Metamath Proof Explorer
		
		
		
		Description:  Eliminate a disjunction in a deduction.  (Contributed by Mario Carneiro, 29-May-2016)
		
			
				
					 | 
					 | 
					Ref | 
					Expression | 
				
					
						 | 
						Hypotheses | 
						jaod.1 | 
						⊢ ( 𝜑  →  ( 𝜓  →  𝜒 ) )  | 
					
					
						 | 
						 | 
						jaod.2 | 
						⊢ ( 𝜑  →  ( 𝜃  →  𝜒 ) )  | 
					
					
						 | 
						 | 
						jaod.3 | 
						⊢ ( 𝜑  →  ( 𝜓  ∨  𝜃 ) )  | 
					
				
					 | 
					Assertion | 
					mpjaod | 
					⊢  ( 𝜑  →  𝜒 )  | 
				
			
		
		
			
				Proof
				
					
						| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							jaod.1 | 
							⊢ ( 𝜑  →  ( 𝜓  →  𝜒 ) )  | 
						
						
							| 2 | 
							
								
							 | 
							jaod.2 | 
							⊢ ( 𝜑  →  ( 𝜃  →  𝜒 ) )  | 
						
						
							| 3 | 
							
								
							 | 
							jaod.3 | 
							⊢ ( 𝜑  →  ( 𝜓  ∨  𝜃 ) )  | 
						
						
							| 4 | 
							
								1 2
							 | 
							jaod | 
							⊢ ( 𝜑  →  ( ( 𝜓  ∨  𝜃 )  →  𝜒 ) )  | 
						
						
							| 5 | 
							
								3 4
							 | 
							mpd | 
							⊢ ( 𝜑  →  𝜒 )  |