Step |
Hyp |
Ref |
Expression |
1 |
|
hvmulcl |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·ℎ 𝐶 ) ∈ ℋ ) |
2 |
1
|
3adant2 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·ℎ 𝐶 ) ∈ ℋ ) |
3 |
|
hvmulcl |
⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐵 ·ℎ 𝐶 ) ∈ ℋ ) |
4 |
3
|
3adant1 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐵 ·ℎ 𝐶 ) ∈ ℋ ) |
5 |
|
hvsubeq0 |
⊢ ( ( ( 𝐴 ·ℎ 𝐶 ) ∈ ℋ ∧ ( 𝐵 ·ℎ 𝐶 ) ∈ ℋ ) → ( ( ( 𝐴 ·ℎ 𝐶 ) −ℎ ( 𝐵 ·ℎ 𝐶 ) ) = 0ℎ ↔ ( 𝐴 ·ℎ 𝐶 ) = ( 𝐵 ·ℎ 𝐶 ) ) ) |
6 |
2 4 5
|
syl2anc |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( ( 𝐴 ·ℎ 𝐶 ) −ℎ ( 𝐵 ·ℎ 𝐶 ) ) = 0ℎ ↔ ( 𝐴 ·ℎ 𝐶 ) = ( 𝐵 ·ℎ 𝐶 ) ) ) |
7 |
6
|
3adant3r |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℋ ∧ 𝐶 ≠ 0ℎ ) ) → ( ( ( 𝐴 ·ℎ 𝐶 ) −ℎ ( 𝐵 ·ℎ 𝐶 ) ) = 0ℎ ↔ ( 𝐴 ·ℎ 𝐶 ) = ( 𝐵 ·ℎ 𝐶 ) ) ) |
8 |
|
hvsubdistr2 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 − 𝐵 ) ·ℎ 𝐶 ) = ( ( 𝐴 ·ℎ 𝐶 ) −ℎ ( 𝐵 ·ℎ 𝐶 ) ) ) |
9 |
8
|
eqeq1d |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( ( 𝐴 − 𝐵 ) ·ℎ 𝐶 ) = 0ℎ ↔ ( ( 𝐴 ·ℎ 𝐶 ) −ℎ ( 𝐵 ·ℎ 𝐶 ) ) = 0ℎ ) ) |
10 |
|
subcl |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ ) |
11 |
|
hvmul0or |
⊢ ( ( ( 𝐴 − 𝐵 ) ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( ( 𝐴 − 𝐵 ) ·ℎ 𝐶 ) = 0ℎ ↔ ( ( 𝐴 − 𝐵 ) = 0 ∨ 𝐶 = 0ℎ ) ) ) |
12 |
10 11
|
stoic3 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( ( 𝐴 − 𝐵 ) ·ℎ 𝐶 ) = 0ℎ ↔ ( ( 𝐴 − 𝐵 ) = 0 ∨ 𝐶 = 0ℎ ) ) ) |
13 |
9 12
|
bitr3d |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( ( 𝐴 ·ℎ 𝐶 ) −ℎ ( 𝐵 ·ℎ 𝐶 ) ) = 0ℎ ↔ ( ( 𝐴 − 𝐵 ) = 0 ∨ 𝐶 = 0ℎ ) ) ) |
14 |
13
|
3adant3r |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℋ ∧ 𝐶 ≠ 0ℎ ) ) → ( ( ( 𝐴 ·ℎ 𝐶 ) −ℎ ( 𝐵 ·ℎ 𝐶 ) ) = 0ℎ ↔ ( ( 𝐴 − 𝐵 ) = 0 ∨ 𝐶 = 0ℎ ) ) ) |
15 |
|
df-ne |
⊢ ( 𝐶 ≠ 0ℎ ↔ ¬ 𝐶 = 0ℎ ) |
16 |
|
biorf |
⊢ ( ¬ 𝐶 = 0ℎ → ( ( 𝐴 − 𝐵 ) = 0 ↔ ( 𝐶 = 0ℎ ∨ ( 𝐴 − 𝐵 ) = 0 ) ) ) |
17 |
|
orcom |
⊢ ( ( 𝐶 = 0ℎ ∨ ( 𝐴 − 𝐵 ) = 0 ) ↔ ( ( 𝐴 − 𝐵 ) = 0 ∨ 𝐶 = 0ℎ ) ) |
18 |
16 17
|
bitrdi |
⊢ ( ¬ 𝐶 = 0ℎ → ( ( 𝐴 − 𝐵 ) = 0 ↔ ( ( 𝐴 − 𝐵 ) = 0 ∨ 𝐶 = 0ℎ ) ) ) |
19 |
15 18
|
sylbi |
⊢ ( 𝐶 ≠ 0ℎ → ( ( 𝐴 − 𝐵 ) = 0 ↔ ( ( 𝐴 − 𝐵 ) = 0 ∨ 𝐶 = 0ℎ ) ) ) |
20 |
19
|
ad2antll |
⊢ ( ( 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℋ ∧ 𝐶 ≠ 0ℎ ) ) → ( ( 𝐴 − 𝐵 ) = 0 ↔ ( ( 𝐴 − 𝐵 ) = 0 ∨ 𝐶 = 0ℎ ) ) ) |
21 |
20
|
3adant1 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℋ ∧ 𝐶 ≠ 0ℎ ) ) → ( ( 𝐴 − 𝐵 ) = 0 ↔ ( ( 𝐴 − 𝐵 ) = 0 ∨ 𝐶 = 0ℎ ) ) ) |
22 |
|
subeq0 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = 0 ↔ 𝐴 = 𝐵 ) ) |
23 |
22
|
3adant3 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℋ ∧ 𝐶 ≠ 0ℎ ) ) → ( ( 𝐴 − 𝐵 ) = 0 ↔ 𝐴 = 𝐵 ) ) |
24 |
14 21 23
|
3bitr2d |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℋ ∧ 𝐶 ≠ 0ℎ ) ) → ( ( ( 𝐴 ·ℎ 𝐶 ) −ℎ ( 𝐵 ·ℎ 𝐶 ) ) = 0ℎ ↔ 𝐴 = 𝐵 ) ) |
25 |
7 24
|
bitr3d |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℋ ∧ 𝐶 ≠ 0ℎ ) ) → ( ( 𝐴 ·ℎ 𝐶 ) = ( 𝐵 ·ℎ 𝐶 ) ↔ 𝐴 = 𝐵 ) ) |