Metamath Proof Explorer
		
		
		
		Description:  If G is an inverse to F , then G is an isomorphism.
         (Contributed by Mario Carneiro, 3-Jan-2017)
		
			
				
					|  |  | Ref | Expression | 
					
						|  | Hypotheses | invfval.b |  | 
					
						|  |  | invfval.n |  | 
					
						|  |  | invfval.c |  | 
					
						|  |  | invfval.x |  | 
					
						|  |  | invfval.y |  | 
					
						|  |  | isoval.n |  | 
					
						|  |  | inviso1.1 |  | 
				
					|  | Assertion | inviso2 |  | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | invfval.b |  | 
						
							| 2 |  | invfval.n |  | 
						
							| 3 |  | invfval.c |  | 
						
							| 4 |  | invfval.x |  | 
						
							| 5 |  | invfval.y |  | 
						
							| 6 |  | isoval.n |  | 
						
							| 7 |  | inviso1.1 |  | 
						
							| 8 | 1 2 3 4 5 | invsym |  | 
						
							| 9 | 7 8 | mpbid |  | 
						
							| 10 | 1 2 3 5 4 6 9 | inviso1 |  |