Metamath Proof Explorer


Theorem sqrlearg

Description: The square compared with its argument. (Contributed by Glauco Siliprandi, 26-Jun-2021)

Ref Expression
Hypothesis sqrlearg.1 ⊢ φ → A ∈ ℝ
Assertion sqrlearg ⊢ φ → A 2 ≤ A ↔ A ∈ 0 1

Proof

Step Hyp Ref Expression
1 sqrlearg.1 ⊢ φ → A ∈ ℝ
2 0re ⊢ 0 ∈ ℝ
3 2 a1i ⊢ φ ∧ A 2 ≤ A → 0 ∈ ℝ
4 simpr ⊢ φ ∧ ¬ A ≤ 1 → ¬ A ≤ 1
5 1red ⊢ φ ∧ ¬ A ≤ 1 → 1 ∈ ℝ
6 1 adantr ⊢ φ ∧ ¬ A ≤ 1 → A ∈ ℝ
7 5 6 ltnled ⊢ φ ∧ ¬ A ≤ 1 → 1 < A ↔ ¬ A ≤ 1
8 4 7 mpbird ⊢ φ ∧ ¬ A ≤ 1 → 1 < A
9 1red ⊢ φ ∧ 1 < A → 1 ∈ ℝ
10 1 adantr ⊢ φ ∧ 1 < A → A ∈ ℝ
11 2 a1i ⊢ φ ∧ 1 < A → 0 ∈ ℝ
12 0lt1 ⊢ 0 < 1
13 12 a1i ⊢ φ ∧ 1 < A → 0 < 1
14 simpr ⊢ φ ∧ 1 < A → 1 < A
15 11 9 10 13 14 lttrd ⊢ φ ∧ 1 < A → 0 < A
16 10 15 elrpd ⊢ φ ∧ 1 < A → A ∈ ℝ +
17 9 10 16 14 ltmul2dd ⊢ φ ∧ 1 < A → A ⋅ 1 < A ⁢ A
18 1 recnd ⊢ φ → A ∈ ℂ
19 18 mulridd ⊢ φ → A ⋅ 1 = A
20 19 adantr ⊢ φ ∧ 1 < A → A ⋅ 1 = A
21 18 sqvald ⊢ φ → A 2 = A ⁢ A
22 21 eqcomd ⊢ φ → A ⁢ A = A 2
23 22 adantr ⊢ φ ∧ 1 < A → A ⁢ A = A 2
24 20 23 breq12d ⊢ φ ∧ 1 < A → A ⋅ 1 < A ⁢ A ↔ A < A 2
25 17 24 mpbid ⊢ φ ∧ 1 < A → A < A 2
26 8 25 syldan ⊢ φ ∧ ¬ A ≤ 1 → A < A 2
27 26 adantlr ⊢ φ ∧ A 2 ≤ A ∧ ¬ A ≤ 1 → A < A 2
28 simpr ⊢ φ ∧ A 2 ≤ A → A 2 ≤ A
29 1 resqcld ⊢ φ → A 2 ∈ ℝ
30 29 adantr ⊢ φ ∧ A 2 ≤ A → A 2 ∈ ℝ
31 1 adantr ⊢ φ ∧ A 2 ≤ A → A ∈ ℝ
32 30 31 lenltd ⊢ φ ∧ A 2 ≤ A → A 2 ≤ A ↔ ¬ A < A 2
33 28 32 mpbid ⊢ φ ∧ A 2 ≤ A → ¬ A < A 2
34 33 adantr ⊢ φ ∧ A 2 ≤ A ∧ ¬ A ≤ 1 → ¬ A < A 2
35 27 34 condan ⊢ φ ∧ A 2 ≤ A → A ≤ 1
36 1red ⊢ φ ∧ A ≤ 1 → 1 ∈ ℝ
37 35 36 syldan ⊢ φ ∧ A 2 ≤ A → 1 ∈ ℝ
38 31 sqge0d ⊢ φ ∧ A 2 ≤ A → 0 ≤ A 2
39 3 30 31 38 28 letrd ⊢ φ ∧ A 2 ≤ A → 0 ≤ A
40 3 37 31 39 35 eliccd ⊢ φ ∧ A 2 ≤ A → A ∈ 0 1
41 40 ex ⊢ φ → A 2 ≤ A → A ∈ 0 1
42 unitssre ⊢ 0 1 ⊆ ℝ
43 42 sseli ⊢ A ∈ 0 1 → A ∈ ℝ
44 1red ⊢ A ∈ 0 1 → 1 ∈ ℝ
45 0xr ⊢ 0 ∈ ℝ *
46 45 a1i ⊢ A ∈ 0 1 → 0 ∈ ℝ *
47 44 rexrd ⊢ A ∈ 0 1 → 1 ∈ ℝ *
48 id ⊢ A ∈ 0 1 → A ∈ 0 1
49 46 47 48 iccgelbd ⊢ A ∈ 0 1 → 0 ≤ A
50 46 47 48 iccleubd ⊢ A ∈ 0 1 → A ≤ 1
51 43 44 43 49 50 lemul2ad ⊢ A ∈ 0 1 → A ⁢ A ≤ A ⋅ 1
52 51 adantl ⊢ φ ∧ A ∈ 0 1 → A ⁢ A ≤ A ⋅ 1
53 22 adantr ⊢ φ ∧ A ∈ 0 1 → A ⁢ A = A 2
54 19 adantr ⊢ φ ∧ A ∈ 0 1 → A ⋅ 1 = A
55 53 54 breq12d ⊢ φ ∧ A ∈ 0 1 → A ⁢ A ≤ A ⋅ 1 ↔ A 2 ≤ A
56 52 55 mpbid ⊢ φ ∧ A ∈ 0 1 → A 2 ≤ A
57 56 ex ⊢ φ → A ∈ 0 1 → A 2 ≤ A
58 41 57 impbid ⊢ φ → A 2 ≤ A ↔ A ∈ 0 1