Metamath Proof Explorer
		
		
		
		Description:  Contrapositive law deduction for inequality.  (Contributed by NM, 9-May-2012)
		
			
				
					 | 
					 | 
					Ref | 
					Expression | 
				
					
						 | 
						Hypothesis | 
						necon4bbid.1 | 
						⊢ ( 𝜑  →  ( ¬  𝜓  ↔  𝐴  ≠  𝐵 ) )  | 
					
				
					 | 
					Assertion | 
					necon4bbid | 
					⊢  ( 𝜑  →  ( 𝜓  ↔  𝐴  =  𝐵 ) )  | 
				
			
		
		
			
				Proof
				
					
						| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							necon4bbid.1 | 
							⊢ ( 𝜑  →  ( ¬  𝜓  ↔  𝐴  ≠  𝐵 ) )  | 
						
						
							| 2 | 
							
								1
							 | 
							bicomd | 
							⊢ ( 𝜑  →  ( 𝐴  ≠  𝐵  ↔  ¬  𝜓 ) )  | 
						
						
							| 3 | 
							
								2
							 | 
							necon4abid | 
							⊢ ( 𝜑  →  ( 𝐴  =  𝐵  ↔  𝜓 ) )  | 
						
						
							| 4 | 
							
								3
							 | 
							bicomd | 
							⊢ ( 𝜑  →  ( 𝜓  ↔  𝐴  =  𝐵 ) )  |