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 ) ) |