Metamath Proof Explorer
		
		
		
		Description:  Equality deduction for a binary relation.  (Contributed by NM, 8-Feb-1996)
		
			
				
					 | 
					 | 
					Ref | 
					Expression | 
				
					
						 | 
						Hypotheses | 
						breq1d.1 | 
						⊢ ( 𝜑  →  𝐴  =  𝐵 )  | 
					
					
						 | 
						 | 
						breqan12i.2 | 
						⊢ ( 𝜓  →  𝐶  =  𝐷 )  | 
					
				
					 | 
					Assertion | 
					breqan12d | 
					⊢  ( ( 𝜑  ∧  𝜓 )  →  ( 𝐴 𝑅 𝐶  ↔  𝐵 𝑅 𝐷 ) )  | 
				
			
		
		
			
				Proof
				
					
						| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							breq1d.1 | 
							⊢ ( 𝜑  →  𝐴  =  𝐵 )  | 
						
						
							| 2 | 
							
								
							 | 
							breqan12i.2 | 
							⊢ ( 𝜓  →  𝐶  =  𝐷 )  | 
						
						
							| 3 | 
							
								
							 | 
							breq12 | 
							⊢ ( ( 𝐴  =  𝐵  ∧  𝐶  =  𝐷 )  →  ( 𝐴 𝑅 𝐶  ↔  𝐵 𝑅 𝐷 ) )  | 
						
						
							| 4 | 
							
								1 2 3
							 | 
							syl2an | 
							⊢ ( ( 𝜑  ∧  𝜓 )  →  ( 𝐴 𝑅 𝐶  ↔  𝐵 𝑅 𝐷 ) )  |