Proof of Theorem shftlem
| Step | Hyp | Ref | Expression | 
|---|
| 1 |  | df-rab 3436 | . 2
⊢ {𝑥 ∈ ℂ ∣ (𝑥 − 𝐴) ∈ 𝐵} = {𝑥 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵)} | 
| 2 |  | npcan 11518 | . . . . . . . . 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 2742 | . . . . . . 7
⊢ ((𝐴 ∈ ℂ ∧ 𝑥 ∈ ℂ) → 𝑥 = ((𝑥 − 𝐴) + 𝐴)) | 
| 5 |  | oveq1 7439 | . . . . . . . . 9
⊢ (𝑦 = (𝑥 − 𝐴) → (𝑦 + 𝐴) = ((𝑥 − 𝐴) + 𝐴)) | 
| 6 | 5 | rspceeqv 3644 | . . . . . . . 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 3977 | . . . . . . . . . 10
⊢ ((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) → 𝑦 ∈ ℂ) | 
| 12 |  | addcl 11238 | . . . . . . . . . 10
⊢ ((𝑦 ∈ ℂ ∧ 𝐴 ∈ ℂ) → (𝑦 + 𝐴) ∈ ℂ) | 
| 13 | 11, 12 | sylan 580 | . . . . . . . . 9
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → (𝑦 + 𝐴) ∈ ℂ) | 
| 14 |  | pncan 11515 | . . . . . . . . . . 11
⊢ ((𝑦 ∈ ℂ ∧ 𝐴 ∈ ℂ) → ((𝑦 + 𝐴) − 𝐴) = 𝑦) | 
| 15 | 11, 14 | sylan 580 | . . . . . . . . . 10
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → ((𝑦 + 𝐴) − 𝐴) = 𝑦) | 
| 16 |  | simplr 768 | . . . . . . . . . 10
⊢ (((𝐵 ⊆ ℂ ∧ 𝑦 ∈ 𝐵) ∧ 𝐴 ∈ ℂ) → 𝑦 ∈ 𝐵) | 
| 17 | 15, 16 | eqeltrd 2840 | . . . . . . . . 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 2828 | . . . . . . 7
⊢ (𝑥 = (𝑦 + 𝐴) → (𝑥 ∈ ℂ ↔ (𝑦 + 𝐴) ∈ ℂ)) | 
| 22 |  | oveq1 7439 | . . . . . . . 8
⊢ (𝑥 = (𝑦 + 𝐴) → (𝑥 − 𝐴) = ((𝑦 + 𝐴) − 𝐴)) | 
| 23 | 22 | eleq1d 2825 | . . . . . . 7
⊢ (𝑥 = (𝑦 + 𝐴) → ((𝑥 − 𝐴) ∈ 𝐵 ↔ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵)) | 
| 24 | 21, 23 | anbi12d 632 | . . . . . 6
⊢ (𝑥 = (𝑦 + 𝐴) → ((𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵) ↔ ((𝑦 + 𝐴) ∈ ℂ ∧ ((𝑦 + 𝐴) − 𝐴) ∈ 𝐵))) | 
| 25 | 20, 24 | syl5ibrcom 247 | . . . . 5
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) ∧ 𝑦 ∈ 𝐵) → (𝑥 = (𝑦 + 𝐴) → (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵))) | 
| 26 | 25 | rexlimdva 3154 | . . . 4
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) →
(∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴) → (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵))) | 
| 27 | 10, 26 | impbid 212 | . . 3
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) → ((𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵) ↔ ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴))) | 
| 28 | 27 | abbidv 2807 | . 2
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) → {𝑥 ∣ (𝑥 ∈ ℂ ∧ (𝑥 − 𝐴) ∈ 𝐵)} = {𝑥 ∣ ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴)}) | 
| 29 | 1, 28 | eqtrid 2788 | 1
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ⊆ ℂ) → {𝑥 ∈ ℂ ∣ (𝑥 − 𝐴) ∈ 𝐵} = {𝑥 ∣ ∃𝑦 ∈ 𝐵 𝑥 = (𝑦 + 𝐴)}) |