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 | nppcan2d | ⊢  ( 𝜑  →  ( ( 𝐴  −  ( 𝐵  +  𝐶 ) )  +  𝐶 )  =  ( 𝐴  −  𝐵 ) ) | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | negidd.1 | ⊢ ( 𝜑  →  𝐴  ∈  ℂ ) | 
						
							| 2 |  | pncand.2 | ⊢ ( 𝜑  →  𝐵  ∈  ℂ ) | 
						
							| 3 |  | subaddd.3 | ⊢ ( 𝜑  →  𝐶  ∈  ℂ ) | 
						
							| 4 |  | nppcan2 | ⊢ ( ( 𝐴  ∈  ℂ  ∧  𝐵  ∈  ℂ  ∧  𝐶  ∈  ℂ )  →  ( ( 𝐴  −  ( 𝐵  +  𝐶 ) )  +  𝐶 )  =  ( 𝐴  −  𝐵 ) ) | 
						
							| 5 | 1 2 3 4 | syl3anc | ⊢ ( 𝜑  →  ( ( 𝐴  −  ( 𝐵  +  𝐶 ) )  +  𝐶 )  =  ( 𝐴  −  𝐵 ) ) |