Metamath Proof Explorer
		
		
		
		Description:  Deduction from equality to inequality.  (Contributed by NM, 2-Jun-2007)
		
			
				
					|  |  | Ref | Expression | 
					
						|  | Hypothesis | necon3bbid.1 | ⊢ ( 𝜑  →  ( 𝜓  ↔  𝐴  =  𝐵 ) ) | 
				
					|  | Assertion | necon3bbid | ⊢  ( 𝜑  →  ( ¬  𝜓  ↔  𝐴  ≠  𝐵 ) ) | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | necon3bbid.1 | ⊢ ( 𝜑  →  ( 𝜓  ↔  𝐴  =  𝐵 ) ) | 
						
							| 2 | 1 | bicomd | ⊢ ( 𝜑  →  ( 𝐴  =  𝐵  ↔  𝜓 ) ) | 
						
							| 3 | 2 | necon3abid | ⊢ ( 𝜑  →  ( 𝐴  ≠  𝐵  ↔  ¬  𝜓 ) ) | 
						
							| 4 | 3 | bicomd | ⊢ ( 𝜑  →  ( ¬  𝜓  ↔  𝐴  ≠  𝐵 ) ) |