| Step | 
						Hyp | 
						Ref | 
						Expression | 
					
						
							| 1 | 
							
								
							 | 
							oldmm1.b | 
							⊢ 𝐵  =  ( Base ‘ 𝐾 )  | 
						
						
							| 2 | 
							
								
							 | 
							oldmm1.j | 
							⊢  ∨   =  ( join ‘ 𝐾 )  | 
						
						
							| 3 | 
							
								
							 | 
							oldmm1.m | 
							⊢  ∧   =  ( meet ‘ 𝐾 )  | 
						
						
							| 4 | 
							
								
							 | 
							oldmm1.o | 
							⊢  ⊥   =  ( oc ‘ 𝐾 )  | 
						
						
							| 5 | 
							
								
							 | 
							olop | 
							⊢ ( 𝐾  ∈  OL  →  𝐾  ∈  OP )  | 
						
						
							| 6 | 
							
								1 4
							 | 
							opoccl | 
							⊢ ( ( 𝐾  ∈  OP  ∧  𝑋  ∈  𝐵 )  →  (  ⊥  ‘ 𝑋 )  ∈  𝐵 )  | 
						
						
							| 7 | 
							
								5 6
							 | 
							sylan | 
							⊢ ( ( 𝐾  ∈  OL  ∧  𝑋  ∈  𝐵 )  →  (  ⊥  ‘ 𝑋 )  ∈  𝐵 )  | 
						
						
							| 8 | 
							
								7
							 | 
							3adant3 | 
							⊢ ( ( 𝐾  ∈  OL  ∧  𝑋  ∈  𝐵  ∧  𝑌  ∈  𝐵 )  →  (  ⊥  ‘ 𝑋 )  ∈  𝐵 )  | 
						
						
							| 9 | 
							
								1 2 3 4
							 | 
							oldmj1 | 
							⊢ ( ( 𝐾  ∈  OL  ∧  (  ⊥  ‘ 𝑋 )  ∈  𝐵  ∧  𝑌  ∈  𝐵 )  →  (  ⊥  ‘ ( (  ⊥  ‘ 𝑋 )  ∨  𝑌 ) )  =  ( (  ⊥  ‘ (  ⊥  ‘ 𝑋 ) )  ∧  (  ⊥  ‘ 𝑌 ) ) )  | 
						
						
							| 10 | 
							
								8 9
							 | 
							syld3an2 | 
							⊢ ( ( 𝐾  ∈  OL  ∧  𝑋  ∈  𝐵  ∧  𝑌  ∈  𝐵 )  →  (  ⊥  ‘ ( (  ⊥  ‘ 𝑋 )  ∨  𝑌 ) )  =  ( (  ⊥  ‘ (  ⊥  ‘ 𝑋 ) )  ∧  (  ⊥  ‘ 𝑌 ) ) )  | 
						
						
							| 11 | 
							
								1 4
							 | 
							opococ | 
							⊢ ( ( 𝐾  ∈  OP  ∧  𝑋  ∈  𝐵 )  →  (  ⊥  ‘ (  ⊥  ‘ 𝑋 ) )  =  𝑋 )  | 
						
						
							| 12 | 
							
								5 11
							 | 
							sylan | 
							⊢ ( ( 𝐾  ∈  OL  ∧  𝑋  ∈  𝐵 )  →  (  ⊥  ‘ (  ⊥  ‘ 𝑋 ) )  =  𝑋 )  | 
						
						
							| 13 | 
							
								12
							 | 
							3adant3 | 
							⊢ ( ( 𝐾  ∈  OL  ∧  𝑋  ∈  𝐵  ∧  𝑌  ∈  𝐵 )  →  (  ⊥  ‘ (  ⊥  ‘ 𝑋 ) )  =  𝑋 )  | 
						
						
							| 14 | 
							
								13
							 | 
							oveq1d | 
							⊢ ( ( 𝐾  ∈  OL  ∧  𝑋  ∈  𝐵  ∧  𝑌  ∈  𝐵 )  →  ( (  ⊥  ‘ (  ⊥  ‘ 𝑋 ) )  ∧  (  ⊥  ‘ 𝑌 ) )  =  ( 𝑋  ∧  (  ⊥  ‘ 𝑌 ) ) )  | 
						
						
							| 15 | 
							
								10 14
							 | 
							eqtrd | 
							⊢ ( ( 𝐾  ∈  OL  ∧  𝑋  ∈  𝐵  ∧  𝑌  ∈  𝐵 )  →  (  ⊥  ‘ ( (  ⊥  ‘ 𝑋 )  ∨  𝑌 ) )  =  ( 𝑋  ∧  (  ⊥  ‘ 𝑌 ) ) )  |