Metamath Proof Explorer
		
		
		
		Description:  Lemma for dath .  Frequently-used utility lemma.  (Contributed by NM, 15-Aug-2012)
		
			
				
					 | 
					 | 
					Ref | 
					Expression | 
				
					
						 | 
						Hypothesis | 
						da.ps0 | 
						⊢ ( 𝜓  ↔  ( ( 𝑐  ∈  𝐴  ∧  𝑑  ∈  𝐴 )  ∧  ¬  𝑐  ≤  𝑌  ∧  ( 𝑑  ≠  𝑐  ∧  ¬  𝑑  ≤  𝑌  ∧  𝐶  ≤  ( 𝑐  ∨  𝑑 ) ) ) )  | 
					
				
					 | 
					Assertion | 
					dalemccnedd | 
					⊢  ( 𝜓  →  𝑐  ≠  𝑑 )  | 
				
			
		
		
			
				Proof
				
					
						| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							da.ps0 | 
							⊢ ( 𝜓  ↔  ( ( 𝑐  ∈  𝐴  ∧  𝑑  ∈  𝐴 )  ∧  ¬  𝑐  ≤  𝑌  ∧  ( 𝑑  ≠  𝑐  ∧  ¬  𝑑  ≤  𝑌  ∧  𝐶  ≤  ( 𝑐  ∨  𝑑 ) ) ) )  | 
						
						
							| 2 | 
							
								
							 | 
							simp31 | 
							⊢ ( ( ( 𝑐  ∈  𝐴  ∧  𝑑  ∈  𝐴 )  ∧  ¬  𝑐  ≤  𝑌  ∧  ( 𝑑  ≠  𝑐  ∧  ¬  𝑑  ≤  𝑌  ∧  𝐶  ≤  ( 𝑐  ∨  𝑑 ) ) )  →  𝑑  ≠  𝑐 )  | 
						
						
							| 3 | 
							
								1 2
							 | 
							sylbi | 
							⊢ ( 𝜓  →  𝑑  ≠  𝑐 )  | 
						
						
							| 4 | 
							
								3
							 | 
							necomd | 
							⊢ ( 𝜓  →  𝑐  ≠  𝑑 )  |