Proof of Theorem shftlem
Step | Hyp | Ref
| Expression |
1 | | df-rab 2457 |
. 2
⊢ {𝑥 ∈ ℂ ∣ (𝑥 − 𝐴) ∈ 𝐵} = {𝑥 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵)} |
2 | | npcan 8128 |
. . . . . . . . 9
⊢ ((𝑥 ∈ ℂ ∧ 𝐴 ∈ ℂ) → ((𝑥 − 𝐴) + 𝐴) = 𝑥) |
3 | 2 | ancoms 266 |
. . . . . . . 8
⊢ ((𝐴 ∈ ℂ ∧ 𝑥 ∈ ℂ) → ((𝑥 − 𝐴) + 𝐴) = 𝑥) |
4 | 3 | eqcomd 2176 |
. . . . . . 7
⊢ ((𝐴 ∈ ℂ ∧ 𝑥 ∈ ℂ) → 𝑥 = ((𝑥 − 𝐴) + 𝐴)) |
5 | | oveq1 5860 |
. . . . . . . . . 10
⊢ (𝑦 = (𝑥 − 𝐴) → (𝑦 + 𝐴) = ((𝑥 − 𝐴) + 𝐴)) |
6 | 5 | eqeq2d 2182 |
. . . . . . . . 9
⊢ (𝑦 = (𝑥 − 𝐴) → (𝑥 = (𝑦 + 𝐴) ↔ 𝑥 = ((𝑥 − 𝐴) + 𝐴))) |
7 | 6 | rspcev 2834 |
. . . . . . . 8
⊢ (((𝑥 − 𝐴) ∈ 𝐵 ∧ 𝑥 = ((𝑥 − 𝐴) + 𝐴)) → ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴)) |
8 | 7 | expcom 115 |
. . . . . . 7
⊢ (𝑥 = ((𝑥 − 𝐴) + 𝐴) → ((𝑥 − 𝐴) ∈ 𝐵 → ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴))) |
9 | 4, 8 | syl 14 |
. . . . . 6
⊢ ((𝐴 ∈ ℂ ∧ 𝑥 ∈ ℂ) → ((𝑥 − 𝐴) ∈ 𝐵 → ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴))) |
10 | 9 | expimpd 361 |
. . . . 5
⊢ (𝐴 ∈ ℂ → ((𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵) → ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴))) |
11 | 10 | adantr 274 |
. . . 4
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) → ((𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵) → ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴))) |
12 | | ssel2 3142 |
. . . . . . . . . 10
⊢ ((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) → 𝑦 ∈ ℂ) |
13 | | addcl 7899 |
. . . . . . . . . 10
⊢ ((𝑦 ∈ ℂ ∧ 𝐴 ∈ ℂ) → (𝑦 + 𝐴) ∈ ℂ) |
14 | 12, 13 | sylan 281 |
. . . . . . . . 9
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → (𝑦 + 𝐴) ∈ ℂ) |
15 | | pncan 8125 |
. . . . . . . . . . 11
⊢ ((𝑦 ∈ ℂ ∧ 𝐴 ∈ ℂ) → ((𝑦 + 𝐴) − 𝐴) = 𝑦) |
16 | 12, 15 | sylan 281 |
. . . . . . . . . 10
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → ((𝑦 + 𝐴) − 𝐴) = 𝑦) |
17 | | simplr 525 |
. . . . . . . . . 10
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → 𝑦 ∈ 𝐵) |
18 | 16, 17 | eqeltrd 2247 |
. . . . . . . . 9
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵) |
19 | 14, 18 | jca 304 |
. . . . . . . 8
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → ((𝑦 + 𝐴) ∈ ℂ ∧ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵)) |
20 | 19 | ancoms 266 |
. . . . . . 7
⊢ ((𝐴 ∈ ℂ ∧ (𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵)) → ((𝑦 + 𝐴) ∈ ℂ ∧ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵)) |
21 | 20 | anassrs 398 |
. . . . . 6
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) ∧ 𝑦 ∈ 𝐵) → ((𝑦 + 𝐴) ∈ ℂ ∧ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵)) |
22 | | eleq1 2233 |
. . . . . . 7
⊢ (𝑥 = (𝑦 + 𝐴) → (𝑥 ∈ ℂ ↔ (𝑦 + 𝐴) ∈ ℂ)) |
23 | | oveq1 5860 |
. . . . . . . 8
⊢ (𝑥 = (𝑦 + 𝐴) → (𝑥 − 𝐴) = ((𝑦 + 𝐴) − 𝐴)) |
24 | 23 | eleq1d 2239 |
. . . . . . 7
⊢ (𝑥 = (𝑦 + 𝐴) → ((𝑥 − 𝐴) ∈ 𝐵 ↔ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵)) |
25 | 22, 24 | anbi12d 470 |
. . . . . 6
⊢ (𝑥 = (𝑦 + 𝐴) → ((𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵) ↔ ((𝑦 + 𝐴) ∈ ℂ ∧ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵))) |
26 | 21, 25 | syl5ibrcom 156 |
. . . . 5
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) ∧ 𝑦 ∈ 𝐵) → (𝑥 = (𝑦 + 𝐴) → (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵))) |
27 | 26 | rexlimdva 2587 |
. . . 4
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) →
(∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴) → (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵))) |
28 | 11, 27 | impbid 128 |
. . 3
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) → ((𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵) ↔ ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴))) |
29 | 28 | abbidv 2288 |
. 2
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) → {𝑥 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵)} = {𝑥 ∣ ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴)}) |
30 | 1, 29 | eqtrid 2215 |
1
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) → {𝑥 ∈ ℂ ∣ (𝑥 − 𝐴) ∈ 𝐵} = {𝑥 ∣ ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴)}) |