Metamath Proof Explorer
		
		
		
		Description:  Membership in a class abstraction, using implicit substitution.
       (Contributed by NM, 13-Sep-1995)
		
			
				
					|  |  | Ref | Expression | 
					
						|  | Hypotheses | elab2g.1 | ⊢ ( 𝑥  =  𝐴  →  ( 𝜑  ↔  𝜓 ) ) | 
					
						|  |  | elab2g.2 | ⊢ 𝐵  =  { 𝑥  ∣  𝜑 } | 
				
					|  | Assertion | elab2g | ⊢  ( 𝐴  ∈  𝑉  →  ( 𝐴  ∈  𝐵  ↔  𝜓 ) ) | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | elab2g.1 | ⊢ ( 𝑥  =  𝐴  →  ( 𝜑  ↔  𝜓 ) ) | 
						
							| 2 |  | elab2g.2 | ⊢ 𝐵  =  { 𝑥  ∣  𝜑 } | 
						
							| 3 | 2 | eleq2i | ⊢ ( 𝐴  ∈  𝐵  ↔  𝐴  ∈  { 𝑥  ∣  𝜑 } ) | 
						
							| 4 | 1 | elabg | ⊢ ( 𝐴  ∈  𝑉  →  ( 𝐴  ∈  { 𝑥  ∣  𝜑 }  ↔  𝜓 ) ) | 
						
							| 5 | 3 4 | bitrid | ⊢ ( 𝐴  ∈  𝑉  →  ( 𝐴  ∈  𝐵  ↔  𝜓 ) ) |