| Step | Hyp | Ref | Expression | 
						
							| 1 |  | mulcom | ⊢ ( ( 𝐴  ∈  ℂ  ∧  𝐶  ∈  ℂ )  →  ( 𝐴  ·  𝐶 )  =  ( 𝐶  ·  𝐴 ) ) | 
						
							| 2 | 1 | 3adant2 | ⊢ ( ( 𝐴  ∈  ℂ  ∧  𝐵  ∈  ℂ  ∧  𝐶  ∈  ℂ )  →  ( 𝐴  ·  𝐶 )  =  ( 𝐶  ·  𝐴 ) ) | 
						
							| 3 |  | mulcom | ⊢ ( ( 𝐵  ∈  ℂ  ∧  𝐶  ∈  ℂ )  →  ( 𝐵  ·  𝐶 )  =  ( 𝐶  ·  𝐵 ) ) | 
						
							| 4 | 3 | 3adant1 | ⊢ ( ( 𝐴  ∈  ℂ  ∧  𝐵  ∈  ℂ  ∧  𝐶  ∈  ℂ )  →  ( 𝐵  ·  𝐶 )  =  ( 𝐶  ·  𝐵 ) ) | 
						
							| 5 | 2 4 | eqeq12d | ⊢ ( ( 𝐴  ∈  ℂ  ∧  𝐵  ∈  ℂ  ∧  𝐶  ∈  ℂ )  →  ( ( 𝐴  ·  𝐶 )  =  ( 𝐵  ·  𝐶 )  ↔  ( 𝐶  ·  𝐴 )  =  ( 𝐶  ·  𝐵 ) ) ) | 
						
							| 6 |  | mulcan1g | ⊢ ( ( 𝐶  ∈  ℂ  ∧  𝐴  ∈  ℂ  ∧  𝐵  ∈  ℂ )  →  ( ( 𝐶  ·  𝐴 )  =  ( 𝐶  ·  𝐵 )  ↔  ( 𝐶  =  0  ∨  𝐴  =  𝐵 ) ) ) | 
						
							| 7 | 6 | 3coml | ⊢ ( ( 𝐴  ∈  ℂ  ∧  𝐵  ∈  ℂ  ∧  𝐶  ∈  ℂ )  →  ( ( 𝐶  ·  𝐴 )  =  ( 𝐶  ·  𝐵 )  ↔  ( 𝐶  =  0  ∨  𝐴  =  𝐵 ) ) ) | 
						
							| 8 |  | orcom | ⊢ ( ( 𝐶  =  0  ∨  𝐴  =  𝐵 )  ↔  ( 𝐴  =  𝐵  ∨  𝐶  =  0 ) ) | 
						
							| 9 | 7 8 | bitrdi | ⊢ ( ( 𝐴  ∈  ℂ  ∧  𝐵  ∈  ℂ  ∧  𝐶  ∈  ℂ )  →  ( ( 𝐶  ·  𝐴 )  =  ( 𝐶  ·  𝐵 )  ↔  ( 𝐴  =  𝐵  ∨  𝐶  =  0 ) ) ) | 
						
							| 10 | 5 9 | bitrd | ⊢ ( ( 𝐴  ∈  ℂ  ∧  𝐵  ∈  ℂ  ∧  𝐶  ∈  ℂ )  →  ( ( 𝐴  ·  𝐶 )  =  ( 𝐵  ·  𝐶 )  ↔  ( 𝐴  =  𝐵  ∨  𝐶  =  0 ) ) ) |