Metamath Proof Explorer
		
		
		
		Description:  Deduction from equality to inequality.  (Contributed by NM, 21-Mar-2007)
		
			
				
					 | 
					 | 
					Ref | 
					Expression | 
				
					
						 | 
						Hypothesis | 
						necon3abid.1 | 
						⊢ ( 𝜑  →  ( 𝐴  =  𝐵  ↔  𝜓 ) )  | 
					
				
					 | 
					Assertion | 
					necon3abid | 
					⊢  ( 𝜑  →  ( 𝐴  ≠  𝐵  ↔  ¬  𝜓 ) )  | 
				
			
		
		
			
				Proof
				
					
						| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							necon3abid.1 | 
							⊢ ( 𝜑  →  ( 𝐴  =  𝐵  ↔  𝜓 ) )  | 
						
						
							| 2 | 
							
								
							 | 
							df-ne | 
							⊢ ( 𝐴  ≠  𝐵  ↔  ¬  𝐴  =  𝐵 )  | 
						
						
							| 3 | 
							
								1
							 | 
							notbid | 
							⊢ ( 𝜑  →  ( ¬  𝐴  =  𝐵  ↔  ¬  𝜓 ) )  | 
						
						
							| 4 | 
							
								2 3
							 | 
							bitrid | 
							⊢ ( 𝜑  →  ( 𝐴  ≠  𝐵  ↔  ¬  𝜓 ) )  |