Metamath Proof Explorer
Description: Absolute value and 'less than' relation. (Contributed by Mario
Carneiro, 29-May-2016)
|
|
Ref |
Expression |
|
Hypotheses |
absltd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ ) |
|
|
absltd.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℝ ) |
|
Assertion |
absltd |
⊢ ( 𝜑 → ( ( abs ‘ 𝐴 ) < 𝐵 ↔ ( - 𝐵 < 𝐴 ∧ 𝐴 < 𝐵 ) ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
absltd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ ) |
2 |
|
absltd.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℝ ) |
3 |
|
abslt |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( abs ‘ 𝐴 ) < 𝐵 ↔ ( - 𝐵 < 𝐴 ∧ 𝐴 < 𝐵 ) ) ) |
4 |
1 2 3
|
syl2anc |
⊢ ( 𝜑 → ( ( abs ‘ 𝐴 ) < 𝐵 ↔ ( - 𝐵 < 𝐴 ∧ 𝐴 < 𝐵 ) ) ) |