Metamath Proof Explorer
		
		
		
		Description:  Square of value of absolute value function.  (Contributed by Mario
       Carneiro, 29-May-2016)
		
			
				
					|  |  | Ref | Expression | 
					
						|  | Hypothesis | abscld.1 | ⊢ ( 𝜑  →  𝐴  ∈  ℂ ) | 
				
					|  | Assertion | absvalsqd | ⊢  ( 𝜑  →  ( ( abs ‘ 𝐴 ) ↑ 2 )  =  ( 𝐴  ·  ( ∗ ‘ 𝐴 ) ) ) | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | abscld.1 | ⊢ ( 𝜑  →  𝐴  ∈  ℂ ) | 
						
							| 2 |  | absvalsq | ⊢ ( 𝐴  ∈  ℂ  →  ( ( abs ‘ 𝐴 ) ↑ 2 )  =  ( 𝐴  ·  ( ∗ ‘ 𝐴 ) ) ) | 
						
							| 3 | 1 2 | syl | ⊢ ( 𝜑  →  ( ( abs ‘ 𝐴 ) ↑ 2 )  =  ( 𝐴  ·  ( ∗ ‘ 𝐴 ) ) ) |