Metamath Proof Explorer


Theorem ceilhalf1

Description: The ceiling of one half is one. (Contributed by AV, 2-Nov-2025)

Ref Expression
Assertion ceilhalf1 ( ⌈ ‘ ( 1 / 2 ) ) = 1

Proof

Step Hyp Ref Expression
1 halfre ( 1 / 2 ) ∈ ℝ
2 1re 1 ∈ ℝ
3 halflt1 ( 1 / 2 ) < 1
4 1 2 3 ltleii ( 1 / 2 ) ≤ 1
5 1m1e0 ( 1 − 1 ) = 0
6 halfgt0 0 < ( 1 / 2 )
7 5 6 eqbrtri ( 1 − 1 ) < ( 1 / 2 )
8 2 2 1 ltsubaddi ( ( 1 − 1 ) < ( 1 / 2 ) ↔ 1 < ( ( 1 / 2 ) + 1 ) )
9 7 8 mpbi 1 < ( ( 1 / 2 ) + 1 )
10 1z 1 ∈ ℤ
11 ceilbi ( ( ( 1 / 2 ) ∈ ℝ ∧ 1 ∈ ℤ ) → ( ( ⌈ ‘ ( 1 / 2 ) ) = 1 ↔ ( ( 1 / 2 ) ≤ 1 ∧ 1 < ( ( 1 / 2 ) + 1 ) ) ) )
12 1 10 11 mp2an ( ( ⌈ ‘ ( 1 / 2 ) ) = 1 ↔ ( ( 1 / 2 ) ≤ 1 ∧ 1 < ( ( 1 / 2 ) + 1 ) ) )
13 4 9 12 mpbir2an ( ⌈ ‘ ( 1 / 2 ) ) = 1