Metamath Proof Explorer
Description: Absolute value and 'less than' relation. (Contributed by NM, 6-Apr-2005)
|
|
Ref |
Expression |
|
Hypotheses |
sqrtthi.1 |
⊢ 𝐴 ∈ ℝ |
|
|
sqr11.1 |
⊢ 𝐵 ∈ ℝ |
|
Assertion |
abslti |
⊢ ( ( abs ‘ 𝐴 ) < 𝐵 ↔ ( - 𝐵 < 𝐴 ∧ 𝐴 < 𝐵 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
sqrtthi.1 |
⊢ 𝐴 ∈ ℝ |
2 |
|
sqr11.1 |
⊢ 𝐵 ∈ ℝ |
3 |
|
abslt |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( abs ‘ 𝐴 ) < 𝐵 ↔ ( - 𝐵 < 𝐴 ∧ 𝐴 < 𝐵 ) ) ) |
4 |
1 2 3
|
mp2an |
⊢ ( ( abs ‘ 𝐴 ) < 𝐵 ↔ ( - 𝐵 < 𝐴 ∧ 𝐴 < 𝐵 ) ) |