Database  
				SUPPLEMENTARY MATERIAL (USERS' MATHBOXES)  
				Mathbox for Peter Mazsa  
				Partition-Equivalence Theorems  
				mpet3  
			 
				
		 
		 Metamath Proof Explorer 
		
			
		 
		 
		
		Description:   Member Partition-Equivalence Theorem.  Together with mpet  mpet2  ,
     mostly in its conventional cpet  and cpet2  form, this is what we used
     to think of as the partition equivalence theorem (but cf. pet2  with
     general R  ).  (Contributed by Peter Mazsa , 4-May-2018)   (Revised by Peter Mazsa , 26-Sep-2021) 
		
			
				
					 
					 
					Ref 
					Expression 
				 
				
					 
					Assertion 
					mpet3  
					   ⊢   ElDisj  A   ∧   ¬   ∅  ∈  A         ↔   CoElEqvRel  A   ∧     ⋃  A    /  ∼  A      =  A            
				 
			
		 
		 
			
				Proof 
				
					
						Step 
						Hyp 
						Ref 
						Expression 
					 
						
							1  
							
								
							 
							eldisjn0elb  
							    ⊢   ElDisj  A   ∧   ¬   ∅  ∈  A         ↔   Disj    E  -1    ↾   A    ∧     dom  ⁡    E  -1    ↾   A      /    E  -1    ↾   A      =  A            
						 
						
							2  
							
								
							 
							eqvrelqseqdisj3  
							    ⊢   EqvRel  ≀    E  -1    ↾   A      ∧     dom  ⁡  ≀    E  -1    ↾   A        /  ≀    E  -1    ↾   A        =  A       →  Disj    E  -1    ↾   A         
						 
						
							3  
							
								2 
							 
							petlem  
							    ⊢   Disj    E  -1    ↾   A    ∧     dom  ⁡    E  -1    ↾   A      /    E  -1    ↾   A      =  A       ↔   EqvRel  ≀    E  -1    ↾   A      ∧     dom  ⁡  ≀    E  -1    ↾   A        /  ≀    E  -1    ↾   A        =  A            
						 
						
							4  
							
								
							 
							eqvreldmqs  
							    ⊢   EqvRel  ≀    E  -1    ↾   A      ∧     dom  ⁡  ≀    E  -1    ↾   A        /  ≀    E  -1    ↾   A        =  A       ↔   CoElEqvRel  A   ∧     ⋃  A    /  ∼  A      =  A            
						 
						
							5  
							
								1  3  4 
							 
							3bitri  
							    ⊢   ElDisj  A   ∧   ¬   ∅  ∈  A         ↔   CoElEqvRel  A   ∧     ⋃  A    /  ∼  A      =  A