Metamath Proof Explorer


Theorem expnlbnd2

Description: The reciprocal of exponentiation with a base greater than 1 has no positive lower bound. (Contributed by NM, 18-Jul-2008) (Proof shortened by Mario Carneiro, 5-Jun-2014)

Ref Expression
Assertion expnlbnd2 ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B → ∃ j ∈ ℕ ∀ k ∈ ℤ ≥ j 1 B k < A

Proof

Step Hyp Ref Expression
1 expnlbnd ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B → ∃ j ∈ ℕ 1 B j < A
2 simpl2 ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → B ∈ ℝ
3 simpl3 ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 1 < B
4 1re ⊢ 1 ∈ ℝ
5 ltle ⊢ 1 ∈ ℝ ∧ B ∈ ℝ → 1 < B → 1 ≤ B
6 4 2 5 sylancr ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 1 < B → 1 ≤ B
7 3 6 mpd ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 1 ≤ B
8 simprr ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → k ∈ ℤ ≥ j
9 leexp2a ⊢ B ∈ ℝ ∧ 1 ≤ B ∧ k ∈ ℤ ≥ j → B j ≤ B k
10 2 7 8 9 syl3anc ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → B j ≤ B k
11 0red ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 0 ∈ ℝ
12 1red ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 1 ∈ ℝ
13 0lt1 ⊢ 0 < 1
14 13 a1i ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 0 < 1
15 11 12 2 14 3 lttrd ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 0 < B
16 2 15 elrpd ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → B ∈ ℝ +
17 nnz ⊢ j ∈ ℕ → j ∈ ℤ
18 17 ad2antrl ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → j ∈ ℤ
19 rpexpcl ⊢ B ∈ ℝ + ∧ j ∈ ℤ → B j ∈ ℝ +
20 16 18 19 syl2anc ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → B j ∈ ℝ +
21 eluzelz ⊢ k ∈ ℤ ≥ j → k ∈ ℤ
22 21 ad2antll ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → k ∈ ℤ
23 rpexpcl ⊢ B ∈ ℝ + ∧ k ∈ ℤ → B k ∈ ℝ +
24 16 22 23 syl2anc ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → B k ∈ ℝ +
25 20 24 lerecd ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → B j ≤ B k ↔ 1 B k ≤ 1 B j
26 10 25 mpbid ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 1 B k ≤ 1 B j
27 24 rprecred ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 1 B k ∈ ℝ
28 20 rprecred ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 1 B j ∈ ℝ
29 simpl1 ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → A ∈ ℝ +
30 29 rpred ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → A ∈ ℝ
31 lelttr ⊢ 1 B k ∈ ℝ ∧ 1 B j ∈ ℝ ∧ A ∈ ℝ → 1 B k ≤ 1 B j ∧ 1 B j < A → 1 B k < A
32 27 28 30 31 syl3anc ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 1 B k ≤ 1 B j ∧ 1 B j < A → 1 B k < A
33 26 32 mpand ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 1 B j < A → 1 B k < A
34 33 anassrs ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ ∧ k ∈ ℤ ≥ j → 1 B j < A → 1 B k < A
35 34 ralrimdva ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B ∧ j ∈ ℕ → 1 B j < A → ∀ k ∈ ℤ ≥ j 1 B k < A
36 35 reximdva ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B → ∃ j ∈ ℕ 1 B j < A → ∃ j ∈ ℕ ∀ k ∈ ℤ ≥ j 1 B k < A
37 1 36 mpd ⊢ A ∈ ℝ + ∧ B ∈ ℝ ∧ 1 < B → ∃ j ∈ ℕ ∀ k ∈ ℤ ≥ j 1 B k < A