Step |
Hyp |
Ref |
Expression |
1 |
|
elspansn |
⊢ ( 𝐵 ∈ ℋ → ( 𝐴 ∈ ( span ‘ { 𝐵 } ) ↔ ∃ 𝑥 ∈ ℂ 𝐴 = ( 𝑥 ·ℎ 𝐵 ) ) ) |
2 |
1
|
adantr |
⊢ ( ( 𝐵 ∈ ℋ ∧ 𝐴 ≠ 0ℎ ) → ( 𝐴 ∈ ( span ‘ { 𝐵 } ) ↔ ∃ 𝑥 ∈ ℂ 𝐴 = ( 𝑥 ·ℎ 𝐵 ) ) ) |
3 |
|
sneq |
⊢ ( 𝐴 = ( 𝑥 ·ℎ 𝐵 ) → { 𝐴 } = { ( 𝑥 ·ℎ 𝐵 ) } ) |
4 |
3
|
fveq2d |
⊢ ( 𝐴 = ( 𝑥 ·ℎ 𝐵 ) → ( span ‘ { 𝐴 } ) = ( span ‘ { ( 𝑥 ·ℎ 𝐵 ) } ) ) |
5 |
4
|
ad2antll |
⊢ ( ( ( 𝐵 ∈ ℋ ∧ 𝐴 ≠ 0ℎ ) ∧ ( 𝑥 ∈ ℂ ∧ 𝐴 = ( 𝑥 ·ℎ 𝐵 ) ) ) → ( span ‘ { 𝐴 } ) = ( span ‘ { ( 𝑥 ·ℎ 𝐵 ) } ) ) |
6 |
|
oveq1 |
⊢ ( 𝑥 = 0 → ( 𝑥 ·ℎ 𝐵 ) = ( 0 ·ℎ 𝐵 ) ) |
7 |
|
ax-hvmul0 |
⊢ ( 𝐵 ∈ ℋ → ( 0 ·ℎ 𝐵 ) = 0ℎ ) |
8 |
6 7
|
sylan9eqr |
⊢ ( ( 𝐵 ∈ ℋ ∧ 𝑥 = 0 ) → ( 𝑥 ·ℎ 𝐵 ) = 0ℎ ) |
9 |
8
|
ex |
⊢ ( 𝐵 ∈ ℋ → ( 𝑥 = 0 → ( 𝑥 ·ℎ 𝐵 ) = 0ℎ ) ) |
10 |
|
eqeq1 |
⊢ ( 𝐴 = ( 𝑥 ·ℎ 𝐵 ) → ( 𝐴 = 0ℎ ↔ ( 𝑥 ·ℎ 𝐵 ) = 0ℎ ) ) |
11 |
10
|
biimprd |
⊢ ( 𝐴 = ( 𝑥 ·ℎ 𝐵 ) → ( ( 𝑥 ·ℎ 𝐵 ) = 0ℎ → 𝐴 = 0ℎ ) ) |
12 |
9 11
|
sylan9 |
⊢ ( ( 𝐵 ∈ ℋ ∧ 𝐴 = ( 𝑥 ·ℎ 𝐵 ) ) → ( 𝑥 = 0 → 𝐴 = 0ℎ ) ) |
13 |
12
|
necon3d |
⊢ ( ( 𝐵 ∈ ℋ ∧ 𝐴 = ( 𝑥 ·ℎ 𝐵 ) ) → ( 𝐴 ≠ 0ℎ → 𝑥 ≠ 0 ) ) |
14 |
13
|
ex |
⊢ ( 𝐵 ∈ ℋ → ( 𝐴 = ( 𝑥 ·ℎ 𝐵 ) → ( 𝐴 ≠ 0ℎ → 𝑥 ≠ 0 ) ) ) |
15 |
14
|
com23 |
⊢ ( 𝐵 ∈ ℋ → ( 𝐴 ≠ 0ℎ → ( 𝐴 = ( 𝑥 ·ℎ 𝐵 ) → 𝑥 ≠ 0 ) ) ) |
16 |
15
|
impd |
⊢ ( 𝐵 ∈ ℋ → ( ( 𝐴 ≠ 0ℎ ∧ 𝐴 = ( 𝑥 ·ℎ 𝐵 ) ) → 𝑥 ≠ 0 ) ) |
17 |
16
|
adantr |
⊢ ( ( 𝐵 ∈ ℋ ∧ 𝑥 ∈ ℂ ) → ( ( 𝐴 ≠ 0ℎ ∧ 𝐴 = ( 𝑥 ·ℎ 𝐵 ) ) → 𝑥 ≠ 0 ) ) |
18 |
|
spansncol |
⊢ ( ( 𝐵 ∈ ℋ ∧ 𝑥 ∈ ℂ ∧ 𝑥 ≠ 0 ) → ( span ‘ { ( 𝑥 ·ℎ 𝐵 ) } ) = ( span ‘ { 𝐵 } ) ) |
19 |
18
|
3expia |
⊢ ( ( 𝐵 ∈ ℋ ∧ 𝑥 ∈ ℂ ) → ( 𝑥 ≠ 0 → ( span ‘ { ( 𝑥 ·ℎ 𝐵 ) } ) = ( span ‘ { 𝐵 } ) ) ) |
20 |
17 19
|
syld |
⊢ ( ( 𝐵 ∈ ℋ ∧ 𝑥 ∈ ℂ ) → ( ( 𝐴 ≠ 0ℎ ∧ 𝐴 = ( 𝑥 ·ℎ 𝐵 ) ) → ( span ‘ { ( 𝑥 ·ℎ 𝐵 ) } ) = ( span ‘ { 𝐵 } ) ) ) |
21 |
20
|
exp4b |
⊢ ( 𝐵 ∈ ℋ → ( 𝑥 ∈ ℂ → ( 𝐴 ≠ 0ℎ → ( 𝐴 = ( 𝑥 ·ℎ 𝐵 ) → ( span ‘ { ( 𝑥 ·ℎ 𝐵 ) } ) = ( span ‘ { 𝐵 } ) ) ) ) ) |
22 |
21
|
com23 |
⊢ ( 𝐵 ∈ ℋ → ( 𝐴 ≠ 0ℎ → ( 𝑥 ∈ ℂ → ( 𝐴 = ( 𝑥 ·ℎ 𝐵 ) → ( span ‘ { ( 𝑥 ·ℎ 𝐵 ) } ) = ( span ‘ { 𝐵 } ) ) ) ) ) |
23 |
22
|
imp43 |
⊢ ( ( ( 𝐵 ∈ ℋ ∧ 𝐴 ≠ 0ℎ ) ∧ ( 𝑥 ∈ ℂ ∧ 𝐴 = ( 𝑥 ·ℎ 𝐵 ) ) ) → ( span ‘ { ( 𝑥 ·ℎ 𝐵 ) } ) = ( span ‘ { 𝐵 } ) ) |
24 |
5 23
|
eqtrd |
⊢ ( ( ( 𝐵 ∈ ℋ ∧ 𝐴 ≠ 0ℎ ) ∧ ( 𝑥 ∈ ℂ ∧ 𝐴 = ( 𝑥 ·ℎ 𝐵 ) ) ) → ( span ‘ { 𝐴 } ) = ( span ‘ { 𝐵 } ) ) |
25 |
24
|
rexlimdvaa |
⊢ ( ( 𝐵 ∈ ℋ ∧ 𝐴 ≠ 0ℎ ) → ( ∃ 𝑥 ∈ ℂ 𝐴 = ( 𝑥 ·ℎ 𝐵 ) → ( span ‘ { 𝐴 } ) = ( span ‘ { 𝐵 } ) ) ) |
26 |
2 25
|
sylbid |
⊢ ( ( 𝐵 ∈ ℋ ∧ 𝐴 ≠ 0ℎ ) → ( 𝐴 ∈ ( span ‘ { 𝐵 } ) → ( span ‘ { 𝐴 } ) = ( span ‘ { 𝐵 } ) ) ) |