Step | Hyp | Ref
| Expression |
1 | | simp2 993 |
. . . . 5
⊢ ((𝐹 ∈ 𝑉 ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → 𝐴 ∈ ℂ) |
2 | | simp1 992 |
. . . . 5
⊢ ((𝐹 ∈ 𝑉 ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → 𝐹 ∈ 𝑉) |
3 | | simp3 994 |
. . . . 5
⊢ ((𝐹 ∈ 𝑉 ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → 𝐵 ∈ ℂ) |
4 | | shftfvalg 10782 |
. . . . . . 7
⊢ ((𝐴 ∈ ℂ ∧ 𝐹 ∈ 𝑉) → (𝐹 shift 𝐴) = {〈𝑥, 𝑦〉 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴)𝐹𝑦)}) |
5 | 4 | breqd 4000 |
. . . . . 6
⊢ ((𝐴 ∈ ℂ ∧ 𝐹 ∈ 𝑉) → (𝐵(𝐹 shift 𝐴)𝑧 ↔ 𝐵{〈𝑥, 𝑦〉 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴)𝐹𝑦)}𝑧)) |
6 | | vex 2733 |
. . . . . . 7
⊢ 𝑧 ∈ V |
7 | | eleq1 2233 |
. . . . . . . . 9
⊢ (𝑥 = 𝐵 → (𝑥 ∈ ℂ ↔ 𝐵 ∈ ℂ)) |
8 | | oveq1 5860 |
. . . . . . . . . 10
⊢ (𝑥 = 𝐵 → (𝑥 − 𝐴) = (𝐵 − 𝐴)) |
9 | 8 | breq1d 3999 |
. . . . . . . . 9
⊢ (𝑥 = 𝐵 → ((𝑥 − 𝐴)𝐹𝑦 ↔ (𝐵 − 𝐴)𝐹𝑦)) |
10 | 7, 9 | anbi12d 470 |
. . . . . . . 8
⊢ (𝑥 = 𝐵 → ((𝑥 ∈ ℂ ∧ (𝑥 − 𝐴)𝐹𝑦) ↔ (𝐵 ∈ ℂ ∧ (𝐵 − 𝐴)𝐹𝑦))) |
11 | | breq2 3993 |
. . . . . . . . 9
⊢ (𝑦 = 𝑧 → ((𝐵 − 𝐴)𝐹𝑦 ↔ (𝐵 − 𝐴)𝐹𝑧)) |
12 | 11 | anbi2d 461 |
. . . . . . . 8
⊢ (𝑦 = 𝑧 → ((𝐵 ∈ ℂ ∧ (𝐵 − 𝐴)𝐹𝑦) ↔ (𝐵 ∈ ℂ ∧ (𝐵 − 𝐴)𝐹𝑧))) |
13 | | eqid 2170 |
. . . . . . . 8
⊢
{〈𝑥, 𝑦〉 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴)𝐹𝑦)} = {〈𝑥, 𝑦〉 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴)𝐹𝑦)} |
14 | 10, 12, 13 | brabg 4254 |
. . . . . . 7
⊢ ((𝐵 ∈ ℂ ∧ 𝑧 ∈ V) → (𝐵{〈𝑥, 𝑦〉 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴)𝐹𝑦)}𝑧 ↔ (𝐵 ∈ ℂ ∧ (𝐵 − 𝐴)𝐹𝑧))) |
15 | 6, 14 | mpan2 423 |
. . . . . 6
⊢ (𝐵 ∈ ℂ → (𝐵{〈𝑥, 𝑦〉 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴)𝐹𝑦)}𝑧 ↔ (𝐵 ∈ ℂ ∧ (𝐵 − 𝐴)𝐹𝑧))) |
16 | 5, 15 | sylan9bb 459 |
. . . . 5
⊢ (((𝐴 ∈ ℂ ∧ 𝐹 ∈ 𝑉) ∧ 𝐵 ∈ ℂ) → (𝐵(𝐹 shift 𝐴)𝑧 ↔ (𝐵 ∈ ℂ ∧ (𝐵 − 𝐴)𝐹𝑧))) |
17 | 1, 2, 3, 16 | syl21anc 1232 |
. . . 4
⊢ ((𝐹 ∈ 𝑉 ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐵(𝐹 shift 𝐴)𝑧 ↔ (𝐵 ∈ ℂ ∧ (𝐵 − 𝐴)𝐹𝑧))) |
18 | 17 | 3anibar 1160 |
. . 3
⊢ ((𝐹 ∈ 𝑉 ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐵(𝐹 shift 𝐴)𝑧 ↔ (𝐵 − 𝐴)𝐹𝑧)) |
19 | 18 | abbidv 2288 |
. 2
⊢ ((𝐹 ∈ 𝑉 ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → {𝑧 ∣ 𝐵(𝐹 shift 𝐴)𝑧} = {𝑧 ∣ (𝐵 − 𝐴)𝐹𝑧}) |
20 | | imasng 4976 |
. . 3
⊢ (𝐵 ∈ ℂ → ((𝐹 shift 𝐴) “ {𝐵}) = {𝑧 ∣ 𝐵(𝐹 shift 𝐴)𝑧}) |
21 | 20 | 3ad2ant3 1015 |
. 2
⊢ ((𝐹 ∈ 𝑉 ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐹 shift 𝐴) “ {𝐵}) = {𝑧 ∣ 𝐵(𝐹 shift 𝐴)𝑧}) |
22 | 3, 1 | subcld 8230 |
. . 3
⊢ ((𝐹 ∈ 𝑉 ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐵 − 𝐴) ∈ ℂ) |
23 | | imasng 4976 |
. . 3
⊢ ((𝐵 − 𝐴) ∈ ℂ → (𝐹 “ {(𝐵 − 𝐴)}) = {𝑧 ∣ (𝐵 − 𝐴)𝐹𝑧}) |
24 | 22, 23 | syl 14 |
. 2
⊢ ((𝐹 ∈ 𝑉 ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐹 “ {(𝐵 − 𝐴)}) = {𝑧 ∣ (𝐵 − 𝐴)𝐹𝑧}) |
25 | 19, 21, 24 | 3eqtr4d 2213 |
1
⊢ ((𝐹 ∈ 𝑉 ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐹 shift 𝐴) “ {𝐵}) = (𝐹 “ {(𝐵 − 𝐴)})) |