Proof of Theorem reupick2
Step | Hyp | Ref
| Expression |
1 | | ancr 547 |
. . . . . 6
⊢ ((𝜓 → 𝜑) → (𝜓 → (𝜑 ∧ 𝜓))) |
2 | 1 | ralimi 3087 |
. . . . 5
⊢
(∀𝑥 ∈
𝐴 (𝜓 → 𝜑) → ∀𝑥 ∈ 𝐴 (𝜓 → (𝜑 ∧ 𝜓))) |
3 | | rexim 3172 |
. . . . 5
⊢
(∀𝑥 ∈
𝐴 (𝜓 → (𝜑 ∧ 𝜓)) → (∃𝑥 ∈ 𝐴 𝜓 → ∃𝑥 ∈ 𝐴 (𝜑 ∧ 𝜓))) |
4 | 2, 3 | syl 17 |
. . . 4
⊢
(∀𝑥 ∈
𝐴 (𝜓 → 𝜑) → (∃𝑥 ∈ 𝐴 𝜓 → ∃𝑥 ∈ 𝐴 (𝜑 ∧ 𝜓))) |
5 | | reupick3 4253 |
. . . . . 6
⊢
((∃!𝑥 ∈
𝐴 𝜑 ∧ ∃𝑥 ∈ 𝐴 (𝜑 ∧ 𝜓) ∧ 𝑥 ∈ 𝐴) → (𝜑 → 𝜓)) |
6 | 5 | 3exp 1118 |
. . . . 5
⊢
(∃!𝑥 ∈
𝐴 𝜑 → (∃𝑥 ∈ 𝐴 (𝜑 ∧ 𝜓) → (𝑥 ∈ 𝐴 → (𝜑 → 𝜓)))) |
7 | 6 | com12 32 |
. . . 4
⊢
(∃𝑥 ∈
𝐴 (𝜑 ∧ 𝜓) → (∃!𝑥 ∈ 𝐴 𝜑 → (𝑥 ∈ 𝐴 → (𝜑 → 𝜓)))) |
8 | 4, 7 | syl6 35 |
. . 3
⊢
(∀𝑥 ∈
𝐴 (𝜓 → 𝜑) → (∃𝑥 ∈ 𝐴 𝜓 → (∃!𝑥 ∈ 𝐴 𝜑 → (𝑥 ∈ 𝐴 → (𝜑 → 𝜓))))) |
9 | 8 | 3imp1 1346 |
. 2
⊢
(((∀𝑥 ∈
𝐴 (𝜓 → 𝜑) ∧ ∃𝑥 ∈ 𝐴 𝜓 ∧ ∃!𝑥 ∈ 𝐴 𝜑) ∧ 𝑥 ∈ 𝐴) → (𝜑 → 𝜓)) |
10 | | rsp 3131 |
. . . 4
⊢
(∀𝑥 ∈
𝐴 (𝜓 → 𝜑) → (𝑥 ∈ 𝐴 → (𝜓 → 𝜑))) |
11 | 10 | 3ad2ant1 1132 |
. . 3
⊢
((∀𝑥 ∈
𝐴 (𝜓 → 𝜑) ∧ ∃𝑥 ∈ 𝐴 𝜓 ∧ ∃!𝑥 ∈ 𝐴 𝜑) → (𝑥 ∈ 𝐴 → (𝜓 → 𝜑))) |
12 | 11 | imp 407 |
. 2
⊢
(((∀𝑥 ∈
𝐴 (𝜓 → 𝜑) ∧ ∃𝑥 ∈ 𝐴 𝜓 ∧ ∃!𝑥 ∈ 𝐴 𝜑) ∧ 𝑥 ∈ 𝐴) → (𝜓 → 𝜑)) |
13 | 9, 12 | impbid 211 |
1
⊢
(((∀𝑥 ∈
𝐴 (𝜓 → 𝜑) ∧ ∃𝑥 ∈ 𝐴 𝜓 ∧ ∃!𝑥 ∈ 𝐴 𝜑) ∧ 𝑥 ∈ 𝐴) → (𝜑 ↔ 𝜓)) |