| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							hlatjcom.j | 
							⊢  ∨   =  ( join ‘ 𝐾 )  | 
						
						
							| 2 | 
							
								
							 | 
							hlatjcom.a | 
							⊢ 𝐴  =  ( Atoms ‘ 𝐾 )  | 
						
						
							| 3 | 
							
								1 2
							 | 
							hlatjcom | 
							⊢ ( ( 𝐾  ∈  HL  ∧  𝑃  ∈  𝐴  ∧  𝑄  ∈  𝐴 )  →  ( 𝑃  ∨  𝑄 )  =  ( 𝑄  ∨  𝑃 ) )  | 
						
						
							| 4 | 
							
								3
							 | 
							3adant3r3 | 
							⊢ ( ( 𝐾  ∈  HL  ∧  ( 𝑃  ∈  𝐴  ∧  𝑄  ∈  𝐴  ∧  𝑅  ∈  𝐴 ) )  →  ( 𝑃  ∨  𝑄 )  =  ( 𝑄  ∨  𝑃 ) )  | 
						
						
							| 5 | 
							
								4
							 | 
							oveq1d | 
							⊢ ( ( 𝐾  ∈  HL  ∧  ( 𝑃  ∈  𝐴  ∧  𝑄  ∈  𝐴  ∧  𝑅  ∈  𝐴 ) )  →  ( ( 𝑃  ∨  𝑄 )  ∨  𝑅 )  =  ( ( 𝑄  ∨  𝑃 )  ∨  𝑅 ) )  | 
						
						
							| 6 | 
							
								1 2
							 | 
							hlatjass | 
							⊢ ( ( 𝐾  ∈  HL  ∧  ( 𝑃  ∈  𝐴  ∧  𝑄  ∈  𝐴  ∧  𝑅  ∈  𝐴 ) )  →  ( ( 𝑃  ∨  𝑄 )  ∨  𝑅 )  =  ( 𝑃  ∨  ( 𝑄  ∨  𝑅 ) ) )  | 
						
						
							| 7 | 
							
								
							 | 
							simpl | 
							⊢ ( ( 𝐾  ∈  HL  ∧  ( 𝑃  ∈  𝐴  ∧  𝑄  ∈  𝐴  ∧  𝑅  ∈  𝐴 ) )  →  𝐾  ∈  HL )  | 
						
						
							| 8 | 
							
								
							 | 
							simpr2 | 
							⊢ ( ( 𝐾  ∈  HL  ∧  ( 𝑃  ∈  𝐴  ∧  𝑄  ∈  𝐴  ∧  𝑅  ∈  𝐴 ) )  →  𝑄  ∈  𝐴 )  | 
						
						
							| 9 | 
							
								
							 | 
							simpr1 | 
							⊢ ( ( 𝐾  ∈  HL  ∧  ( 𝑃  ∈  𝐴  ∧  𝑄  ∈  𝐴  ∧  𝑅  ∈  𝐴 ) )  →  𝑃  ∈  𝐴 )  | 
						
						
							| 10 | 
							
								
							 | 
							simpr3 | 
							⊢ ( ( 𝐾  ∈  HL  ∧  ( 𝑃  ∈  𝐴  ∧  𝑄  ∈  𝐴  ∧  𝑅  ∈  𝐴 ) )  →  𝑅  ∈  𝐴 )  | 
						
						
							| 11 | 
							
								1 2
							 | 
							hlatjass | 
							⊢ ( ( 𝐾  ∈  HL  ∧  ( 𝑄  ∈  𝐴  ∧  𝑃  ∈  𝐴  ∧  𝑅  ∈  𝐴 ) )  →  ( ( 𝑄  ∨  𝑃 )  ∨  𝑅 )  =  ( 𝑄  ∨  ( 𝑃  ∨  𝑅 ) ) )  | 
						
						
							| 12 | 
							
								7 8 9 10 11
							 | 
							syl13anc | 
							⊢ ( ( 𝐾  ∈  HL  ∧  ( 𝑃  ∈  𝐴  ∧  𝑄  ∈  𝐴  ∧  𝑅  ∈  𝐴 ) )  →  ( ( 𝑄  ∨  𝑃 )  ∨  𝑅 )  =  ( 𝑄  ∨  ( 𝑃  ∨  𝑅 ) ) )  | 
						
						
							| 13 | 
							
								5 6 12
							 | 
							3eqtr3d | 
							⊢ ( ( 𝐾  ∈  HL  ∧  ( 𝑃  ∈  𝐴  ∧  𝑄  ∈  𝐴  ∧  𝑅  ∈  𝐴 ) )  →  ( 𝑃  ∨  ( 𝑄  ∨  𝑅 ) )  =  ( 𝑄  ∨  ( 𝑃  ∨  𝑅 ) ) )  |