Step |
Hyp |
Ref |
Expression |
1 |
|
remet.1 |
⊢ 𝐷 = ( ( abs ∘ − ) ↾ ( ℝ × ℝ ) ) |
2 |
|
df-ov |
⊢ ( 𝐴 𝐷 𝐵 ) = ( 𝐷 ‘ 〈 𝐴 , 𝐵 〉 ) |
3 |
1
|
fveq1i |
⊢ ( 𝐷 ‘ 〈 𝐴 , 𝐵 〉 ) = ( ( ( abs ∘ − ) ↾ ( ℝ × ℝ ) ) ‘ 〈 𝐴 , 𝐵 〉 ) |
4 |
2 3
|
eqtri |
⊢ ( 𝐴 𝐷 𝐵 ) = ( ( ( abs ∘ − ) ↾ ( ℝ × ℝ ) ) ‘ 〈 𝐴 , 𝐵 〉 ) |
5 |
|
opelxpi |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → 〈 𝐴 , 𝐵 〉 ∈ ( ℝ × ℝ ) ) |
6 |
5
|
fvresd |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( ( abs ∘ − ) ↾ ( ℝ × ℝ ) ) ‘ 〈 𝐴 , 𝐵 〉 ) = ( ( abs ∘ − ) ‘ 〈 𝐴 , 𝐵 〉 ) ) |
7 |
|
df-ov |
⊢ ( 𝐴 ( abs ∘ − ) 𝐵 ) = ( ( abs ∘ − ) ‘ 〈 𝐴 , 𝐵 〉 ) |
8 |
|
recn |
⊢ ( 𝐴 ∈ ℝ → 𝐴 ∈ ℂ ) |
9 |
|
recn |
⊢ ( 𝐵 ∈ ℝ → 𝐵 ∈ ℂ ) |
10 |
|
eqid |
⊢ ( abs ∘ − ) = ( abs ∘ − ) |
11 |
10
|
cnmetdval |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 ( abs ∘ − ) 𝐵 ) = ( abs ‘ ( 𝐴 − 𝐵 ) ) ) |
12 |
8 9 11
|
syl2an |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 ( abs ∘ − ) 𝐵 ) = ( abs ‘ ( 𝐴 − 𝐵 ) ) ) |
13 |
7 12
|
eqtr3id |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( abs ∘ − ) ‘ 〈 𝐴 , 𝐵 〉 ) = ( abs ‘ ( 𝐴 − 𝐵 ) ) ) |
14 |
6 13
|
eqtrd |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( ( abs ∘ − ) ↾ ( ℝ × ℝ ) ) ‘ 〈 𝐴 , 𝐵 〉 ) = ( abs ‘ ( 𝐴 − 𝐵 ) ) ) |
15 |
4 14
|
syl5eq |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 𝐷 𝐵 ) = ( abs ‘ ( 𝐴 − 𝐵 ) ) ) |