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Metamath Proof Explorer

Theorem absltd

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 ) ) )