Metamath Proof Explorer


Theorem nn0opthlem1

Description: A rather pretty lemma for nn0opthi . (Contributed by Raph Levien, 10-Dec-2002)

Ref Expression
Hypotheses nn0opthlem1.1 ⊢ A ∈ ℕ 0
nn0opthlem1.2 ⊢ C ∈ ℕ 0
Assertion nn0opthlem1 ⊢ A < C ↔ A ⁢ A + 2 ⁢ A < C ⁢ C

Proof

Step Hyp Ref Expression
1 nn0opthlem1.1 ⊢ A ∈ ℕ 0
2 nn0opthlem1.2 ⊢ C ∈ ℕ 0
3 1nn0 ⊢ 1 ∈ ℕ 0
4 1 3 nn0addcli ⊢ A + 1 ∈ ℕ 0
5 4 2 nn0le2msqi ⊢ A + 1 ≤ C ↔ A + 1 ⁢ A + 1 ≤ C ⁢ C
6 nn0ltp1le ⊢ A ∈ ℕ 0 ∧ C ∈ ℕ 0 → A < C ↔ A + 1 ≤ C
7 1 2 6 mp2an ⊢ A < C ↔ A + 1 ≤ C
8 1 1 nn0mulcli ⊢ A ⁢ A ∈ ℕ 0
9 2nn0 ⊢ 2 ∈ ℕ 0
10 9 1 nn0mulcli ⊢ 2 ⁢ A ∈ ℕ 0
11 8 10 nn0addcli ⊢ A ⁢ A + 2 ⁢ A ∈ ℕ 0
12 2 2 nn0mulcli ⊢ C ⁢ C ∈ ℕ 0
13 nn0ltp1le ⊢ A ⁢ A + 2 ⁢ A ∈ ℕ 0 ∧ C ⁢ C ∈ ℕ 0 → A ⁢ A + 2 ⁢ A < C ⁢ C ↔ A ⁢ A + 2 ⁢ A + 1 ≤ C ⁢ C
14 11 12 13 mp2an ⊢ A ⁢ A + 2 ⁢ A < C ⁢ C ↔ A ⁢ A + 2 ⁢ A + 1 ≤ C ⁢ C
15 1 nn0cni ⊢ A ∈ ℂ
16 ax-1cn ⊢ 1 ∈ ℂ
17 15 16 binom2i ⊢ A + 1 2 = A 2 + 2 ⁢ A ⋅ 1 + 1 2
18 15 16 addcli ⊢ A + 1 ∈ ℂ
19 18 sqvali ⊢ A + 1 2 = A + 1 ⁢ A + 1
20 15 sqvali ⊢ A 2 = A ⁢ A
21 20 oveq1i ⊢ A 2 + 2 ⁢ A ⋅ 1 = A ⁢ A + 2 ⁢ A ⋅ 1
22 16 sqvali ⊢ 1 2 = 1 ⋅ 1
23 21 22 oveq12i ⊢ A 2 + 2 ⁢ A ⋅ 1 + 1 2 = A ⁢ A + 2 ⁢ A ⋅ 1 + 1 ⋅ 1
24 17 19 23 3eqtr3i ⊢ A + 1 ⁢ A + 1 = A ⁢ A + 2 ⁢ A ⋅ 1 + 1 ⋅ 1
25 15 mulridi ⊢ A ⋅ 1 = A
26 25 oveq2i ⊢ 2 ⁢ A ⋅ 1 = 2 ⁢ A
27 26 oveq2i ⊢ A ⁢ A + 2 ⁢ A ⋅ 1 = A ⁢ A + 2 ⁢ A
28 16 mulridi ⊢ 1 ⋅ 1 = 1
29 27 28 oveq12i ⊢ A ⁢ A + 2 ⁢ A ⋅ 1 + 1 ⋅ 1 = A ⁢ A + 2 ⁢ A + 1
30 24 29 eqtri ⊢ A + 1 ⁢ A + 1 = A ⁢ A + 2 ⁢ A + 1
31 30 breq1i ⊢ A + 1 ⁢ A + 1 ≤ C ⁢ C ↔ A ⁢ A + 2 ⁢ A + 1 ≤ C ⁢ C
32 14 31 bitr4i ⊢ A ⁢ A + 2 ⁢ A < C ⁢ C ↔ A + 1 ⁢ A + 1 ≤ C ⁢ C
33 5 7 32 3bitr4i ⊢ A < C ↔ A ⁢ A + 2 ⁢ A < C ⁢ C