Metamath Proof Explorer
		
		
		
		Description:  Membership in a class abstraction, using implicit substitution.
       Deduction version of elab .  (Contributed by GG, 12-Oct-2024)
		
			
				
					 | 
					 | 
					Ref | 
					Expression | 
				
					
						 | 
						Hypotheses | 
						elabd3.ex | 
						⊢ ( 𝜑  →  𝐴  ∈  𝑉 )  | 
					
					
						 | 
						 | 
						elabd3.is | 
						⊢ ( ( 𝜑  ∧  𝑥  =  𝐴 )  →  ( 𝜓  ↔  𝜒 ) )  | 
					
				
					 | 
					Assertion | 
					elabd3 | 
					⊢  ( 𝜑  →  ( 𝐴  ∈  { 𝑥  ∣  𝜓 }  ↔  𝜒 ) )  | 
				
			
		
		
			
				Proof
				
					
						| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							elabd3.ex | 
							⊢ ( 𝜑  →  𝐴  ∈  𝑉 )  | 
						
						
							| 2 | 
							
								
							 | 
							elabd3.is | 
							⊢ ( ( 𝜑  ∧  𝑥  =  𝐴 )  →  ( 𝜓  ↔  𝜒 ) )  | 
						
						
							| 3 | 
							
								
							 | 
							eqidd | 
							⊢ ( 𝜑  →  { 𝑥  ∣  𝜓 }  =  { 𝑥  ∣  𝜓 } )  | 
						
						
							| 4 | 
							
								1 3 2
							 | 
							elabd2 | 
							⊢ ( 𝜑  →  ( 𝐴  ∈  { 𝑥  ∣  𝜓 }  ↔  𝜒 ) )  |