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Description: The floor function satisfies |_ ( x ) = x + O(1) . (Contributed by Mario Carneiro, 21-May-2016)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | flo1 | ⊢ ( 𝑥 ∈ ℝ ↦ ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ) ∈ 𝑂(1) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | ssidd | ⊢ ( ⊤ → ℝ ⊆ ℝ ) | |
| 2 | reflcl | ⊢ ( 𝑥 ∈ ℝ → ( ⌊ ‘ 𝑥 ) ∈ ℝ ) | |
| 3 | resubcl | ⊢ ( ( 𝑥 ∈ ℝ ∧ ( ⌊ ‘ 𝑥 ) ∈ ℝ ) → ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ∈ ℝ ) | |
| 4 | 2 3 | mpdan | ⊢ ( 𝑥 ∈ ℝ → ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ∈ ℝ ) |
| 5 | 4 | recnd | ⊢ ( 𝑥 ∈ ℝ → ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ∈ ℂ ) |
| 6 | 5 | adantl | ⊢ ( ( ⊤ ∧ 𝑥 ∈ ℝ ) → ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ∈ ℂ ) |
| 7 | 1red | ⊢ ( ⊤ → 1 ∈ ℝ ) | |
| 8 | id | ⊢ ( 𝑥 ∈ ℝ → 𝑥 ∈ ℝ ) | |
| 9 | flle | ⊢ ( 𝑥 ∈ ℝ → ( ⌊ ‘ 𝑥 ) ≤ 𝑥 ) | |
| 10 | 2 8 9 | abssubge0d | ⊢ ( 𝑥 ∈ ℝ → ( abs ‘ ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ) = ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ) |
| 11 | fracle1 | ⊢ ( 𝑥 ∈ ℝ → ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ≤ 1 ) | |
| 12 | 10 11 | eqbrtrd | ⊢ ( 𝑥 ∈ ℝ → ( abs ‘ ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ) ≤ 1 ) |
| 13 | 12 | ad2antrl | ⊢ ( ( ⊤ ∧ ( 𝑥 ∈ ℝ ∧ 1 ≤ 𝑥 ) ) → ( abs ‘ ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ) ≤ 1 ) |
| 14 | 1 6 7 7 13 | elo1d | ⊢ ( ⊤ → ( 𝑥 ∈ ℝ ↦ ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ) ∈ 𝑂(1) ) |
| 15 | 14 | mptru | ⊢ ( 𝑥 ∈ ℝ ↦ ( 𝑥 − ( ⌊ ‘ 𝑥 ) ) ) ∈ 𝑂(1) |