| Step | Hyp | Ref | Expression | 
						
							| 1 |  | neg1cn | ⊢ - 1  ∈  ℂ | 
						
							| 2 |  | hvmulcl | ⊢ ( ( - 1  ∈  ℂ  ∧  𝐶  ∈   ℋ )  →  ( - 1  ·ℎ  𝐶 )  ∈   ℋ ) | 
						
							| 3 | 1 2 | mpan | ⊢ ( 𝐶  ∈   ℋ  →  ( - 1  ·ℎ  𝐶 )  ∈   ℋ ) | 
						
							| 4 |  | hvadd12 | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ  ∧  ( - 1  ·ℎ  𝐶 )  ∈   ℋ )  →  ( 𝐴  +ℎ  ( 𝐵  +ℎ  ( - 1  ·ℎ  𝐶 ) ) )  =  ( 𝐵  +ℎ  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐶 ) ) ) ) | 
						
							| 5 | 3 4 | syl3an3 | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( 𝐴  +ℎ  ( 𝐵  +ℎ  ( - 1  ·ℎ  𝐶 ) ) )  =  ( 𝐵  +ℎ  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐶 ) ) ) ) | 
						
							| 6 |  | hvsubval | ⊢ ( ( 𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( 𝐵  −ℎ  𝐶 )  =  ( 𝐵  +ℎ  ( - 1  ·ℎ  𝐶 ) ) ) | 
						
							| 7 | 6 | oveq2d | ⊢ ( ( 𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( 𝐴  +ℎ  ( 𝐵  −ℎ  𝐶 ) )  =  ( 𝐴  +ℎ  ( 𝐵  +ℎ  ( - 1  ·ℎ  𝐶 ) ) ) ) | 
						
							| 8 | 7 | 3adant1 | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( 𝐴  +ℎ  ( 𝐵  −ℎ  𝐶 ) )  =  ( 𝐴  +ℎ  ( 𝐵  +ℎ  ( - 1  ·ℎ  𝐶 ) ) ) ) | 
						
							| 9 |  | hvsubval | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( 𝐴  −ℎ  𝐶 )  =  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐶 ) ) ) | 
						
							| 10 | 9 | oveq2d | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( 𝐵  +ℎ  ( 𝐴  −ℎ  𝐶 ) )  =  ( 𝐵  +ℎ  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐶 ) ) ) ) | 
						
							| 11 | 10 | 3adant2 | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( 𝐵  +ℎ  ( 𝐴  −ℎ  𝐶 ) )  =  ( 𝐵  +ℎ  ( 𝐴  +ℎ  ( - 1  ·ℎ  𝐶 ) ) ) ) | 
						
							| 12 | 5 8 11 | 3eqtr4d | ⊢ ( ( 𝐴  ∈   ℋ  ∧  𝐵  ∈   ℋ  ∧  𝐶  ∈   ℋ )  →  ( 𝐴  +ℎ  ( 𝐵  −ℎ  𝐶 ) )  =  ( 𝐵  +ℎ  ( 𝐴  −ℎ  𝐶 ) ) ) |