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Metamath Proof Explorer

Theorem ico01fl0

Description: The floor of a real number in [ 0 , 1 ) is 0. Remark: may shorten the proof of modid or a version of it where the antecedent is membership in an interval. (Contributed by BJ, 29-Jun-2019)

Ref Expression
Assertion ico01fl0 ( 𝐴 ∈ ( 0 [,) 1 ) → ( ⌊ ‘ 𝐴 ) = 0 )

Proof

Step Hyp Ref Expression
1 0re ⊢ 0 ∈ ℝ
2 1xr ⊢ 1 ∈ ℝ*
3 icossre ⊢ ( ( 0 ∈ ℝ ∧ 1 ∈ ℝ* ) → ( 0 [,) 1 ) ⊆ ℝ )
4 1 2 3 mp2an ⊢ ( 0 [,) 1 ) ⊆ ℝ
5 4 sseli ⊢ ( 𝐴 ∈ ( 0 [,) 1 ) → 𝐴 ∈ ℝ )
6 0xr ⊢ 0 ∈ ℝ*
7 elico1 ⊢ ( ( 0 ∈ ℝ* ∧ 1 ∈ ℝ* ) → ( 𝐴 ∈ ( 0 [,) 1 ) ↔ ( 𝐴 ∈ ℝ* ∧ 0 ≤ 𝐴 ∧ 𝐴 < 1 ) ) )
8 6 2 7 mp2an ⊢ ( 𝐴 ∈ ( 0 [,) 1 ) ↔ ( 𝐴 ∈ ℝ* ∧ 0 ≤ 𝐴 ∧ 𝐴 < 1 ) )
9 8 simp2bi ⊢ ( 𝐴 ∈ ( 0 [,) 1 ) → 0 ≤ 𝐴 )
10 8 simp3bi ⊢ ( 𝐴 ∈ ( 0 [,) 1 ) → 𝐴 < 1 )
11 recn ⊢ ( 𝐴 ∈ ℝ → 𝐴 ∈ ℂ )
12 11 addlidd ⊢ ( 𝐴 ∈ ℝ → ( 0 + 𝐴 ) = 𝐴 )
13 12 fveqeq2d ⊢ ( 𝐴 ∈ ℝ → ( ( ⌊ ‘ ( 0 + 𝐴 ) ) = 0 ↔ ( ⌊ ‘ 𝐴 ) = 0 ) )
14 0z ⊢ 0 ∈ ℤ
15 flbi2 ⊢ ( ( 0 ∈ ℤ ∧ 𝐴 ∈ ℝ ) → ( ( ⌊ ‘ ( 0 + 𝐴 ) ) = 0 ↔ ( 0 ≤ 𝐴 ∧ 𝐴 < 1 ) ) )
16 14 15 mpan ⊢ ( 𝐴 ∈ ℝ → ( ( ⌊ ‘ ( 0 + 𝐴 ) ) = 0 ↔ ( 0 ≤ 𝐴 ∧ 𝐴 < 1 ) ) )
17 13 16 bitr3d ⊢ ( 𝐴 ∈ ℝ → ( ( ⌊ ‘ 𝐴 ) = 0 ↔ ( 0 ≤ 𝐴 ∧ 𝐴 < 1 ) ) )
18 17 biimpar ⊢ ( ( 𝐴 ∈ ℝ ∧ ( 0 ≤ 𝐴 ∧ 𝐴 < 1 ) ) → ( ⌊ ‘ 𝐴 ) = 0 )
19 5 9 10 18 syl12anc ⊢ ( 𝐴 ∈ ( 0 [,) 1 ) → ( ⌊ ‘ 𝐴 ) = 0 )