Metamath Proof Explorer
		
		
		Theorem fn0
		Description:  A function with empty domain is empty.  (Contributed by NM, 15-Apr-1998)
     (Proof shortened by Andrew Salmon, 17-Sep-2011)
		
			
				
					 | 
					 | 
					Ref | 
					Expression | 
				
				
					 | 
					Assertion | 
					fn0 | 
					⊢  ( 𝐹  Fn  ∅  ↔  𝐹  =  ∅ )  | 
				
			
		
		
			
				Proof
				
					
						| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							fnrel | 
							⊢ ( 𝐹  Fn  ∅  →  Rel  𝐹 )  | 
						
						
							| 2 | 
							
								
							 | 
							fndm | 
							⊢ ( 𝐹  Fn  ∅  →  dom  𝐹  =  ∅ )  | 
						
						
							| 3 | 
							
								
							 | 
							reldm0 | 
							⊢ ( Rel  𝐹  →  ( 𝐹  =  ∅  ↔  dom  𝐹  =  ∅ ) )  | 
						
						
							| 4 | 
							
								3
							 | 
							biimpar | 
							⊢ ( ( Rel  𝐹  ∧  dom  𝐹  =  ∅ )  →  𝐹  =  ∅ )  | 
						
						
							| 5 | 
							
								1 2 4
							 | 
							syl2anc | 
							⊢ ( 𝐹  Fn  ∅  →  𝐹  =  ∅ )  | 
						
						
							| 6 | 
							
								
							 | 
							fun0 | 
							⊢ Fun  ∅  | 
						
						
							| 7 | 
							
								
							 | 
							dm0 | 
							⊢ dom  ∅  =  ∅  | 
						
						
							| 8 | 
							
								
							 | 
							df-fn | 
							⊢ ( ∅  Fn  ∅  ↔  ( Fun  ∅  ∧  dom  ∅  =  ∅ ) )  | 
						
						
							| 9 | 
							
								6 7 8
							 | 
							mpbir2an | 
							⊢ ∅  Fn  ∅  | 
						
						
							| 10 | 
							
								
							 | 
							fneq1 | 
							⊢ ( 𝐹  =  ∅  →  ( 𝐹  Fn  ∅  ↔  ∅  Fn  ∅ ) )  | 
						
						
							| 11 | 
							
								9 10
							 | 
							mpbiri | 
							⊢ ( 𝐹  =  ∅  →  𝐹  Fn  ∅ )  | 
						
						
							| 12 | 
							
								5 11
							 | 
							impbii | 
							⊢ ( 𝐹  Fn  ∅  ↔  𝐹  =  ∅ )  |