Proof of Theorem ceqsralt
Step | Hyp | Ref
| Expression |
1 | | df-ral 3066 |
. . . 4
⊢
(∀𝑥 ∈
𝐵 (𝑥 = 𝐴 → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 = 𝐴 → 𝜑))) |
2 | | eleq1 2825 |
. . . . . . . 8
⊢ (𝑥 = 𝐴 → (𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵)) |
3 | 2 | pm5.32ri 579 |
. . . . . . 7
⊢ ((𝑥 ∈ 𝐵 ∧ 𝑥 = 𝐴) ↔ (𝐴 ∈ 𝐵 ∧ 𝑥 = 𝐴)) |
4 | 3 | imbi1i 353 |
. . . . . 6
⊢ (((𝑥 ∈ 𝐵 ∧ 𝑥 = 𝐴) → 𝜑) ↔ ((𝐴 ∈ 𝐵 ∧ 𝑥 = 𝐴) → 𝜑)) |
5 | | impexp 454 |
. . . . . 6
⊢ (((𝑥 ∈ 𝐵 ∧ 𝑥 = 𝐴) → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 = 𝐴 → 𝜑))) |
6 | | impexp 454 |
. . . . . 6
⊢ (((𝐴 ∈ 𝐵 ∧ 𝑥 = 𝐴) → 𝜑) ↔ (𝐴 ∈ 𝐵 → (𝑥 = 𝐴 → 𝜑))) |
7 | 4, 5, 6 | 3bitr3i 304 |
. . . . 5
⊢ ((𝑥 ∈ 𝐵 → (𝑥 = 𝐴 → 𝜑)) ↔ (𝐴 ∈ 𝐵 → (𝑥 = 𝐴 → 𝜑))) |
8 | 7 | albii 1827 |
. . . 4
⊢
(∀𝑥(𝑥 ∈ 𝐵 → (𝑥 = 𝐴 → 𝜑)) ↔ ∀𝑥(𝐴 ∈ 𝐵 → (𝑥 = 𝐴 → 𝜑))) |
9 | | 19.21v 1947 |
. . . 4
⊢
(∀𝑥(𝐴 ∈ 𝐵 → (𝑥 = 𝐴 → 𝜑)) ↔ (𝐴 ∈ 𝐵 → ∀𝑥(𝑥 = 𝐴 → 𝜑))) |
10 | 1, 8, 9 | 3bitri 300 |
. . 3
⊢
(∀𝑥 ∈
𝐵 (𝑥 = 𝐴 → 𝜑) ↔ (𝐴 ∈ 𝐵 → ∀𝑥(𝑥 = 𝐴 → 𝜑))) |
11 | 10 | a1i 11 |
. 2
⊢
((Ⅎ𝑥𝜓 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) ∧ 𝐴 ∈ 𝐵) → (∀𝑥 ∈ 𝐵 (𝑥 = 𝐴 → 𝜑) ↔ (𝐴 ∈ 𝐵 → ∀𝑥(𝑥 = 𝐴 → 𝜑)))) |
12 | | biimt 364 |
. . 3
⊢ (𝐴 ∈ 𝐵 → (∀𝑥(𝑥 = 𝐴 → 𝜑) ↔ (𝐴 ∈ 𝐵 → ∀𝑥(𝑥 = 𝐴 → 𝜑)))) |
13 | 12 | 3ad2ant3 1137 |
. 2
⊢
((Ⅎ𝑥𝜓 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) ∧ 𝐴 ∈ 𝐵) → (∀𝑥(𝑥 = 𝐴 → 𝜑) ↔ (𝐴 ∈ 𝐵 → ∀𝑥(𝑥 = 𝐴 → 𝜑)))) |
14 | | ceqsalt 3438 |
. 2
⊢
((Ⅎ𝑥𝜓 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) ∧ 𝐴 ∈ 𝐵) → (∀𝑥(𝑥 = 𝐴 → 𝜑) ↔ 𝜓)) |
15 | 11, 13, 14 | 3bitr2d 310 |
1
⊢
((Ⅎ𝑥𝜓 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) ∧ 𝐴 ∈ 𝐵) → (∀𝑥 ∈ 𝐵 (𝑥 = 𝐴 → 𝜑) ↔ 𝜓)) |