| 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 |
|
ax-hvass |
⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ ( - 1 ·ℎ 𝐶 ) ∈ ℋ ) → ( ( 𝐴 +ℎ 𝐵 ) +ℎ ( - 1 ·ℎ 𝐶 ) ) = ( 𝐴 +ℎ ( 𝐵 +ℎ ( - 1 ·ℎ 𝐶 ) ) ) ) |
| 5 |
3 4
|
syl3an3 |
⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 +ℎ 𝐵 ) +ℎ ( - 1 ·ℎ 𝐶 ) ) = ( 𝐴 +ℎ ( 𝐵 +ℎ ( - 1 ·ℎ 𝐶 ) ) ) ) |
| 6 |
|
hvaddcl |
⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( 𝐴 +ℎ 𝐵 ) ∈ ℋ ) |
| 7 |
|
hvsubval |
⊢ ( ( ( 𝐴 +ℎ 𝐵 ) ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 +ℎ 𝐵 ) −ℎ 𝐶 ) = ( ( 𝐴 +ℎ 𝐵 ) +ℎ ( - 1 ·ℎ 𝐶 ) ) ) |
| 8 |
6 7
|
stoic3 |
⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 +ℎ 𝐵 ) −ℎ 𝐶 ) = ( ( 𝐴 +ℎ 𝐵 ) +ℎ ( - 1 ·ℎ 𝐶 ) ) ) |
| 9 |
|
hvsubval |
⊢ ( ( 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐵 −ℎ 𝐶 ) = ( 𝐵 +ℎ ( - 1 ·ℎ 𝐶 ) ) ) |
| 10 |
9
|
3adant1 |
⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐵 −ℎ 𝐶 ) = ( 𝐵 +ℎ ( - 1 ·ℎ 𝐶 ) ) ) |
| 11 |
10
|
oveq2d |
⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 +ℎ ( 𝐵 −ℎ 𝐶 ) ) = ( 𝐴 +ℎ ( 𝐵 +ℎ ( - 1 ·ℎ 𝐶 ) ) ) ) |
| 12 |
5 8 11
|
3eqtr4d |
⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 +ℎ 𝐵 ) −ℎ 𝐶 ) = ( 𝐴 +ℎ ( 𝐵 −ℎ 𝐶 ) ) ) |