Proof of Theorem shftlem
Step | Hyp | Ref
| Expression |
1 | | df-rab 3072 |
. 2
⊢ {𝑥 ∈ ℂ ∣ (𝑥 − 𝐴) ∈ 𝐵} = {𝑥 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵)} |
2 | | npcan 11160 |
. . . . . . . . 9
⊢ ((𝑥 ∈ ℂ ∧ 𝐴 ∈ ℂ) → ((𝑥 − 𝐴) + 𝐴) = 𝑥) |
3 | 2 | ancoms 458 |
. . . . . . . 8
⊢ ((𝐴 ∈ ℂ ∧ 𝑥 ∈ ℂ) → ((𝑥 − 𝐴) + 𝐴) = 𝑥) |
4 | 3 | eqcomd 2744 |
. . . . . . 7
⊢ ((𝐴 ∈ ℂ ∧ 𝑥 ∈ ℂ) → 𝑥 = ((𝑥 − 𝐴) + 𝐴)) |
5 | | oveq1 7262 |
. . . . . . . . 9
⊢ (𝑦 = (𝑥 − 𝐴) → (𝑦 + 𝐴) = ((𝑥 − 𝐴) + 𝐴)) |
6 | 5 | rspceeqv 3567 |
. . . . . . . 8
⊢ (((𝑥 − 𝐴) ∈ 𝐵 ∧ 𝑥 = ((𝑥 − 𝐴) + 𝐴)) → ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴)) |
7 | 6 | expcom 413 |
. . . . . . 7
⊢ (𝑥 = ((𝑥 − 𝐴) + 𝐴) → ((𝑥 − 𝐴) ∈ 𝐵 → ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴))) |
8 | 4, 7 | syl 17 |
. . . . . 6
⊢ ((𝐴 ∈ ℂ ∧ 𝑥 ∈ ℂ) → ((𝑥 − 𝐴) ∈ 𝐵 → ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴))) |
9 | 8 | expimpd 453 |
. . . . 5
⊢ (𝐴 ∈ ℂ → ((𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵) → ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴))) |
10 | 9 | adantr 480 |
. . . 4
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) → ((𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵) → ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴))) |
11 | | ssel2 3912 |
. . . . . . . . . 10
⊢ ((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) → 𝑦 ∈ ℂ) |
12 | | addcl 10884 |
. . . . . . . . . 10
⊢ ((𝑦 ∈ ℂ ∧ 𝐴 ∈ ℂ) → (𝑦 + 𝐴) ∈ ℂ) |
13 | 11, 12 | sylan 579 |
. . . . . . . . 9
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → (𝑦 + 𝐴) ∈ ℂ) |
14 | | pncan 11157 |
. . . . . . . . . . 11
⊢ ((𝑦 ∈ ℂ ∧ 𝐴 ∈ ℂ) → ((𝑦 + 𝐴) − 𝐴) = 𝑦) |
15 | 11, 14 | sylan 579 |
. . . . . . . . . 10
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → ((𝑦 + 𝐴) − 𝐴) = 𝑦) |
16 | | simplr 765 |
. . . . . . . . . 10
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → 𝑦 ∈ 𝐵) |
17 | 15, 16 | eqeltrd 2839 |
. . . . . . . . 9
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵) |
18 | 13, 17 | jca 511 |
. . . . . . . 8
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → ((𝑦 + 𝐴) ∈ ℂ ∧ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵)) |
19 | 18 | ancoms 458 |
. . . . . . 7
⊢ ((𝐴 ∈ ℂ ∧ (𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵)) → ((𝑦 + 𝐴) ∈ ℂ ∧ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵)) |
20 | 19 | anassrs 467 |
. . . . . 6
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) ∧ 𝑦 ∈ 𝐵) → ((𝑦 + 𝐴) ∈ ℂ ∧ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵)) |
21 | | eleq1 2826 |
. . . . . . 7
⊢ (𝑥 = (𝑦 + 𝐴) → (𝑥 ∈ ℂ ↔ (𝑦 + 𝐴) ∈ ℂ)) |
22 | | oveq1 7262 |
. . . . . . . 8
⊢ (𝑥 = (𝑦 + 𝐴) → (𝑥 − 𝐴) = ((𝑦 + 𝐴) − 𝐴)) |
23 | 22 | eleq1d 2823 |
. . . . . . 7
⊢ (𝑥 = (𝑦 + 𝐴) → ((𝑥 − 𝐴) ∈ 𝐵 ↔ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵)) |
24 | 21, 23 | anbi12d 630 |
. . . . . 6
⊢ (𝑥 = (𝑦 + 𝐴) → ((𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵) ↔ ((𝑦 + 𝐴) ∈ ℂ ∧ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵))) |
25 | 20, 24 | syl5ibrcom 246 |
. . . . 5
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) ∧ 𝑦 ∈ 𝐵) → (𝑥 = (𝑦 + 𝐴) → (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵))) |
26 | 25 | rexlimdva 3212 |
. . . 4
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) →
(∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴) → (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵))) |
27 | 10, 26 | impbid 211 |
. . 3
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) → ((𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵) ↔ ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴))) |
28 | 27 | abbidv 2808 |
. 2
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) → {𝑥 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵)} = {𝑥 ∣ ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴)}) |
29 | 1, 28 | eqtrid 2790 |
1
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) → {𝑥 ∈ ℂ ∣ (𝑥 − 𝐴) ∈ 𝐵} = {𝑥 ∣ ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴)}) |