| Step | Hyp | Ref | Expression | 
						
							| 1 |  | hvsubval | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ )  →  ( 𝐴  −ℎ  𝐵 )  =  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐵 ) ) ) | 
						
							| 2 | 1 | 3adant3 | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( 𝐴  −ℎ  𝐵 )  =  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐵 ) ) ) | 
						
							| 3 |  | hvsubval | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( 𝐴  −ℎ  𝐶 )  =  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐶 ) ) ) | 
						
							| 4 | 3 | 3adant2 | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( 𝐴  −ℎ  𝐶 )  =  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐶 ) ) ) | 
						
							| 5 | 2 4 | eqeq12d | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( ( 𝐴  −ℎ  𝐵 )  =  ( 𝐴  −ℎ  𝐶 )  ↔  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐵 ) )  =  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐶 ) ) ) ) | 
						
							| 6 |  | neg1cn | ⊢ - 1  ∈  ℂ | 
						
							| 7 |  | hvmulcl | ⊢ ( ( - 1  ∈  ℂ  ∧  𝐵  ∈   ℋ )  →  ( - 1  ·ℎ  𝐵 )  ∈   ℋ ) | 
						
							| 8 | 6 7 | mpan | ⊢ ( 𝐵  ∈   ℋ  →  ( - 1  ·ℎ  𝐵 )  ∈   ℋ ) | 
						
							| 9 |  | hvmulcl | ⊢ ( ( - 1  ∈  ℂ  ∧  𝐶  ∈   ℋ )  →  ( - 1  ·ℎ  𝐶 )  ∈   ℋ ) | 
						
							| 10 | 6 9 | mpan | ⊢ ( 𝐶  ∈   ℋ  →  ( - 1  ·ℎ  𝐶 )  ∈   ℋ ) | 
						
							| 11 |  | hvaddcan | ⊢ ( ( 𝐴  ∈   ℋ  ∧  ( - 1  ·ℎ  𝐵 )  ∈   ℋ  ∧  ( - 1  ·ℎ  𝐶 )  ∈   ℋ )  →  ( ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐵 ) )  =  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐶 ) )  ↔  ( - 1  ·ℎ  𝐵 )  =  ( - 1  ·ℎ  𝐶 ) ) ) | 
						
							| 12 | 10 11 | syl3an3 | ⊢ ( ( 𝐴  ∈   ℋ  ∧  ( - 1  ·ℎ  𝐵 )  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐵 ) )  =  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐶 ) )  ↔  ( - 1  ·ℎ  𝐵 )  =  ( - 1  ·ℎ  𝐶 ) ) ) | 
						
							| 13 | 8 12 | syl3an2 | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐵 ) )  =  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐶 ) )  ↔  ( - 1  ·ℎ  𝐵 )  =  ( - 1  ·ℎ  𝐶 ) ) ) | 
						
							| 14 |  | neg1ne0 | ⊢ - 1  ≠  0 | 
						
							| 15 | 6 14 | pm3.2i | ⊢ ( - 1  ∈  ℂ  ∧  - 1  ≠  0 ) | 
						
							| 16 |  | hvmulcan | ⊢ ( ( ( - 1  ∈  ℂ  ∧  - 1  ≠  0 )  ∧  𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( ( - 1  ·ℎ  𝐵 )  =  ( - 1  ·ℎ  𝐶 )  ↔  𝐵  =  𝐶 ) ) | 
						
							| 17 | 15 16 | mp3an1 | ⊢ ( ( 𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( ( - 1  ·ℎ  𝐵 )  =  ( - 1  ·ℎ  𝐶 )  ↔  𝐵  =  𝐶 ) ) | 
						
							| 18 | 17 | 3adant1 | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( ( - 1  ·ℎ  𝐵 )  =  ( - 1  ·ℎ  𝐶 )  ↔  𝐵  =  𝐶 ) ) | 
						
							| 19 | 5 13 18 | 3bitrd | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( ( 𝐴  −ℎ  𝐵 )  =  ( 𝐴  −ℎ  𝐶 )  ↔  𝐵  =  𝐶 ) ) |