Metamath Proof Explorer
		
		
		
		Description:  Detach truth from conjunction in biconditional.  (Contributed by Glauco
       Siliprandi, 3-Mar-2021)
		
			
				
					 | 
					 | 
					Ref | 
					Expression | 
				
					
						 | 
						Hypotheses | 
						mpbirand.1 | 
						⊢ ( 𝜑  →  𝜒 )  | 
					
					
						 | 
						 | 
						mpbirand.2 | 
						⊢ ( 𝜑  →  ( 𝜓  ↔  ( 𝜒  ∧  𝜃 ) ) )  | 
					
				
					 | 
					Assertion | 
					mpbirand | 
					⊢  ( 𝜑  →  ( 𝜓  ↔  𝜃 ) )  | 
				
			
		
		
			
				Proof
				
					
						| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							mpbirand.1 | 
							⊢ ( 𝜑  →  𝜒 )  | 
						
						
							| 2 | 
							
								
							 | 
							mpbirand.2 | 
							⊢ ( 𝜑  →  ( 𝜓  ↔  ( 𝜒  ∧  𝜃 ) ) )  | 
						
						
							| 3 | 
							
								1
							 | 
							biantrurd | 
							⊢ ( 𝜑  →  ( 𝜃  ↔  ( 𝜒  ∧  𝜃 ) ) )  | 
						
						
							| 4 | 
							
								2 3
							 | 
							bitr4d | 
							⊢ ( 𝜑  →  ( 𝜓  ↔  𝜃 ) )  |