Proof of Theorem rexbidar
Step | Hyp | Ref
| Expression |
1 | | ralbidar.1 |
. . . . 5
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 𝜑) |
2 | | ralbidar.2 |
. . . . . . 7
⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒)) |
3 | 2 | ex 412 |
. . . . . 6
⊢ (𝜑 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) |
4 | 3 | ralimi 3086 |
. . . . 5
⊢
(∀𝑥 ∈
𝐴 𝜑 → ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) |
5 | 1, 4 | syl 17 |
. . . 4
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) |
6 | | df-ral 3068 |
. . . 4
⊢
(∀𝑥 ∈
𝐴 (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)) ↔ ∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))) |
7 | 5, 6 | sylib 217 |
. . 3
⊢ (𝜑 → ∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))) |
8 | | pm2.43 56 |
. . . . 5
⊢ ((𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) |
9 | 8 | pm5.32d 576 |
. . . 4
⊢ ((𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒))) |
10 | 9 | alimi 1815 |
. . 3
⊢
(∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒))) |
11 | | exbi 1850 |
. . 3
⊢
(∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)) → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜓) ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜒))) |
12 | 7, 10, 11 | 3syl 18 |
. 2
⊢ (𝜑 → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜓) ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜒))) |
13 | | df-rex 3069 |
. 2
⊢
(∃𝑥 ∈
𝐴 𝜓 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜓)) |
14 | | df-rex 3069 |
. 2
⊢
(∃𝑥 ∈
𝐴 𝜒 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜒)) |
15 | 12, 13, 14 | 3bitr4g 313 |
1
⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒)) |