Metamath Proof Explorer
		
		
		
		Description:  Contrapositive law deduction for inequality.  (Contributed by NM, 2-Apr-2007)  (Proof shortened by Andrew Salmon, 25-May-2011)
		
			
				
					|  |  | Ref | Expression | 
					
						|  | Hypothesis | necon3bd.1 | ⊢ ( 𝜑  →  ( 𝐴  =  𝐵  →  𝜓 ) ) | 
				
					|  | Assertion | necon3bd | ⊢  ( 𝜑  →  ( ¬  𝜓  →  𝐴  ≠  𝐵 ) ) | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | necon3bd.1 | ⊢ ( 𝜑  →  ( 𝐴  =  𝐵  →  𝜓 ) ) | 
						
							| 2 |  | nne | ⊢ ( ¬  𝐴  ≠  𝐵  ↔  𝐴  =  𝐵 ) | 
						
							| 3 | 2 1 | biimtrid | ⊢ ( 𝜑  →  ( ¬  𝐴  ≠  𝐵  →  𝜓 ) ) | 
						
							| 4 | 3 | con1d | ⊢ ( 𝜑  →  ( ¬  𝜓  →  𝐴  ≠  𝐵 ) ) |