Metamath Proof Explorer
		
		
		
		Description:  Equality deduction for intersection of two classes.  (Contributed by NM, 7-Feb-2007)
		
			
				
					|  |  | Ref | Expression | 
					
						|  | Hypotheses | ineq1d.1 | ⊢ ( 𝜑  →  𝐴  =  𝐵 ) | 
					
						|  |  | ineqan12d.2 | ⊢ ( 𝜓  →  𝐶  =  𝐷 ) | 
				
					|  | Assertion | ineqan12d | ⊢  ( ( 𝜑  ∧  𝜓 )  →  ( 𝐴  ∩  𝐶 )  =  ( 𝐵  ∩  𝐷 ) ) | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | ineq1d.1 | ⊢ ( 𝜑  →  𝐴  =  𝐵 ) | 
						
							| 2 |  | ineqan12d.2 | ⊢ ( 𝜓  →  𝐶  =  𝐷 ) | 
						
							| 3 |  | ineq12 | ⊢ ( ( 𝐴  =  𝐵  ∧  𝐶  =  𝐷 )  →  ( 𝐴  ∩  𝐶 )  =  ( 𝐵  ∩  𝐷 ) ) | 
						
							| 4 | 1 2 3 | syl2an | ⊢ ( ( 𝜑  ∧  𝜓 )  →  ( 𝐴  ∩  𝐶 )  =  ( 𝐵  ∩  𝐷 ) ) |