Metamath Proof Explorer


Theorem 01sqrexlem7

Description: Lemma for 01sqrex . (Contributed by Mario Carneiro, 10-Jul-2013) (Proof shortened by AV, 9-Jul-2022)

Ref Expression
Hypotheses 01sqrexlem1.1 ⊢ S = x ∈ ℝ + | x 2 ≤ A
01sqrexlem1.2 ⊢ B = sup S ℝ <
01sqrexlem5.3 ⊢ T = y | ∃ a ∈ S ∃ b ∈ S y = a ⁢ b
Assertion 01sqrexlem7 ⊢ A ∈ ℝ + ∧ A ≤ 1 → B 2 = A

Proof

Step Hyp Ref Expression
1 01sqrexlem1.1 ⊢ S = x ∈ ℝ + | x 2 ≤ A
2 01sqrexlem1.2 ⊢ B = sup S ℝ <
3 01sqrexlem5.3 ⊢ T = y | ∃ a ∈ S ∃ b ∈ S y = a ⁢ b
4 1 2 3 01sqrexlem6 ⊢ A ∈ ℝ + ∧ A ≤ 1 → B 2 ≤ A
5 1 2 01sqrexlem3 ⊢ A ∈ ℝ + ∧ A ≤ 1 → S ⊆ ℝ ∧ S ≠ ∅ ∧ ∃ y ∈ ℝ ∀ z ∈ S z ≤ y
6 5 adantr ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → S ⊆ ℝ ∧ S ≠ ∅ ∧ ∃ y ∈ ℝ ∀ z ∈ S z ≤ y
7 1 2 01sqrexlem4 ⊢ A ∈ ℝ + ∧ A ≤ 1 → B ∈ ℝ + ∧ B ≤ 1
8 7 adantr ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B ∈ ℝ + ∧ B ≤ 1
9 8 simpld ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B ∈ ℝ +
10 rpre ⊢ A ∈ ℝ + → A ∈ ℝ
11 10 adantr ⊢ A ∈ ℝ + ∧ A ≤ 1 → A ∈ ℝ
12 rpre ⊢ B ∈ ℝ + → B ∈ ℝ
13 12 adantr ⊢ B ∈ ℝ + ∧ B ≤ 1 → B ∈ ℝ
14 7 13 syl ⊢ A ∈ ℝ + ∧ A ≤ 1 → B ∈ ℝ
15 14 resqcld ⊢ A ∈ ℝ + ∧ A ≤ 1 → B 2 ∈ ℝ
16 11 15 resubcld ⊢ A ∈ ℝ + ∧ A ≤ 1 → A − B 2 ∈ ℝ
17 16 adantr ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 ∈ ℝ
18 15 11 posdifd ⊢ A ∈ ℝ + ∧ A ≤ 1 → B 2 < A ↔ 0 < A − B 2
19 18 biimpa ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 0 < A − B 2
20 17 19 elrpd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 ∈ ℝ +
21 3rp ⊢ 3 ∈ ℝ +
22 rpdivcl ⊢ A − B 2 ∈ ℝ + ∧ 3 ∈ ℝ + → A − B 2 3 ∈ ℝ +
23 20 21 22 sylancl ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 3 ∈ ℝ +
24 9 23 rpaddcld ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B + A − B 2 3 ∈ ℝ +
25 14 adantr ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B ∈ ℝ
26 25 recnd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B ∈ ℂ
27 3nn ⊢ 3 ∈ ℕ
28 nndivre ⊢ A − B 2 ∈ ℝ ∧ 3 ∈ ℕ → A − B 2 3 ∈ ℝ
29 16 27 28 sylancl ⊢ A ∈ ℝ + ∧ A ≤ 1 → A − B 2 3 ∈ ℝ
30 29 adantr ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 3 ∈ ℝ
31 30 recnd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 3 ∈ ℂ
32 binom2 ⊢ B ∈ ℂ ∧ A − B 2 3 ∈ ℂ → B + A − B 2 3 2 = B 2 + 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2
33 26 31 32 syl2anc ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B + A − B 2 3 2 = B 2 + 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2
34 15 adantr ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B 2 ∈ ℝ
35 34 recnd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B 2 ∈ ℂ
36 2re ⊢ 2 ∈ ℝ
37 25 30 remulcld ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B ⁢ A − B 2 3 ∈ ℝ
38 remulcl ⊢ 2 ∈ ℝ ∧ B ⁢ A − B 2 3 ∈ ℝ → 2 ⁢ B ⁢ A − B 2 3 ∈ ℝ
39 36 37 38 sylancr ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ⁢ A − B 2 3 ∈ ℝ
40 39 recnd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ⁢ A − B 2 3 ∈ ℂ
41 30 resqcld ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 3 2 ∈ ℝ
42 41 recnd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 3 2 ∈ ℂ
43 35 40 42 addassd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B 2 + 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2 = B 2 + 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2
44 33 43 eqtrd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B + A − B 2 3 2 = B 2 + 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2
45 2cn ⊢ 2 ∈ ℂ
46 mulass ⊢ 2 ∈ ℂ ∧ B ∈ ℂ ∧ A − B 2 3 ∈ ℂ → 2 ⁢ B ⁢ A − B 2 3 = 2 ⁢ B ⁢ A − B 2 3
47 45 26 31 46 mp3an2i ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ⁢ A − B 2 3 = 2 ⁢ B ⁢ A − B 2 3
48 47 eqcomd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ⁢ A − B 2 3 = 2 ⁢ B ⁢ A − B 2 3
49 31 sqvald ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 3 2 = A − B 2 3 ⁢ A − B 2 3
50 48 49 oveq12d ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2 = 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 ⁢ A − B 2 3
51 remulcl ⊢ 2 ∈ ℝ ∧ B ∈ ℝ → 2 ⁢ B ∈ ℝ
52 36 25 51 sylancr ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ∈ ℝ
53 52 recnd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ∈ ℂ
54 53 31 31 adddird ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B + A − B 2 3 ⁢ A − B 2 3 = 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 ⁢ A − B 2 3
55 50 54 eqtr4d ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2 = 2 ⁢ B + A − B 2 3 ⁢ A − B 2 3
56 7 simprd ⊢ A ∈ ℝ + ∧ A ≤ 1 → B ≤ 1
57 1red ⊢ A ∈ ℝ + ∧ A ≤ 1 → 1 ∈ ℝ
58 2rp ⊢ 2 ∈ ℝ +
59 58 a1i ⊢ A ∈ ℝ + ∧ A ≤ 1 → 2 ∈ ℝ +
60 14 57 59 lemul2d ⊢ A ∈ ℝ + ∧ A ≤ 1 → B ≤ 1 ↔ 2 ⁢ B ≤ 2 ⋅ 1
61 56 60 mpbid ⊢ A ∈ ℝ + ∧ A ≤ 1 → 2 ⁢ B ≤ 2 ⋅ 1
62 61 adantr ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ≤ 2 ⋅ 1
63 2t1e2 ⊢ 2 ⋅ 1 = 2
64 62 63 breqtrdi ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ≤ 2
65 11 adantr ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A ∈ ℝ
66 1red ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 1 ∈ ℝ
67 25 sqge0d ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 0 ≤ B 2
68 65 34 addge01d ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 0 ≤ B 2 ↔ A ≤ A + B 2
69 67 68 mpbid ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A ≤ A + B 2
70 65 34 65 lesubaddd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 ≤ A ↔ A ≤ A + B 2
71 69 70 mpbird ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 ≤ A
72 simplr ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A ≤ 1
73 17 65 66 71 72 letrd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 ≤ 1
74 1le3 ⊢ 1 ≤ 3
75 1re ⊢ 1 ∈ ℝ
76 3re ⊢ 3 ∈ ℝ
77 letr ⊢ A − B 2 ∈ ℝ ∧ 1 ∈ ℝ ∧ 3 ∈ ℝ → A − B 2 ≤ 1 ∧ 1 ≤ 3 → A − B 2 ≤ 3
78 75 76 77 mp3an23 ⊢ A − B 2 ∈ ℝ → A − B 2 ≤ 1 ∧ 1 ≤ 3 → A − B 2 ≤ 3
79 17 78 syl ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 ≤ 1 ∧ 1 ≤ 3 → A − B 2 ≤ 3
80 74 79 mpan2i ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 ≤ 1 → A − B 2 ≤ 3
81 73 80 mpd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 ≤ 3
82 3t1e3 ⊢ 3 ⋅ 1 = 3
83 81 82 breqtrrdi ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 ≤ 3 ⋅ 1
84 3pos ⊢ 0 < 3
85 ledivmul ⊢ A − B 2 ∈ ℝ ∧ 1 ∈ ℝ ∧ 3 ∈ ℝ ∧ 0 < 3 → A − B 2 3 ≤ 1 ↔ A − B 2 ≤ 3 ⋅ 1
86 75 85 mp3an2 ⊢ A − B 2 ∈ ℝ ∧ 3 ∈ ℝ ∧ 0 < 3 → A − B 2 3 ≤ 1 ↔ A − B 2 ≤ 3 ⋅ 1
87 76 84 86 mpanr12 ⊢ A − B 2 ∈ ℝ → A − B 2 3 ≤ 1 ↔ A − B 2 ≤ 3 ⋅ 1
88 17 87 syl ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 3 ≤ 1 ↔ A − B 2 ≤ 3 ⋅ 1
89 83 88 mpbird ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 3 ≤ 1
90 le2add ⊢ 2 ⁢ B ∈ ℝ ∧ A − B 2 3 ∈ ℝ ∧ 2 ∈ ℝ ∧ 1 ∈ ℝ → 2 ⁢ B ≤ 2 ∧ A − B 2 3 ≤ 1 → 2 ⁢ B + A − B 2 3 ≤ 2 + 1
91 36 75 90 mpanr12 ⊢ 2 ⁢ B ∈ ℝ ∧ A − B 2 3 ∈ ℝ → 2 ⁢ B ≤ 2 ∧ A − B 2 3 ≤ 1 → 2 ⁢ B + A − B 2 3 ≤ 2 + 1
92 52 30 91 syl2anc ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ≤ 2 ∧ A − B 2 3 ≤ 1 → 2 ⁢ B + A − B 2 3 ≤ 2 + 1
93 64 89 92 mp2and ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B + A − B 2 3 ≤ 2 + 1
94 df-3 ⊢ 3 = 2 + 1
95 93 94 breqtrrdi ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B + A − B 2 3 ≤ 3
96 52 30 readdcld ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B + A − B 2 3 ∈ ℝ
97 76 a1i ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 3 ∈ ℝ
98 96 97 23 lemul1d ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B + A − B 2 3 ≤ 3 ↔ 2 ⁢ B + A − B 2 3 ⁢ A − B 2 3 ≤ 3 ⁢ A − B 2 3
99 95 98 mpbid ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B + A − B 2 3 ⁢ A − B 2 3 ≤ 3 ⁢ A − B 2 3
100 17 recnd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → A − B 2 ∈ ℂ
101 3cn ⊢ 3 ∈ ℂ
102 3ne0 ⊢ 3 ≠ 0
103 divcan2 ⊢ A − B 2 ∈ ℂ ∧ 3 ∈ ℂ ∧ 3 ≠ 0 → 3 ⁢ A − B 2 3 = A − B 2
104 101 102 103 mp3an23 ⊢ A − B 2 ∈ ℂ → 3 ⁢ A − B 2 3 = A − B 2
105 100 104 syl ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 3 ⁢ A − B 2 3 = A − B 2
106 99 105 breqtrd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B + A − B 2 3 ⁢ A − B 2 3 ≤ A − B 2
107 55 106 eqbrtrd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2 ≤ A − B 2
108 39 41 readdcld ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2 ∈ ℝ
109 34 108 65 leaddsub2d ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B 2 + 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2 ≤ A ↔ 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2 ≤ A − B 2
110 107 109 mpbird ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B 2 + 2 ⁢ B ⁢ A − B 2 3 + A − B 2 3 2 ≤ A
111 44 110 eqbrtrd ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B + A − B 2 3 2 ≤ A
112 oveq1 ⊢ y = B + A − B 2 3 → y 2 = B + A − B 2 3 2
113 112 breq1d ⊢ y = B + A − B 2 3 → y 2 ≤ A ↔ B + A − B 2 3 2 ≤ A
114 oveq1 ⊢ x = y → x 2 = y 2
115 114 breq1d ⊢ x = y → x 2 ≤ A ↔ y 2 ≤ A
116 115 cbvrabv ⊢ x ∈ ℝ + | x 2 ≤ A = y ∈ ℝ + | y 2 ≤ A
117 1 116 eqtri ⊢ S = y ∈ ℝ + | y 2 ≤ A
118 113 117 elrab2 ⊢ B + A − B 2 3 ∈ S ↔ B + A − B 2 3 ∈ ℝ + ∧ B + A − B 2 3 2 ≤ A
119 24 111 118 sylanbrc ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B + A − B 2 3 ∈ S
120 suprub ⊢ S ⊆ ℝ ∧ S ≠ ∅ ∧ ∃ y ∈ ℝ ∀ z ∈ S z ≤ y ∧ B + A − B 2 3 ∈ S → B + A − B 2 3 ≤ sup S ℝ <
121 120 2 breqtrrdi ⊢ S ⊆ ℝ ∧ S ≠ ∅ ∧ ∃ y ∈ ℝ ∀ z ∈ S z ≤ y ∧ B + A − B 2 3 ∈ S → B + A − B 2 3 ≤ B
122 6 119 121 syl2anc ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → B + A − B 2 3 ≤ B
123 23 rpgt0d ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → 0 < A − B 2 3
124 29 14 ltaddposd ⊢ A ∈ ℝ + ∧ A ≤ 1 → 0 < A − B 2 3 ↔ B < B + A − B 2 3
125 14 29 readdcld ⊢ A ∈ ℝ + ∧ A ≤ 1 → B + A − B 2 3 ∈ ℝ
126 14 125 ltnled ⊢ A ∈ ℝ + ∧ A ≤ 1 → B < B + A − B 2 3 ↔ ¬ B + A − B 2 3 ≤ B
127 124 126 bitrd ⊢ A ∈ ℝ + ∧ A ≤ 1 → 0 < A − B 2 3 ↔ ¬ B + A − B 2 3 ≤ B
128 127 biimpa ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ 0 < A − B 2 3 → ¬ B + A − B 2 3 ≤ B
129 123 128 syldan ⊢ A ∈ ℝ + ∧ A ≤ 1 ∧ B 2 < A → ¬ B + A − B 2 3 ≤ B
130 122 129 pm2.65da ⊢ A ∈ ℝ + ∧ A ≤ 1 → ¬ B 2 < A
131 15 11 eqleltd ⊢ A ∈ ℝ + ∧ A ≤ 1 → B 2 = A ↔ B 2 ≤ A ∧ ¬ B 2 < A
132 4 130 131 mpbir2and ⊢ A ∈ ℝ + ∧ A ≤ 1 → B 2 = A