Metamath Proof Explorer
		
		
		
		Description:  Cancellation law for subtraction.  (Contributed by Mario Carneiro, 27-May-2016)
		
			
				
					 | 
					 | 
					Ref | 
					Expression | 
				
					
						 | 
						Hypotheses | 
						negidd.1 | 
						⊢ ( 𝜑  →  𝐴  ∈  ℂ )  | 
					
					
						 | 
						 | 
						pncand.2 | 
						⊢ ( 𝜑  →  𝐵  ∈  ℂ )  | 
					
					
						 | 
						 | 
						subaddd.3 | 
						⊢ ( 𝜑  →  𝐶  ∈  ℂ )  | 
					
				
					 | 
					Assertion | 
					npncand | 
					⊢  ( 𝜑  →  ( ( 𝐴  −  𝐵 )  +  ( 𝐵  −  𝐶 ) )  =  ( 𝐴  −  𝐶 ) )  | 
				
			
		
		
			
				Proof
				
					
						| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							negidd.1 | 
							⊢ ( 𝜑  →  𝐴  ∈  ℂ )  | 
						
						
							| 2 | 
							
								
							 | 
							pncand.2 | 
							⊢ ( 𝜑  →  𝐵  ∈  ℂ )  | 
						
						
							| 3 | 
							
								
							 | 
							subaddd.3 | 
							⊢ ( 𝜑  →  𝐶  ∈  ℂ )  | 
						
						
							| 4 | 
							
								
							 | 
							npncan | 
							⊢ ( ( 𝐴  ∈  ℂ  ∧  𝐵  ∈  ℂ  ∧  𝐶  ∈  ℂ )  →  ( ( 𝐴  −  𝐵 )  +  ( 𝐵  −  𝐶 ) )  =  ( 𝐴  −  𝐶 ) )  | 
						
						
							| 5 | 
							
								1 2 3 4
							 | 
							syl3anc | 
							⊢ ( 𝜑  →  ( ( 𝐴  −  𝐵 )  +  ( 𝐵  −  𝐶 ) )  =  ( 𝐴  −  𝐶 ) )  |