Metamath Proof Explorer
		
		
		
		Description:  A cancellation law for division.  (Contributed by Mario Carneiro, 27-May-2016)
		
			
				
					 | 
					 | 
					Ref | 
					Expression | 
				
					
						 | 
						Hypotheses | 
						div1d.1 | 
						⊢ ( 𝜑  →  𝐴  ∈  ℂ )  | 
					
					
						 | 
						 | 
						divcld.2 | 
						⊢ ( 𝜑  →  𝐵  ∈  ℂ )  | 
					
					
						 | 
						 | 
						divcld.3 | 
						⊢ ( 𝜑  →  𝐵  ≠  0 )  | 
					
				
					 | 
					Assertion | 
					divcan4d | 
					⊢  ( 𝜑  →  ( ( 𝐴  ·  𝐵 )  /  𝐵 )  =  𝐴 )  | 
				
			
		
		
			
				Proof
				
					
						| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							div1d.1 | 
							⊢ ( 𝜑  →  𝐴  ∈  ℂ )  | 
						
						
							| 2 | 
							
								
							 | 
							divcld.2 | 
							⊢ ( 𝜑  →  𝐵  ∈  ℂ )  | 
						
						
							| 3 | 
							
								
							 | 
							divcld.3 | 
							⊢ ( 𝜑  →  𝐵  ≠  0 )  | 
						
						
							| 4 | 
							
								
							 | 
							divcan4 | 
							⊢ ( ( 𝐴  ∈  ℂ  ∧  𝐵  ∈  ℂ  ∧  𝐵  ≠  0 )  →  ( ( 𝐴  ·  𝐵 )  /  𝐵 )  =  𝐴 )  | 
						
						
							| 5 | 
							
								1 2 3 4
							 | 
							syl3anc | 
							⊢ ( 𝜑  →  ( ( 𝐴  ·  𝐵 )  /  𝐵 )  =  𝐴 )  |