| Step | Hyp | Ref | Expression | 
						
							| 1 |  | chjcom | ⊢ ( ( 𝐴  ∈   Cℋ   ∧  𝐵  ∈   Cℋ  )  →  ( 𝐴  ∨ℋ  𝐵 )  =  ( 𝐵  ∨ℋ  𝐴 ) ) | 
						
							| 2 | 1 | 3adant3 | ⊢ ( ( 𝐴  ∈   Cℋ   ∧  𝐵  ∈   Cℋ   ∧  𝐶  ∈   Cℋ  )  →  ( 𝐴  ∨ℋ  𝐵 )  =  ( 𝐵  ∨ℋ  𝐴 ) ) | 
						
							| 3 | 2 | oveq1d | ⊢ ( ( 𝐴  ∈   Cℋ   ∧  𝐵  ∈   Cℋ   ∧  𝐶  ∈   Cℋ  )  →  ( ( 𝐴  ∨ℋ  𝐵 )  ∨ℋ  𝐶 )  =  ( ( 𝐵  ∨ℋ  𝐴 )  ∨ℋ  𝐶 ) ) | 
						
							| 4 |  | chjass | ⊢ ( ( 𝐴  ∈   Cℋ   ∧  𝐵  ∈   Cℋ   ∧  𝐶  ∈   Cℋ  )  →  ( ( 𝐴  ∨ℋ  𝐵 )  ∨ℋ  𝐶 )  =  ( 𝐴  ∨ℋ  ( 𝐵  ∨ℋ  𝐶 ) ) ) | 
						
							| 5 |  | chjass | ⊢ ( ( 𝐵  ∈   Cℋ   ∧  𝐴  ∈   Cℋ   ∧  𝐶  ∈   Cℋ  )  →  ( ( 𝐵  ∨ℋ  𝐴 )  ∨ℋ  𝐶 )  =  ( 𝐵  ∨ℋ  ( 𝐴  ∨ℋ  𝐶 ) ) ) | 
						
							| 6 | 5 | 3com12 | ⊢ ( ( 𝐴  ∈   Cℋ   ∧  𝐵  ∈   Cℋ   ∧  𝐶  ∈   Cℋ  )  →  ( ( 𝐵  ∨ℋ  𝐴 )  ∨ℋ  𝐶 )  =  ( 𝐵  ∨ℋ  ( 𝐴  ∨ℋ  𝐶 ) ) ) | 
						
							| 7 | 3 4 6 | 3eqtr3d | ⊢ ( ( 𝐴  ∈   Cℋ   ∧  𝐵  ∈   Cℋ   ∧  𝐶  ∈   Cℋ  )  →  ( 𝐴  ∨ℋ  ( 𝐵  ∨ℋ  𝐶 ) )  =  ( 𝐵  ∨ℋ  ( 𝐴  ∨ℋ  𝐶 ) ) ) |