e0p1 ⊢ 1 = 0 + 1
76 74 75 eqtr4i ⊢ 0 + 4 3 - 1 3 = 1
77 52 76 eqtrdi ⊢ φ → j − j + 4 3 - 1 3 = 1
78 77 oveq1d ⊢ φ → j − j + 4 3 - 1 3 ⁢ E = 1 ⁢ E
79 50 78 breqtrd ⊢ φ → X − Y < 1 ⁢ E
80 1lt2 ⊢ 1 < 2
81 10 a1i ⊢ φ → 2 ∈ ℝ
82 18 81 1 ltmul1d ⊢ φ → 1 < 2 ↔ 1 ⁢ E < 2 ⁢ E
83 80 82 mpbii ⊢ φ → 1 ⁢ E < 2 ⁢ E
84 17 19 13 79 83 lttrd ⊢ φ → X − Y < 2 ⁢ E
85 16 84 eqbrtrd ⊢ φ → − Y − X < 2 ⁢ E
86 9 13 85 ltnegcon1d ⊢ φ → − 2 ⁢ E < Y − X
87 5re ⊢ 5 ∈ ℝ
88 87 a1i ⊢ φ → 5 ∈ ℝ
89 20 a1i ⊢ φ → 3 ∈ ℝ
90 21 a1i ⊢ φ → 3 ≠ 0
91 88 89 90 redivcld ⊢ φ → 5 3 ∈ ℝ
92 91 11 remulcld ⊢ φ → 5 3 ⁢ E ∈ ℝ
93 2 renegcld ⊢ φ → − X ∈ ℝ
94 4 23 readdcld ⊢ φ → j + 1 3 ∈ ℝ
95 94 11 remulcld ⊢ φ → j + 1 3 ⁢ E ∈ ℝ
96 32 renegcld ⊢ φ → − j − 4 3 ⁢ E ∈ ℝ
97 32 2 ltnegd ⊢ φ → j − 4 3 ⁢ E < X ↔ − X < − j − 4 3 ⁢ E
98 5 97 mpbid ⊢ φ → − X < − j − 4 3 ⁢ E
99 3 93 95 96 8 98 lt2addd ⊢ φ → Y + − X < j + 1 3 ⁢ E + − j − 4 3 ⁢ E
100 14 15 negsubd ⊢ φ → Y + − X = Y − X
101 37 38 42 adddird ⊢ φ → j + 1 3 ⁢ E = j ⁢ E + 1 3 ⁢ E
102 37 40 negsubd ⊢ φ → j + − 4 3 = j − 4 3
103 102 eqcomd ⊢ φ → j − 4 3 = j + − 4 3
104 103 oveq1d ⊢ φ → j − 4 3 ⁢ E = j + − 4 3 ⁢ E
105 40 negcld ⊢ φ → − 4 3 ∈ ℂ
106 37 105 42 adddird ⊢ φ → j + − 4 3 ⁢ E = j ⁢ E + − 4 3 ⁢ E
107 40 42 mulneg1d ⊢ φ → − 4 3 ⁢ E = − 4 3 ⁢ E
108 107 oveq2d ⊢ φ → j ⁢ E + − 4 3 ⁢ E = j ⁢ E + − 4 3 ⁢ E
109 104 106 108 3eqtrd ⊢ φ → j − 4 3 ⁢ E = j ⁢ E + − 4 3 ⁢ E
110 109 negeqd ⊢ φ → − j − 4 3 ⁢ E = − j ⁢ E + − 4 3 ⁢ E
111 37 42 mulcld ⊢ φ → j ⁢ E ∈ ℂ
112 40 42 mulcld ⊢ φ → 4 3 ⁢ E ∈ ℂ
113 112 negcld ⊢ φ → − 4 3 ⁢ E ∈ ℂ
114 111 113 negdid ⊢ φ → − j ⁢ E + − 4 3 ⁢ E = - j ⁢ E + − − 4 3 ⁢ E
115 112 negnegd ⊢ φ → − − 4 3 ⁢ E = 4 3 ⁢ E
116 115 oveq2d ⊢ φ → - j ⁢ E + − − 4 3 ⁢ E = - j ⁢ E + 4 3 ⁢ E
117 110 114 116 3eqtrd ⊢ φ → − j − 4 3 ⁢ E = - j ⁢ E + 4 3 ⁢ E
118 101 117 oveq12d ⊢ φ → j + 1 3 ⁢ E + − j − 4 3 ⁢ E = j ⁢ E + 1 3 ⁢ E + − j ⁢ E + 4 3 ⁢ E
119 38 42 mulcld ⊢ φ → 1 3 ⁢ E ∈ ℂ
120 111 negcld ⊢ φ → − j ⁢ E ∈ ℂ
121 111 119 120 112 add4d ⊢ φ → j ⁢ E + 1 3 ⁢ E + − j ⁢ E + 4 3 ⁢ E = j ⁢ E + − j ⁢ E + 1 3 ⁢ E + 4 3 ⁢ E
122 111 negidd ⊢ φ → j ⁢ E + − j ⁢ E = 0
123 122 oveq1d ⊢ φ → j ⁢ E + − j ⁢ E + 1 3 ⁢ E + 4 3 ⁢ E = 0 + 1 3 ⁢ E + 4 3 ⁢ E
124 119 112 addcld ⊢ φ → 1 3 ⁢ E + 4 3 ⁢ E ∈ ℂ
125 124 addlidd ⊢ φ → 0 + 1 3 ⁢ E + 4 3 ⁢ E = 1 3 ⁢ E + 4 3 ⁢ E
126 121 123 125 3eqtrd ⊢ φ → j ⁢ E + 1 3 ⁢ E + − j ⁢ E + 4 3 ⁢ E = 1 3 ⁢ E + 4 3 ⁢ E
127 38 40 42 adddird ⊢ φ → 1 3 + 4 3 ⁢ E = 1 3 ⁢ E + 4 3 ⁢ E
128 89 recnd ⊢ φ → 3 ∈ ℂ
129 38 40 addcld ⊢ φ → 1 3 + 4 3 ∈ ℂ
130 128 38 40 adddid ⊢ φ → 3 ⁢ 1 3 + 4 3 = 3 ⁢ 1 3 + 3 ⁢ 4 3
131 56 53 addcomi ⊢ 1 + 4 = 4 + 1
132 56 54 21 divcan2i ⊢ 3 ⁢ 1 3 = 1
133 53 54 21 divcan2i ⊢ 3 ⁢ 4 3 = 4
134 132 133 oveq12i ⊢ 3 ⁢ 1 3 + 3 ⁢ 4 3 = 1 + 4
135 df-5 ⊢ 5 = 4 + 1
136 131 134 135 3eqtr4i ⊢ 3 ⁢ 1 3 + 3 ⁢ 4 3 = 5
137 130 136 eqtrdi ⊢ φ → 3 ⁢ 1 3 + 4 3 = 5
138 128 129 90 137 mvllmuld ⊢ φ → 1 3 + 4 3 = 5 3
139 138 oveq1d ⊢ φ → 1 3 + 4 3 ⁢ E = 5 3 ⁢ E
140 126 127 139 3eqtr2d ⊢ φ → j ⁢ E + 1 3 ⁢ E + − j ⁢ E + 4 3 ⁢ E = 5 3 ⁢ E
141 118 140 eqtrd ⊢ φ → j + 1 3 ⁢ E + − j − 4 3 ⁢ E = 5 3 ⁢ E
142 99 100 141 3brtr3d ⊢ φ → Y − X < 5 3 ⁢ E
143 5lt6 ⊢ 5 < 6
144 3t2e6 ⊢ 3 ⋅ 2 = 6
145 143 144 breqtrri ⊢ 5 < 3 ⋅ 2
146 3pos ⊢ 0 < 3
147 20 146 pm3.2i ⊢ 3 ∈ ℝ ∧ 0 < 3
148 ltdivmul ⊢ 5 ∈ ℝ ∧ 2 ∈ ℝ ∧ 3 ∈ ℝ ∧ 0 < 3 → 5 3 < 2 ↔ 5 < 3 ⋅ 2
149 87 10 147 148 mp3an ⊢ 5 3 < 2 ↔ 5 < 3 ⋅ 2
150 145 149 mpbir ⊢ 5 3 < 2
151 150 a1i ⊢ φ → 5 3 < 2
152 91 81 1 151 ltmul1dd ⊢ φ → 5 3 ⁢ E < 2 ⁢ E
153 9 92 13 142 152 lttrd ⊢ φ → Y − X < 2 ⁢ E
154 9 13 absltd ⊢ φ → Y − X < 2 ⁢ E ↔ − 2 ⁢ E < Y − X ∧ Y − X < 2 ⁢ E
155 86 153 154 mpbir2and ⊢ φ → Y − X < 2 ⁢ E