Metamath Proof Explorer
		
		
		
		Description:  Equality deduction for product.  (Contributed by Scott Fenton, 4-Dec-2017)
		
			
				
					 | 
					 | 
					Ref | 
					Expression | 
				
					
						 | 
						Hypothesis | 
						prodeq2dv.1 | 
						⊢ ( ( 𝜑  ∧  𝑘  ∈  𝐴 )  →  𝐵  =  𝐶 )  | 
					
				
					 | 
					Assertion | 
					prodeq2dv | 
					⊢  ( 𝜑  →  ∏ 𝑘  ∈  𝐴 𝐵  =  ∏ 𝑘  ∈  𝐴 𝐶 )  | 
				
			
		
		
			
				Proof
				
					
						| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							prodeq2dv.1 | 
							⊢ ( ( 𝜑  ∧  𝑘  ∈  𝐴 )  →  𝐵  =  𝐶 )  | 
						
						
							| 2 | 
							
								1
							 | 
							ralrimiva | 
							⊢ ( 𝜑  →  ∀ 𝑘  ∈  𝐴 𝐵  =  𝐶 )  | 
						
						
							| 3 | 
							
								2
							 | 
							prodeq2d | 
							⊢ ( 𝜑  →  ∏ 𝑘  ∈  𝐴 𝐵  =  ∏ 𝑘  ∈  𝐴 𝐶 )  |