Metamath Proof Explorer


Theorem reccn2

Description: The reciprocal function is continuous. (Contributed by Mario Carneiro, 9-Feb-2014) (Revised by Mario Carneiro, 22-Sep-2014)

Ref Expression
Hypothesis reccn2.t ⊢ T = if 1 ≤ A ⁢ B 1 A ⁢ B ⁢ A 2
Assertion reccn2 ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + → ∃ y ∈ ℝ + ∀ z ∈ ℂ ∖ 0 z − A < y → 1 z − 1 A < B

Proof

Step Hyp Ref Expression
1 reccn2.t ⊢ T = if 1 ≤ A ⁢ B 1 A ⁢ B ⁢ A 2
2 1rp ⊢ 1 ∈ ℝ +
3 eldifsn ⊢ A ∈ ℂ ∖ 0 ↔ A ∈ ℂ ∧ A ≠ 0
4 3 birani ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + → A ∈ ℂ ∧ A ≠ 0
5 absrpcl ⊢ A ∈ ℂ ∧ A ≠ 0 → A ∈ ℝ +
6 4 5 syl ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + → A ∈ ℝ +
7 rpmulcl ⊢ A ∈ ℝ + ∧ B ∈ ℝ + → A ⁢ B ∈ ℝ +
8 6 7 sylancom ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + → A ⁢ B ∈ ℝ +
9 ifcl ⊢ 1 ∈ ℝ + ∧ A ⁢ B ∈ ℝ + → if 1 ≤ A ⁢ B 1 A ⁢ B ∈ ℝ +
10 2 8 9 sylancr ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + → if 1 ≤ A ⁢ B 1 A ⁢ B ∈ ℝ +
11 6 rphalfcld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + → A 2 ∈ ℝ +
12 10 11 rpmulcld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + → if 1 ≤ A ⁢ B 1 A ⁢ B ⁢ A 2 ∈ ℝ +
13 1 12 eqeltrid ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + → T ∈ ℝ +
14 4 adantr ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ∈ ℂ ∧ A ≠ 0
15 14 simpld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ∈ ℂ
16 simprl ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z ∈ ℂ ∖ 0
17 eldifsn ⊢ z ∈ ℂ ∖ 0 ↔ z ∈ ℂ ∧ z ≠ 0
18 16 17 sylib ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z ∈ ℂ ∧ z ≠ 0
19 18 simpld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z ∈ ℂ
20 15 19 mulcld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ z ∈ ℂ
21 mulne0 ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ z ∈ ℂ ∧ z ≠ 0 → A ⁢ z ≠ 0
22 14 18 21 syl2anc ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ z ≠ 0
23 15 19 20 22 divsubdird ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z A ⁢ z = A A ⁢ z − z A ⁢ z
24 15 mulridd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⋅ 1 = A
25 24 oveq1d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⋅ 1 A ⁢ z = A A ⁢ z
26 1cnd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → 1 ∈ ℂ
27 divcan5 ⊢ 1 ∈ ℂ ∧ z ∈ ℂ ∧ z ≠ 0 ∧ A ∈ ℂ ∧ A ≠ 0 → A ⋅ 1 A ⁢ z = 1 z
28 26 18 14 27 syl3anc ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⋅ 1 A ⁢ z = 1 z
29 25 28 eqtr3d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A A ⁢ z = 1 z
30 19 mulridd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z ⋅ 1 = z
31 19 15 mulcomd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z ⁢ A = A ⁢ z
32 30 31 oveq12d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z ⋅ 1 z ⁢ A = z A ⁢ z
33 divcan5 ⊢ 1 ∈ ℂ ∧ A ∈ ℂ ∧ A ≠ 0 ∧ z ∈ ℂ ∧ z ≠ 0 → z ⋅ 1 z ⁢ A = 1 A
34 26 14 18 33 syl3anc ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z ⋅ 1 z ⁢ A = 1 A
35 32 34 eqtr3d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z A ⁢ z = 1 A
36 29 35 oveq12d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A A ⁢ z − z A ⁢ z = 1 z − 1 A
37 23 36 eqtrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z A ⁢ z = 1 z − 1 A
38 37 fveq2d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z A ⁢ z = 1 z − 1 A
39 15 19 subcld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z ∈ ℂ
40 39 20 22 absdivd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z A ⁢ z = A − z A ⁢ z
41 38 40 eqtr3d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → 1 z − 1 A = A − z A ⁢ z
42 15 19 abssubd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z = z − A
43 19 15 subcld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z − A ∈ ℂ
44 43 abscld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z − A ∈ ℝ
45 42 44 eqeltrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z ∈ ℝ
46 13 adantr ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → T ∈ ℝ +
47 46 rpred ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → T ∈ ℝ
48 20 abscld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ z ∈ ℝ
49 rpre ⊢ B ∈ ℝ + → B ∈ ℝ
50 49 ad2antlr ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → B ∈ ℝ
51 48 50 remulcld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ z ⁢ B ∈ ℝ
52 simprr ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z − A < T
53 42 52 eqbrtrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z < T
54 8 adantr ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ B ∈ ℝ +
55 54 rpred ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ B ∈ ℝ
56 11 adantr ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A 2 ∈ ℝ +
57 56 rpred ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A 2 ∈ ℝ
58 55 57 remulcld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ B ⁢ A 2 ∈ ℝ
59 1re ⊢ 1 ∈ ℝ
60 min2 ⊢ 1 ∈ ℝ ∧ A ⁢ B ∈ ℝ → if 1 ≤ A ⁢ B 1 A ⁢ B ≤ A ⁢ B
61 59 55 60 sylancr ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → if 1 ≤ A ⁢ B 1 A ⁢ B ≤ A ⁢ B
62 10 adantr ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → if 1 ≤ A ⁢ B 1 A ⁢ B ∈ ℝ +
63 62 rpred ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → if 1 ≤ A ⁢ B 1 A ⁢ B ∈ ℝ
64 63 55 56 lemul1d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → if 1 ≤ A ⁢ B 1 A ⁢ B ≤ A ⁢ B ↔ if 1 ≤ A ⁢ B 1 A ⁢ B ⁢ A 2 ≤ A ⁢ B ⁢ A 2
65 61 64 mpbid ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → if 1 ≤ A ⁢ B 1 A ⁢ B ⁢ A 2 ≤ A ⁢ B ⁢ A 2
66 1 65 eqbrtrid ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → T ≤ A ⁢ B ⁢ A 2
67 19 abscld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z ∈ ℝ
68 15 abscld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ∈ ℝ
69 68 recnd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ∈ ℂ
70 69 2halvesd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A 2 + A 2 = A
71 68 67 resubcld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z ∈ ℝ
72 15 19 abs2difd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z ≤ A − z
73 min1 ⊢ 1 ∈ ℝ ∧ A ⁢ B ∈ ℝ → if 1 ≤ A ⁢ B 1 A ⁢ B ≤ 1
74 59 55 73 sylancr ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → if 1 ≤ A ⁢ B 1 A ⁢ B ≤ 1
75 1red ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → 1 ∈ ℝ
76 63 75 56 lemul1d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → if 1 ≤ A ⁢ B 1 A ⁢ B ≤ 1 ↔ if 1 ≤ A ⁢ B 1 A ⁢ B ⁢ A 2 ≤ 1 ⁢ A 2
77 74 76 mpbid ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → if 1 ≤ A ⁢ B 1 A ⁢ B ⁢ A 2 ≤ 1 ⁢ A 2
78 1 77 eqbrtrid ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → T ≤ 1 ⁢ A 2
79 57 recnd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A 2 ∈ ℂ
80 79 mullidd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → 1 ⁢ A 2 = A 2
81 78 80 breqtrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → T ≤ A 2
82 45 47 57 53 81 ltletrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z < A 2
83 71 45 57 72 82 lelttrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z < A 2
84 68 67 57 ltsubadd2d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z < A 2 ↔ A < z + A 2
85 83 84 mpbid ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A < z + A 2
86 70 85 eqbrtrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A 2 + A 2 < z + A 2
87 57 67 57 ltadd1d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A 2 < z ↔ A 2 + A 2 < z + A 2
88 86 87 mpbird ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A 2 < z
89 57 67 54 88 ltmul2dd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ B ⁢ A 2 < A ⁢ B ⁢ z
90 15 19 absmuld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ z = A ⁢ z
91 90 oveq1d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ z ⁢ B = A ⁢ z ⁢ B
92 67 recnd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → z ∈ ℂ
93 50 recnd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → B ∈ ℂ
94 69 92 93 mul32d ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ z ⁢ B = A ⁢ B ⁢ z
95 91 94 eqtrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ z ⁢ B = A ⁢ B ⁢ z
96 89 95 breqtrrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ B ⁢ A 2 < A ⁢ z ⁢ B
97 47 58 51 66 96 lelttrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → T < A ⁢ z ⁢ B
98 45 47 51 53 97 lttrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z < A ⁢ z ⁢ B
99 20 22 absrpcld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A ⁢ z ∈ ℝ +
100 45 50 99 ltdivmuld ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z A ⁢ z < B ↔ A − z < A ⁢ z ⁢ B
101 98 100 mpbird ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → A − z A ⁢ z < B
102 41 101 eqbrtrd ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 ∧ z − A < T → 1 z − 1 A < B
103 102 expr ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + ∧ z ∈ ℂ ∖ 0 → z − A < T → 1 z − 1 A < B
104 103 ralrimiva ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + → ∀ z ∈ ℂ ∖ 0 z − A < T → 1 z − 1 A < B
105 breq2 ⊢ y = T → z − A < y ↔ z − A < T
106 105 rspceaimv ⊢ T ∈ ℝ + ∧ ∀ z ∈ ℂ ∖ 0 z − A < T → 1 z − 1 A < B → ∃ y ∈ ℝ + ∀ z ∈ ℂ ∖ 0 z − A < y → 1 z − 1 A < B
107 13 104 106 syl2anc ⊢ A ∈ ℂ ∖ 0 ∧ B ∈ ℝ + → ∃ y ∈ ℝ + ∀ z ∈ ℂ ∖ 0 z − A < y → 1 z − 1 A < B